555839476-Pu2-Physics-Part-2-Complete-Notes (1).pdf

Notes

1. Dushyantha Rao mob:[phone] Page 1

OPTICS AND OPTICAL INSTRUMENTS
Optics: It is a branch of physics which deals with the properties and nature and source of light.
Optics is broadly classified into two types:
 Geometrical optics
 Physical optics
Geometrical optics: It is a branch if optics which deals with the concept of light.
This explains reflection, refraction and dispersion of light.
Physical optics: It is a branch of optics which deals with the wave nature of light.
This explains interference, diffraction and polarization.
Homogenous medium: is one which has an uniform composition and density. Ex: Glass, water.
Heterogeneous medium: is one which has a variable density and different composition. Ex: Crystal etc.
Isotropic medium: Is one which velocity is same in all direction. Ex: Glass, water, air.
Anisotropic medium: Is one which speed of light is different in different direction. Ex: Calcite.
Ray of light: It is a straight line along which light energy is transmitted is called ray of light.
Beam of light: A collection of rays of light is called beam of light.
Parallel beam of light: A beam of light in which separation between consequent rays remains constant.
Image: The point in which the rays of light form an object get converged or appear to be diverged after a
refraction or reflection is called image of an object.
Real image: If the rays originating from a point actually meet at another point after reflection or refraction,
that point is called real image point.
Virtual image: The point from which the rays of light form an object appear to diverge is called as virtual
image.
Reflection of light: When a light is incident on the surface a part of light is turned back to the same medium
this is called reflection.
Laws of reflection of light
First law: The incident ray, reflected ray and normal to the reflecting surface lie in the same plane.
Second law: Angle incidence is equal to angle of reflection.

2. Dushyantha Rao mob:[phone] Page 2

Reflection at spherical mirror
Spherical mirror: A spherical mirror is a part of reflecting surface.
Concave mirror: It is a part of a hollow sphere having outer part silvered and the inner part as a reflecting
surface.
Convex mirror: It is a part of a hollow sphere having inner part silvered and the outer part as a reflecting
surface.

Important terms related to thin spherical mirrors
Pole (P): The middle point of spherical mirror is called the pole.
Radius of curvature (R): Radius of curvature of a spherical mirror is the radius of sphere of which the mirror
forms a part.

Principal axis: The straight line joining the pole [P] and centre of curvature [C] of the spherical mirror is
called principal axis.
Normal: The normal to the spherical mirror at any point is the line joining that point to the centre of
curvature.
Principal focus: A narrow parallel beam of light incident on a mirror close to a fixed point or appear to
diverge from a fixed point in the principal axis this point is called principal focus.
Focal length(f): Focal length of the spherical mirror is the distance of the principal focus [f] from pole [p] is
called focal length.

Focal plane: The plane through the principal focus & perpendicular the principal axis is known as focal
plane.
Relation between focal length and radius if curvature (or) show that R
f=
2
Consider a spherical concave mirror.
R  is the radius of curvature
C  be the centre of curvature.
Consider a paraxial ray striking the mirror at M
CM  normal to the mirror at M
θ  be the angle of incidence
From the figure ∠ MCP= θ , ∠ MFP=2θ
From  MCP, 
MD
tan 1
CD
   From ∆ MFP, 
MD
tan2 2
FD
  
For small angle, tan??????≈ θ and tan2θ ≈ 2θ

3. Dushyantha Rao mob:[phone] Page 3

Then (1)& (2) becomes MD
CD
MD MD MD
2 2 2FD CD
FD CD FD
CD
FD
2
R
f
2

     



Sign convention for spherical mirrors (OR) Cartesian sign convention:

1. All distances are measured from the pole of the spherical mirror along the principal axis.
2. The distance measured in the same direction as the incident light are taken as positive & distance
measured opposite to the direction of incident light are taken as negative.
3. The height measured upward with respect to x-axis and normal to the principal axis of the mirror are
taken as positive the height measured downward are taken negative.
Image formation in case of mirrors:
1. The ray from point which is parallel to the principal axis, reflected ray goes through focus of the mirror.
2. The ray incident at any angle at the pole, the reflected ray follows law of reflection.
3. The ray passing through the focus, the reflected ray parallel to the principal axis.
4. The ray passing through the centre of curvature, the reflected ray retraces the path.

Relation between u, v & f (OR) Mirror equation :
Consider a concave mirror, the image formation of an object AB as shown in the figure.
MN aperture AB 
image of an object AB
P pole
PF = f focal length
PC = R radius of curvature PB
= V image distance
PB = U object distance
Δ A B F & MPF are similar.

4. Dushyantha Rao mob:[phone] Page 4

  
A B B F
MP FP
A B B F
1 PM AB
AB FP
  

  
  
Since ABP A B P   , A B P and ABP are

also similar.

A B B P
2
AB BP
  
 B F B P
Compare (1) & (2), we get
FP BP
B P FP B P
FP BP




 
From the figure B P v, FP f,BP u substitute in th e above equation
v f v
fu
v f v
fu
vv
1 by v
fu
    
  




  
1 1 1
=+
f v u

Linear magnification: It is the ratio of the height of the image to the height of the object.
i
o
h v
m
hu


Note:
(a) For concave mirror forming virtual image (b) The convex mirror which is always form virtual image
Note: When image is real m is negative, When image is virtual m is positive.
If |m|>1 image formed is enlarged.
If |m|<1 image formed is diminished.
Practical applications of spherical mirrors
1. A convex mirror is used as a reflector in street lamps.
2. A convex mirror is used as a drivers mirror in all vehicles
3. A concave mirror is used as a reflector in search lights, head lights of vehicles etc.,
4. A concave mirror is used as shaving mirror and makeup mirror.

5. Dushyantha Rao mob:[phone] Page 5

Refraction of light at plane surface:
A ray of light bends when it travels from one medium to another medium of different optical densities is
called refraction of light.
IO=Incident ray, N = Normal, OR=Refracted ray
i  angle of incidence
r  angle of refraction d i r
 angle of deviation


Note: For normal incidence there is no refraction i=0, r=0
Laws of refraction:
First law: Incident ray, refracted ray & normal drawn to the surface at the point of incidences, all are lie in
the same plane.
Second law (OR) Snell’s law: Ratio of sine of angle of incidence to sine of angle of refraction is constant for
a given pair of media and for a given wavelength.
12
sin
constant
sin
sin
n
sin


i
r
i
r
Where 12
n relative refractive index of med(2) w.r.t med(1)
Note: Limitations of Snell’s law
 Snell’s law is not holds good for normal incidence For i = 0, r = 0
 It is not applicable for double refraction.

12
2
1
12
sin
n
sin
nsin
sin n
n sin n sin 1



i
r
i
r
i = r
Note :

If med (1) is air then, 12
n 1, n n
(1) becomes sin
n
sin

i
r where, n  Absolute refractive index

Relative refractive index : It is defined as the ratio of the velocity of light in med (1) to the velocity of light
in med (2)
2
12
1
n
n
n


6. Dushyantha Rao mob:[phone] Page 6

Absolute refractive index: It is defined as the ratio of the velocity of light in air to the velocity of light in a
given medium.
c
n
v

Note:
1) When light travels from one medium to another medium its speed & wave length
changes but frequency doesn’t changes.
2) 2 1 1
1 2 2
nv
nv




Consequences of refraction:
1) Apparent position of sun: The sun is visible little before the actual sunrise & little after actual the
sunset due to refraction of light through the atmosphere.
2) Twinkling of stars: The rays of light from the stars enters at earth atmosphere it passes through medium
change in R I the layers of atmosphere are not stationary. Hence image of the star keep changing position.
Thus gives the impression that star is twinkling.

3) Lateral Shift : It is the perpendicular distance b/n the incident ray & the emergent ray.

s
tsin i r
L
cosr

 This is the expression for the lateral shift.

Note: Factors affecting the Lateral shift
1. Lateral shift increases with increase in thickness(t)
2. Lateral shift increases with angle of incidence(i)
3. It increases with relative refractive index (n)
Condition for lateral shift :
1. Two refracting surfaces should be parallel.
2. Slab should be homogenous.

4) Normal Shift: The distance through which an object appears to be shifted when placed in one medium &
observed from another medium normally.
n
1
S t 1
n



 This is the expression for normal shift
Note: 1. real depth
apparent depth
n
2. Apparent depth of an object when it is viewed normally through a combination of media of
refractive indices n1 and n2 respectively of thickness t1 and t2 is given by, AD =1
1
t
n +2
2
t
n

7. Dushyantha Rao mob:[phone] Page 7

A B

dens
er
rar
er
C
90
o

Note: Factors affecting Normal shift
1. It increases with increase in thickness.
2. It increases with increase in RI.

Total Internal Reflection :

When the ray of light travels from denser to rarer medium it bends away from normal. For a particular angle
of incidence the ray just grazes the surface (0
r 90 ) corresponding angle of incidence called critical angle.
If the angle of incidence increased beyond critical angle, the ray gets reflected back into the denser medium.
This phenomenon is called Total Internal reflection (T I R).
Critical angle: It is the angle of incidence at which the angle of refraction is 0
90

Conditions for T I R :
1. Light should travel from denser medium to rarer medium.
2. The angle of incidence must be greater than critical angle.
Relation between refractive index and critical angle:
Consider a ray of light travelling from a denser medium to a rarer medium.
Let C be the critical angle for the given pair of media.
From the law of refraction, n1 sin i = n2 sin r,
n1 sinC = n2 sin 90
o

2
1
n
sin C
n
 [sin 90
o
=1]
For a denser medium of refractive index n, n1 = n and n2 = 1 for air.
1
n
sin C

Hence refractive index of a medium is equal to reciprocal of the sine of the critical angle.

8. Dushyantha Rao mob:[phone] Page 8

Applications/ Illustrations of total internal reflection:
1. Mirage: During hot summer days the air near the ground becomes hotter and less dense and has smaller
refractive index than the cooler air. The ray of light from the top
of the tall tree goes from denser to rarer medium bends away from
the normal & undergoes TIR. To a distant observer the light
appear to be coming from somewhere below the ground, the
inverted image of the tree creates the impress on reflection from a
pool water.

2. Sparking of diamond: The brilliance of a diamond is due to the TIR of light inside it.
3. Total reflecting prism: These are isosceles right angle prism made of glass they are used to turn the
incident beam of light through 00
90 ,180 & also to invert the image.




4. Optical fiber: An optical fiber is thin transparent fiber of a glass or plastics, which consists of mainly two
parts core and cladding. Core is made up of higher RI, cladding is made up of lower RI.
The ray entering one end of the fiber and emergence from the other end travelling zigzag path due to series
of TIR the ray follows a fiber even if the curved

Application of optical fiber:
1.They are used to observe internal parts of the human body and mechanism.
2.They are used in computers for data transfer.
3. Used to diminish enlarged images.
4. Used for decorate purposes.
5. They are used to transmit light signals from one point to another point in communication.

Refraction of light through a spherical surface
Spherical surface: It is a refracting medium whose curved surface is a part of a sphere.
Concave Spherical surface: A spherical refracting surface which is concave towards rarer medium.
Convex Spherical surface: A spherical refracting surface which is convex towards rarer medium.
Real object: If the object rays on the spherical surface are actually diverge from a point that point is called
real object.
Virtual object: If the incident rays on the spherical surface are appears to be converged to a point that point
is called virtual object.
Real image: If a refracted rays actually converged to a point then that point is called real image.
Virtual image: If the refracted rays from the spherical surface are appears to be diverged from a point that
point is called virtual image.

9. Dushyantha Rao mob:[phone] Page 9

Object space: The region or space on one side of spherical surface in which incident rays are present is
called object space.
Image space: The region or space on one side of spherical surface in which refracted rays are present is
called image space.

Refraction at a single spherical surface:
Figure shows the geometry of formation of image I of an object O on the principal axis of a spherical
surface.
C  centre of curvature
R  radius of curvature
i  angle of incidence
r  angle of refraction
The rays are incident from a medium of refractive
index n1, to another of refractive index n2.
NM will be taken to be nearly equal to the length of
the perpendicular from the point N on the principal axis.
We have, for small angles, MN MN MN
tan NOP tan NCM tan NIM
OM MC MI
     

Now for  NOC, i is the exterior angle.
i NOM NCM
MN MN
i
OM MC
Similarly,
r NCM NIM
MN MN
r
MC MI
    

   

According to Snell’s law
n1 sin i = n2 sin r [For small angles sin i ≈ i and sin r ≈ r]
 n1 i = n2 r
12
1 1 2 2
1 1 2 2
1 2 2 1
1 2 2 1
MN MN MN MN
nn
OM MC MC MI
n n n n
OM MC MC MI
n n n n
OM MC MC MI
n n n n
OM MI MC MC
n n n n
OM MI MC
  
  
  
   

  

10. Dushyantha Rao mob:[phone] Page 10

Apply sign convention (OM= -u, MI= +v and MC= +R) ,we get
1 2 2 1
n n n n
u v R



2 1 2 1
n n n - n
-=
v u R
RI of image space RI of object space Difference in RI
image distance object distance Radius of curveture





Note: In the above equation 12
nn
R is the power the spherical surface.
Lens : It is an optical medium bonded by two spherical surfaces or one spherical & one plane surface.
Convex lens : A lens, which is thicker at the center & thin at the edges.
It converges a parallel beam of light incident on it, It is also known as converging lens.
Concave lens : A lens, which is thinner at the center & thick at the edges is called Concave lens.
It diverges a parallel beam of light incident on it. It is also known as diverging lens.
Principal axis : of a lens is a straight line passing through centre of curvature of the spherical surfaces.
Optic centre : of a lens is a fixed point situated on the principal axis & inside the lens such that any ray
passing through it will have the emergent rays parallel to its corresponding incident rays.
Principal focus (??????) : when a narrow beam of light is incident on a lens close & parallel to its principal axis,
after refraction the rays converges to a fixed point (convex lens) or appears to diverge from a fixed point
(concave lens) on the principal axis of a lens. This fixed point is called principal focus of the lens.
Principal focus is real for convex & virtual for concave lens.
Focal length (f) : It is the distance b/n its optic centre & its principal focus.

Note: focal length is (+
????????????
) for convex, (−
????????????
) for concave.
Thin lens : If the thickness of the lens is negligible to its radii of curvature the lens is called thin lens.

Refraction through thin lens (or) Len’s makers formula :
Consider a thin lens of placed in air 12
R &R
radii of curvature of the two surfaces ABC &
ADC.
O point object
OA paraxial ray incident on the surface ABC undergoes
refraction & emerges along AI .
I real image 12
n &n
refractive index of the medium on the right and left side of ADC

11. Dushyantha Rao mob:[phone] Page 11

The formation of the image I is consider in two stages.
(i) Refraction at ABC : In the absence of ADC surface, the
refracted ray would have gone straight to meet the principal axis
at I1 
2 1 2 1
1 1 2 2 1
2 1 2 1
1
n n n n
Using the formula,
v u R
n n , n n , u u, v v , R R
n n n n
1
v u R




    

  


(ii) Refraction at ADC : For this surface, I may be regarded as a virtual object & its real image formed in the
medium of RI 1
n  

1 2 2 1 2
1 2 1 2
1
2112
1
n n , n n , u v , v v, R R
n n n n
v v R
nnnn
1
v v R
    




  


Adding (1) & (2)
 
 

212 1 1 2 2 1
11
11
21
12
2
1 1 2
nnn n n n n n
v u v v R R
nn 11
nn
u v R R
n1 1 1 1
11
u v n R R
     
        
     

    


  
      
  
If the object is at infinity in air medium, then 12
n 1, n n, v f & u    
 


12
1 1 1 1
n 1 2
f R R
This equation is known as lens makers formula.
From 1 & 2 Thisequationiscalled thinlensformu la

     





12
1 1 1
= n - 1 -
f R R
1 1 1
=-
f v u

Linear magnification: It is the ratio size of the image to the size of the object
i.e. i
o
hv
m = =
hu
Note: 1. m is positive for virtual image.
2. m is negative for real image.

12. Dushyantha Rao mob:[phone] Page 12

To find the image of an object by a lens:
It is convenient to choose any two of the following rays:
1. A ray from the object parallel to the principal axis of the lens after refraction passes through the second
principal focus F′ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus
F.
2. A ray of light, passing through the optical centre of the lens, emerges without any deviation after
refraction.
3. A ray of light passing through the first principal focus (for a convex lens) or appearing to meet at it (for a
concave lens) emerges parallel to the principal axis after refraction.





Power of a lens: It is defined as the tangent of the angle by which it converges a beam of light falling at
unit distant from the optical centre.

Power, 1
P=
f
The SI unit for power of a lens is dioptre (D)
Define 1 diopte: The power of a lens of focal length of 1 metre is one dioptre.
Note: Power of a lens is positive for a converging lens and negative for a diverging lens.

Equivalent focal length of Two lenses in contact:

Consider two thin lenses are in contact
12
L &L two lenses
12
f &f focal length of 12
L &L
O object on the principal axis.
u  object distance.
v  image distance

Image formation takes place in two stages.
(i) Refraction at 1
L : In the absence of 2
L , I1 is the real image of the object at a distance v1. 
11
11
1 1 1
Using the formula
f v u
u u, v v , & f f
1 1 1
1
f v u




  
  

13. Dushyantha Rao mob:[phone] Page 13

(ii) Refraction at2
L : for lens 2
L , I1 is the virtual object & I is the real image after refraction.
Therefore 12
u v , v v, & f f  





21
1 1 1 2
12
12
1 1 1
2
f v v
adding 1 & 2 , weget
1 1 1 1 1 1
v u v v f f
1 1 1 1
3
v u f f
Forequivalent lens,
1 1 1
4
v u f
From 3 & 4
Power, P P P
  
    
   
  

12
1 1 1
=+
f f f
Thus reciprocal of the equivalent focal length of two thin lens in contact is equal to the sum of the reciprocals of the
individual focal length.

Note: Equivalent focal length of n thin lenses in contact, ...
1 2 3 n
1 1 1 1 1
= + + +
f f f f f
Equivalent power of n thin lenses in contact, 1 2 3 n
P P P P ... P   

Refraction through prism
Expression for n in terms of A & D (OR) Derive 





A + D
sin
2
n=
A
sin
2
ABC  Principal section of the prism
A  angle of the prism
n  refractive index of the prism
PQ  incident ray
RS  Emergent ray
At the surface AB 11
i &r
angle of incidence and refraction
At the surface AC 22
i &r
angle of incidence and emergence

In the cyclic quadrilateral AQNR 0ˆˆ
A QNR 180
In  QNR 0
11
ˆ
r r QNR 180  

14. Dushyantha Rao mob:[phone] Page 14


Comparing the above equations, we can write 
12
A r r 1  

Total deviation = deviation at Q + deviation at R.
  
  

1 1 2 2
1 2 1 2
12
d i r i r
d i i r r
d i i A 2
   
   
   

As the angle of incidence is gradually increase, the deviation first decreases, becomes minimum & then
increases again. For any deviation d there are two values of 12
i &i
At minimum deviation d = D
1 2 1 2
i i i & r r r   
Then, (1) becomes A
A 2r r
2
  
(2) becomes AD
D 2i A i
2

   
sini
FromSnell'slaw n
sin r







A+ D
sin
2
n=
A
sin
2

Thin prism : A prism of small refracting angle0
A 10 is called thin prism.
Deviation produced by thin prism : D A n 1
where D Minimum deviation
A Angle of the prism
n refractive index
Angular dispersion (δ) : between any two colures is the difference b/n their deviations.
 
vR
A n n  
A Angle of the prism
n refractive index

Mean deviation : The mean deviation of any two colures is the mean of their deviation.

vR
mean
nn
d A n 1 where n
2

  

Dispersive power ( ): of a medium for a pair of colures is defined as the ratio of angular dispersion to the
mean deviation b/n two colures.
Dispersive power, mean
angulardispersion
meandeviation d

   VR
VR
nn
n1

  


15. Dushyantha Rao mob:[phone] Page 15

Some natural phenomena due to sunlight

Rainbow: The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere.
This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical
water droplets of rain.

Scattering of light: It is defined as the absorption and re-emerges of light by the particles of the medium.

Types of scattering:
Coherent scattering: The wavelength of the scattered light is same as of the incident light.
The energy of photon remain unaltered and hence it is called elastic scattering.
ex: Rayleigh scattering.

Incoherent scattering: The wavelength of scattered light changes is not same as the incident light is called
incoherent scattering.
ex: Raman effect and Compton effect.
Rayleigh scattering: The amount of scattering is inversely proportional to the fourth power of the
wavelength. This is known as Rayleigh scattering.

Blue colour of sky: Blue has a shorter wavelength than red and is scattered much more strongly. In fact,
violet gets scattered even more than blue, having a shorter wavelength. But since our eyes are more sensitive
to blue than violet, we see the sky blue.

Sun appears red: At sunset or sunrise, the sun’s rays have to pass through a larger distance in the
atmosphere. Most of the blue and other shorter wavelengths are removed by scattering. The least scattered
light reaching our eyes, therefore, the sun looks reddish. This explains the reddish appearance of the sun
and full moon near the horizon.


Optical instruments

Simple microscope: Simple microscope is a converging lens of a small focal length.

PRINCIPAL Simple microscope: An object placed between the optical center and the focus of the convex
lens, forms a virtual, erect and magnified image on the same side of the lens.

Magnifying power of simple microscope:
1. For distinct vision:
Magnifying power, D
m1
f

Where D  Near point
f  focal length
2. For normal vision (Eye focused at infinity):
Magnifying power, D
m
f

Where D  Near point
f  focal length

16. Dushyantha Rao mob:[phone] Page 16

Compound microscope:
It consists of two lenses, one compounding the effect of the other.

PRINCIPAL: When an object is placed in front of a convex lens at a distance between Fo and 2Fo; the real,
inverted and magnified image is formed on the other side of this lens. If this image lies within the focal
length of another convex lens E of large aperture then the image acts as an object for this lens. The final
image produced by this lens is virtual, inverted and highly magnified.

Magnifying power of compound microscope:
1. When the final image formed at the near point:
0e
LD
m1
ff
   
   
   
Where L  length of tube
fo  focal length of objective
fe  focal length of eyepiece
D  near point

1. When the final image formed at infinity:
0e
LD
m
ff
   
   
   
Where L  length of tube
fo  focal length of objective
fe  focal length of eyepiece
D  near point

Telescope: Telescope is an optical instrument used to observe the distant objects.

Types of telescopes: (a) Refracting telescope (b) Reflecting telescope

1. Astronomical telescope (Refracting telescope):
Construction: It consist of two convex lenses placed co-axially . The lens facing the object is objective lens
has large focal length. The other lens through which the image is observed is eyepiece. It has small aperture
and has small focal length. The tube having eyepiece can be moved in and out of the tube holding objective
lens with the help of rack and pinion arrangement.

PRINCIPAL : PRINCIPAL of astronomical telescope can be discussed by considering two extreme cases
(a) Distinct vision: When the objective lens of the telescope is directed towards the object, a real and
inverted image is formed at the focal plane of the objective lens. The position of the eyepiece (E) is adjusted
in such a way that the final image is formed at a distance of least of distinct vision.
(b) Normal vision: when the final image is formed at infinity, the telescope is said to be in normal
adjustment. The position of the eyepiece is adjusted in such a way that the final image of the object is
virtual, inverted and highly magnified as shown in fig.

17. Dushyantha Rao mob:[phone] Page 17


Magnifying power of astronomical telescope:
0
e
f
m
f

Where fo  focal length of objective
fe  focal length of eyepiece
Length of telescope is, 0e
L f f

Limitations of refracting type telescope:
1. The refracting type telescope (using lenses) suffers from spherical and chromatic aberrations. Due to these
aberrations, the final image of the object is coloured and blurred.
2. The objective lenses of very large aperture are very difficult to manufacture.

Cassegrain type telescope (Reflecting type telescope):
It consists of a concave mirror of large aperture with a
circular hole in its centre. A small convex mirror is placed
in front of the objective of the telescope. The final image is
observed through an eyepiece E placed in front of the hole
of the objective.
A parallel beam of light from the distant star falls on the
objective which converges it towards its principal focus.
The reflected beam of light is intercepted by the convex
mirror. The convex mirror forms an inverted image which
is seen through the eyepiece.

Advantages and disadvantages of telescope:
Advantages:
(1) Reflecting type telescopes using parabolic mirrors are free from chromatic and spherical aberrations.
Hence sharp image of the object is formed.
(2) Mirror weighs much less than a lens of similar quality.
(3) Full back of a mirror can be supported on its rim only.
(4) Since the mirrors used in reflecting type telescopes reflect the whole light falling on them, so the image
formed by these telescope is quite bright.
(5) The parabolic mirrors of large aperture can be easily manufactured.

Disadvantages:
(1) These type of telescopes need frequent adjustments and hence inconvenient to use.
(2) They cannot be used for general purposes.
(3) The viewer has to sit either near the focal point of the main mirror or another mirror is required to deflect
the light to the viewer.

18. DUSHYANTHA RAO BB MOB: [phone] Page 1

NUMERICAL PROBLEMS ON RAY OPTICS
PROBLEMS ON REFLECTION
1] An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm.
Find the position, nature, and magnification of the image in each case.
[NCERT] Ans: (i) v = -30cm, m = -3, The image is magnified, real and inverted.
(ii) v = 15cm, m = 3, The image is magnified, virtual and erect.
2] A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature
36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image?
Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the
screen have to be moved?
[NCERT] Ans: v = –54 cm. The image is real, inverted and magnified. The size of the image is 5.0 cm.
3] A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location
of the image and the magnification. Describe what happens as the needle is moved farther from the
mirror.
[NCERT] Ans: v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u → ∞; v → f (but
never beyond) while m → 0.
PROBLEMS ON REFRACTION THROUGH CURVED SURFACEN, SHIFT
4] Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature =
20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is
formed? [NCERT] Ans: v = +100 cm
5] A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance
would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab
held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of
the slab? [NCERT] Ans: The pin appears raised by 5.0 cm the answer is independent of the
location of the slab
6] A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom
of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is
replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the
microscope have to be moved to focus on the needle again? [NCERT] Ans: 1.33; 1.7 cm
7] A transparent cube of side 15 cm contains an air bubble in it. When viewed normally through one
face, the bubble appears to be at 6 cm from the surface. When viewed normally through the opposite
face the distance appears to be 4 cm. Find the actual distance of the bubble from the second face and
the refractive index of the material of the cube. Ans: [ 6 cm, n = 1.5]
PROBLEMS ON TIR
8] A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area
of the surface of water through which light from the bulb can emerge out? Refractive index of water is
1.33. (Consider the bulb to be a point source.) [NCERT] Ans: 2.6 m
2

19. DUSHYANTHA RAO BB MOB: [phone] Page 2

9] Light from a luminous point at the bottom of a glass slab of thickness 3cm strikes the upper surface.
The rays which are totally reflected at the top surface outline a circle of radius 2.4 cm. Find RI of glass.
PROBLRMS ON LENS
10] (i) If, f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the
faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive
index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water?
(Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)
[NCERT] Ans: (i) +2 dioptre (ii)n = 1.5 (iii)+ 78.2 cm.
11] Find the position of the image formed by the lens combination given in the Fig.





[NCERT] Ans: 30 cm to the right of the third lens.
12] Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces
of the same radius of curvature. What is the radius of curvature required if the focal length is to be
20cm? [NCERT] Ans: R = 22 cm
13] A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12
cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm,
and (b) a concave lens of focal length 16cm?
[NCERT] Ans: (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side
(b) f = –16 cm. Image is real and at 48 cm from the lens on its right side
14] An object of size 3.0 cm is placed 14cm in front of a concave lens of focal length 21 cm. Describe
the image produced by the lens. What happens if the object is moved further away from the lens?
[NCERT] Ans: v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u → ∞, v → f
(but never beyond f while m → 0).
15] What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of
focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
[NCERT] Ans: A diverging lens of focal length 60 cm
16] The radii of curvature of two surfaces of a convex lens are 0.2m and 0.22m. Find the focal length of
the lens if refractive index of the material of lenses 1.5. Also find the change in focal length, if it is
immersed in water of refractive index 1.33. [J- 18]
17] Two lenses of focal lengths 0.20m and 0.30m are kept in contact. Find the focal length of the
combination. Calculate the powers of two lenses and combination. [M-14]

20. DUSHYANTHA RAO BB MOB: [phone] Page 3

18] A converging lens of refractive index 3/2 and of focal length 15 cm in air, has the same radii of
curvature for both sides. If it is immersed in a water of refractive index 4/3. Find the focal length.
19] Two Plano- convex lenses are placed in contact such that their curved surfaces are facing each
other. The radius of curvature of the curved surfaces is 0.10cm and 0.15m respectively. The space
between them is filled with water of refractive index 1.33. If the refractive index of glass is 1.5 find the
focal length of the combination.
20] A converging lens of refractive index 1.5 and of focal length 15 cm in air, has the same radii of
curvature for both sides. If it is immersed in a water of refractive index 1.33. Find the focal length.
PROBLEMS ON PRISM
21] A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face
of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of
the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water
(refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
[NCERT] Ans: n ≅ 1.53 and Dm for prism in water ≅ 10º
22] At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that
it just suffers total internal reflection at the other face? The refractive index of the material of the
prism is 1.524. [NCERT] Ans: 30°
23] Two convex lenses of focal lengths 0.20 m and 0.30 m are kept in contact. Find the focal length of
the combination. Calculate powers of two lenses and combination. [M-14]
24] An equilateral prism produces a minimum deviation of 40
0
. What is the R.I of the material of the
prism? Calculate the angle of incidence. [J-14]
25] At what angle should a ray of light be incident on the face of an equilateral prism, so that it just
suffers total internal reflection at the other face? The refractive index of the material of the prism is
1.5 [M-20]
26] Calculate the angle of minimum deviation produced by an equilateral prism of refractive index
1.65.
27] At what angle should a ray of light be incident on the face of a prism of refracting angle 60
0
, so that
it first suffers total internal reflection at other face? The refractive index of the prism is 1.524.
28] A prism of angle 60
0
produces angle of minimum deviation of 40
0
. What is its R.I? Calculate angle
of incidence.
29] A ray of light is incident on one face of an equivalent prism of glass having refractive index 1.55 at
an angle of 40°; calculate the angle of deviation produced by the prism.

21. Dushyantha Rao mob:[phone] Page 1

WAVE OPTICS

Huygens principle
1. Every point on the given wave front (called primary wave
front) acts as a source of secondary wavelets, which travel in
all directions with the velocity of light in the medium.
2. A surface touching these secondary wavelets, tangentially in
the forward direction at any instant gives the new wavefront at
that instant. This is called secondary wave front.

Wavefront
It is the locus of all points in phase at a distance from a light source.
Wavefronts can be classified into
(i) Spherical wavefront
For a point source of light, at small distances the wavefront is said to be spherical.
(ii) Cylindrical wavefront
For a linear source, at small distances the wavefront is said to be cylindrical.
(iii) Plane wavefront
For a point source or linear source, at large distances any wavefront is considered to be plane.
Figures (a), (b) and (c) represent wave front and rays of light corresponding to plane wave from, diverging
spherical wave front and converging spherical wave front respectively.







Reflection on the basis of wave theory
In the figure, AB is a plane wave front incident on a reflector MN at an angle i. Let the secondary wavelets
from B strike MN at C in t seconds. 
BC = v × t  (1)
where v  velocity of light in the medium.
The secondary wavelets form A will travel the same distance
(v × t) in the same time. Therefore, with A as centre and (v × t) as
radius, draw an arc at E, so that AE = (v× t)

22. Dushyantha Rao mob:[phone] Page 2

CE is the secondary wavefront propagating after t seconds.
Angle of incidence, i =BAC and angle of reflection, r ECA
In ACEand ACB, AC is common, BC = AE = (v × t) and 0
B E 90    are congruent
BAC ECA
ir

   
   

This proves the first law of reflection.

Refraction on the basis of wave theory
XY is a plane surface that separates a denser medium of
refractive index n from a rarer medium 12
V &V
are the velocity of light in rarer and denser medium 1
2
v
n
v
 
refractive index
AB is a plane wave front incident on XY at BAC i   .
Let the secondary wavelets from B strike XY at C in t seconds.
 B C = 
1
Vt
The secondary wavelets from A travel in the denser medium with a velocity 2
V and would cover a distance  
2
Vt
in t seconds. Therefore, with A as centre and radius equal to  
2
Vt , draw an arc at E to obtain the
secondary wavefront EC.
Let ACE r , angle of refraction.
In 1
vtBC
ACB, sini
AC AC

  
In 2
vtAE
ACE, sin r
AC AC

  
1
2
vsini
n
sin r v
  
This proves Snell’s law of refraction.

INTERFERENCE
Interference: The phenomenon of modification in the distribution of light energy when two or more light
waves superpose on one another is called interference.
Eg: Coloured pattern on a soap bubble, coloured patch on wet roads due to oil spill etc.

23. Dushyantha Rao mob:[phone] Page 3

Coherent sources: The two sources emit light waves of same frequency, same wavelength and same phase
or constant phase difference are said to be coherent sources. Interference can occur only with coherent
sources.
Condition for sustained interference
Well defined and observable interference pattern can be obtained if the following conditions are satisfied:
1. The two superposing waves must be in phase or must have constant phase difference.
2. The two light waves must have same wavelength.
3. The amplitude of the waves must be equal or almost equal.
4. The two sources producing the waves must be very narrow.
5. The two sources producing the waves must be very close to each other.

Theory of interference
Consider two light waves of same wavelength travelling in the same direction. Let a1 and a2 be their
amplitudes. Their displacements at any instant of time t is given by
 
1 1 2 2
y a sin t and y a sin t     
where   is the phase difference between the two waves
2f   angular frequency f  is frequency of the waves.
According to the principle of superposition of waves, the resultant displacement of the waves is
 
 
 
  
  

1 2 1 2
12
1 2 2
1 2 2
1 2 2
y y y a sin t a sin t
y a sin t a sin t cos cos t sin
y a sin t a sin t cos a cos t sin
y a a cos sin t a cos t sin
Take R cos a a cos 1 and R sin a sin 2
the aboveequation becomes
y R cos sin t R sin
       
        

       
      
        

    
 
 
cos t
y R cos sin t sin cos t
y R sin t

       

   
Where R  the amplitude of the resultant wave

Expression for the amplitude of the resultant wave
Squaring and adding equations (1) and (2), we get
R
2
cos
2
θ + R
2
sin
2
θ = (a1+ a2cos  )
2
+ (a2sin  )
2

R
2
(cos
2
θ + sin
2
θ) = a1
2
+ a2
2
cos
2
 + 2 a1 a2cos  + a2
2
sin
2

i.e., R
2
= a1
2
+ a2
2
+ 2 a1 a 2 cos 
or 22
1 2 1 2
R= a +a +2a a cos resultant amplitude

24. Dushyantha Rao mob:[phone] Page 4

Dividing equation (2) by (1), we get
2
12
a sin
tanθ=
a +a cos


where θ  is the phase difference between the resultant wave and the first wave.
Note: The relationship between intensity I and amplitude A is given by
2
IA
i) Condition for constructive interference
When crest of one wave superpose on crest of another wave constructive interference takes place. Hence in
constructive interference amplitude of the resultant wave is maximum.
From 22
1 2 1 2
R= a +a +2a a cos
R is maximum when cos  = +1, or  = 0, 2, 4, 6…….2n.
Phase difference 2n where n = 0, 1, 2, 3………
Path difference xn  
22
max 1 2 1 2
max 1 2
The maximum amplitude is given by
R = a +a +2a a
R a +a

ii) Condition for destructive interference
Amplitude of the resultant wave is minimum during destructive interference. This occurs when crest of a
wave superpose on trough of another wave.
From 22
1 2 1 2
R= a +a +2a a cos
R is minimum when cos  = -1, i.e.  = , 3, 5…….
Phase difference  21n where n = 0, 1, 2, 3………
Path difference  x 2n 1
2

  
22
min 1 2 1 2
min 1 2
The minimum amplitude is given by
R = a +a -2a a
R a -a

25. Dushyantha Rao mob:[phone] Page 5

Young’s double slit experiment
It consists of a narrow slit S which is illuminated by a
monochromatic light. Two narrow slits A and B are placed
equidistant from S. Screen S is placed at a certain distance from the
slits as shown. The wave front sent by S reaches the slits A and B
simultaneously. The two slits act as coherent sources. These waves
superpose gives rise to interference pattern on the screen. When crest
of one wave superpose on crest of another produces a bright fringe.
When crest of one wave superpose on trough of another or vice-
versa, produces a dark fringe on the screen. Thus the interference pattern consists of alternate dark and
bright fringes of equal thickness.
The condition for constructive interference is path difference = n
The condition for destructive interference is path difference = (2n + 1) λ
2
Intensity distribution curve







Fringe width: The distance of separation between two consecutive bright or dark fringes is called fringe
width.
Expression for fringe width
Consider the experimental arrangement of young’s double slit.
A and B  two slits distance
d Distance between the slits
D the distance between the slits and the screen.
O  point on the screen which is equidistant from A and B
A bright fringe is formed at O.
P a point on the screen at a distance x from O. 
= (BP – AP)  The path difference between the waves
arriving from A and B at P


S
1

crest
trough 2
  2
0
Imax.
maxima.
minima
D
N
O
M
P
d
A
B
x

26. Dushyantha Rao mob:[phone] Page 6

From the ∆AMP,
AP
2
= AM
2
+ MP
2
= 2
2
2 2 2dd
D x D x xd
24

     


From the ∆BNP,
BP
2
= BN
2
+ NP
2
=2
2
2 2 2dd
D x D x xd
24

     


BP
2
- AP
2
= 2xd
(BP + AP) (BP - AP) = 2xd
Since d is very small compared to D and the point P is very close to O, BP≈ AP≈ D,
2D (BP - AP) = 2xd

2xd
(BP AP)
2D
xd
path difference 1
D

  
For a bright fringe formed at P,
The path difference is δ = nλ
xd
D = nλ (from (1))
nλD
d
x
For n
th
bright fringe nλD
d

n
x
For (n -1)
th
bright fringe 
1
λD
1
d


n
xn
Fringe width, n n 1
xx

  
D
d

 This is the expression for fringe width
Similarly, for n
th
dark fringe δ = (2n + 1) λ
2
The distance of n
th
dark fringe from O is X n = (2n+1) λD
2d
For (n -1)
th
dark fringe X

n-1 = [2(n -1)+1] λD
2d
Fringe width, β = X n– X n-1
β = λD
d
From the above equations it is clear that the bright and dark fringes are equally spaced.

27. Dushyantha Rao mob:[phone] Page 7

Note:
1. The expression for the resultant intensity when two waves of intensities I1 and I2 superpose on one
another is I = I1 + I2 + 212
I I cos 
2. Imax = (1
I +2
I )
2

3. Imin = (1
I -2
I )
2

4. The ratio of width of two slits is given by 1
2
w
w = 1
2
I
I = 1
2
2
2
a
a
5. In Young’s double slit experiment, for a n
th
bright fringe, the order of the fringe is n [eg., for a 2
nd

fringe, n= 2] whereas for n
th
dark fringe, the order of the fringe is (n-1) [ eg. for a 2
nd
fringe, n = 1]
6. Coherent sources are realized in practice are using i) Lloyd’s mirror ii) Fresnel’s biprism
7. Angular width of a fringe, θ = β
D = λ
d

DIFFRACTION
The phenomenon of bending of light around small obstacles and hence its spreading into the geometrical
shadow region is called diffraction.

Diffraction at a single slit
The experimental set up to obtain
diffraction at a single slit is as
shown. A monochromatic source
S is placed at the principal focus
of the convex lens L1. The rays
from the source are made to fall
on the lens and the emergent plane wavefront is made to incident on the slit AB. When the wavefront arrives
at AB, secondary wavelets emerging from it move towards the lens L2. Due to superposition of these waves,
diffraction pattern is obtained on the screen which is placed at the focal plane of the converging lens L2.
The diffraction pattern consists of a central bright band called central maximum. Alternate dark and bright
bands are formed on both sides of the central maximum. The dark bands are called minima and the less
bright bands are called secondary maxima. The brightness of the bright bands is not uniform. The intensity
of the bright bands goes on decreasing on both the sides of the central maximum. Hence at O always a bright
band is produced. At P band formed can be either bright or dark that depends on the path difference. If the
path difference between AP and BP is (2n + 1) 2
 , a bright band is formed. If the path difference between
AP and BP is nλ dark band is formed.
O
Screen
S
Slit
L1
L2
P
A
B

28. Dushyantha Rao mob:[phone] Page 8

minima
I
o P Q R
Principal maximum
Secondary maxima

The intensity distribution curve of the diffraction pattern






Analysis of diffraction pattern
Let d be the width of the slit AB. Consider a plane wavefront moving towards the slit. Each point on the
wavefront acts as the source of secondary waves. The
superposition of these waves produce diffraction pattern on the
screen.
1. Intensity at O: Let O be the mid point on the screen. It is
equidistant from A and B. Therefore, the secondary waves
proceeding from A and B arrive at O in phase i.e., the waves interfere constructively. Hence O will
be the point of maximum intensity. This is called central maximum.
2. Intensity at P : Let P be a point on the screen such
that the path difference between BP and AP be λ
i.e. d sinθ1 = λ.
Consider the slit AB be made of two equal parts
AC and CB. The secondary waves arriving from A
and C has a path difference of2
 . For every point
between A and C there is a corresponding point between C and B. The waves from these points
interfere destructively giving rise to the first minimum at P.
3. Intensity at Q : Let Q be a point on
the screen such that the path difference
between BQ and AQ be 3
2
 . The slit is
imagined to be made of three equal
parts. The waves emerging from AC
will be out of phase with the waves
emerging from CD. Hence they
superpose destructively. However the waves emerging from the region DB produce some
illumination at Q. At Q first secondary maximum is formed.

O
A
B
S
P
O S
λ

A
B
C
T
θ1
Q
S O
32




A
B
C
T
D
θ2

29. Dushyantha Rao mob:[phone] Page 9

Note:
1) Condition for diffraction minima
Path difference, d Sinθ = n λ
Where n = 0, 1,2,3……
d is the width of the slit
θ  is the angular position of the point
2) Condition for diffraction maxima
Path difference, d Sinθ = (2n + 1) 2

3) Width of central maximum = 2D
d

4) Angular width of central maximum, 

= 2θ = 2
d

Difference between interference and diffraction
Interference Diffraction
1. It is the phenomenon in which the modification
in the distribution of light energy due to the
superposition of two or more waves.
1. It is the phenomenon of bending of light around
small obstacles and hence its encroachment into the
geometrical shadow region
2. Interference is due to the superposition of two
waves emerging from two coherent sources.
2. Diffraction is due to the superposition of
secondary wavelets emerging from different parts
of same wavefront.
3. Interference fringes are of equal width. 3. Diffraction bands are of unequal width.
4. Intensity of all bright fringes is uniform. 4. Intensity of each bright band varies.
5. Condition for bright fringe is path difference
= n λ, for a dark fringe, path difference =  2n 1
2


5.Condition for bright band is
path difference = 2n 1
2

 and for a dark fringe,
path difference = nλ

Resolving power
Resolving power of an optical instrument is its ability to show two closely lying point objects separately.

Limit of resolution of a microscope (dx): is defined as the minimum distance of separation
between two point objects when they are just resolved. dx
2nsin





dx
θ

30. Dushyantha Rao mob:[phone] Page 10

Resolving power of a microscope: It is the reciprocal of the minimum distance of separation between two
point objects when they are just resolved.
Resolving power of a microscope = 11
Limitof resolution dx

2n Sin
RP



Where n is the refractive index of the medium between the object and the objective
θ is the semi vertical angle of the cone of light rays entering the objective
λ is the wavelength of the light used
Note: Resolving power of a microscope can be increased by increasing the refractive index of the medium
and by decreasing the wavelength of the light used.

Limit of resolution of a telescope (dθ): is defined as the minimum angular separation between two point
objects when they are just resolved. 1.22
d
d


Resolving power of a telescope: is the reciprocal of the minimum angular separation between two point
objects when they are just resolved.
Resolving power of a telescope = 11
Limitof resolution d


d
RP
1.22


where d is the diameter the objective
λ is the wavelength of the light used.
Note: Resolving power of a telescope can be increased by increasing the diameter the objective and by
decreasing the wavelength of the light used.
The validity of ray optics: Ray optics is valid when characteristic dimension are much larger than
wavelength of light. The size of the obstacle must be much larger than the wavelength of light. If the
wavelength is comparable to the size of object, then diffraction could happen, but it cannot be explained
using ray optics
Fresnel distance: It is the minimum distance a light ray has to travel before it bends from the original path. 2
F
a
Z

Where ZF  Fresnel distance, a slit width and   wavelength of light\

31. Dushyantha Rao mob:[phone] Page 11

POLARISATION
The phenomenon of restricting the electric field vibrations of light in a single plane is called polarization.
Light being an electromagnetic radiation, is transverse in nature.

Experiment to demonstrate transverse nature of light
Experimental arrangement as shown. Consider two tourmaline crystals A and B cut parallel to their
crystallographic axes and placed parallel to each other. Unpolarised light from the monochromatic source is
made to fall normally on the crystal A. The following observations are made.
1. No change in the intensity of light transmitted from A when it is rotated about the direction of propagation
of light.
2. No change in the intensity of light transmitted from B when both A and B is rotated together about the
direction of propagation of light.
3. Keeping A fixed when B is rotated, the intensity is maximum only when both A and B are parallel to each
other. When B is perpendicular to A the intensity becomes zero. In between the intensity varies from zero to
a maximum.
If the light is longitudinal in nature, there would have been no change in the intensity of the emergent light
when B was rotated. The above observations confirm that light is transverse in nature.

Representation of unpolarised and plane polarised light



(i) (ii) (iii) (iv)
Fig (i) and (ii) represent unpolarised light
Fig (iii) represents plane polarised light with its vibrations in the plane of paper
Fig (iv) represents plane polarised light with its vibrations perpendicular to the plane of paper

A B Unpolarised light Plane polarised light
polarizer analyser
A B Unpolarised light
No light
polarizer analyser

32. Dushyantha Rao mob:[phone] Page 12

Malus’ law
A beam of plane polarized light is incident on the analyzer, the intensity of light transmitted from the
analyzer is directly proportional to the square of the cosine of the angle between the planes of transmission
of the polarizer and analyzer.
According to Malus’ law
Intensity of light transmitted from the analyzer 2
o
I I cos
where Io  is the intensity of the light entering the analyzer
 angle between the planes of polarizer and analyser

Polarisation by reflection
When a beam of ordinary light is reflected by the surface of a
transparent medium like glass or water, the reflected light is partially
polarised. The degree of polarization depends on the angle of
incidence. As the angle of incidence is gradually increased from a
small value, the degree of polarisation also increases. At a particular
angle of incidence the reflected light is completely plane polarised.
This angle of incidence is called polarising angle. The vibrations of the
plane polarised light are found to be perpendicular to the plane of incidence.

Polarising angle (θp)
It is the angle of incidence for which the reflected light is completely plane polarised.

Brewster’s law
It states that the refractive index of the medium (n) is equal to the tangent of polarizing angle(θp) i.e.,
p
n tan
Where n refractive index of the medium
p
 polarizing angle
Note: At polarization, the angle between reflected ray and refracted ray is 90
o
.

To show that n = tan θp (To prove Brewster’s law)
Consider an unpolarised light beam AO incident on a medium of refractive index n. At polarizing angle of
incidence, the reflected light is completely plane polarized. Let r be the angle of refraction.
From the law of reflection P
ˆˆ
AOM MOB  
θ
p
θ
p
r

33. Dushyantha Rao mob:[phone] Page 13

From the figure, 
0
0
P
0
P
ˆˆ
MOB + NOR = 90
r 90
r 90 1
  
   

From Snell’s law, 
 

p
0
P
sin sin i
1
sin r sin 90



n from


p
p
p
sin
n
cos
n tan




Hence the Brewster’s law is proved.

Polarisation by scattering
The light from a clear blue portion of the sky shows a rise and fall of
intensity when viewed through a polaroid which is rotated. This is
due to the scattering of sunlight by the air molecules in different
directions.
The scattered light is polarized in a direction perpendicular to the
direction of incidence of unpolarized light which is represented as
shown.

Polaroids: Polaroids are the films used to obtain plane polarised light.
Tourmaline crystal, calcite, quartz etc., are natural polarizing materials.

Artificial polaroids are made of microcrystals of iodosulphate of quinine. These crystals are aligned with
their optic axis parallel to each other between two sheets of plastic. Such a sheet serves as a Polaroid.

Uses of polaroids
1. Polaroids are used to produce and analyse plane polarised light.
2. They are used in sunglasses.
3. They are used to view the three dimensional pictures.
4. They are used in headlights of vehicles to reduce glare.
5. They are used in the window panes of trains and aeroplanes to control the intensity of light.
6. They are used to cut off the dazzling light of the approaching vehicles.
7. Used in the study of optical properties of certain metals.
8. They are used to increase the colour contrast in old paintings.
******************

θ
p
θ
p
r
A
O
B
M
N R

34. DUSHYANTHA RAO BB MOB:[phone] Page 1

NUMERICALS PROBLEMS ON WAVE OPTICS
Problems on Huygens principle
1] Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the
wavelength, frequency and speed of
(a) reflected, and (b) refracted light? Refractive index of water is 1.33
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth. [NCERT]
Problems on Young's double slit experiment (fringe width and distance of n
th
fringe)
2] Two slits are made one millimetre apart and the screen is placed one metre away. What is the
fringe separation when blue-green light of wavelength 500 nm is used? [NCERT] Ans: 0.5mm
3] What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the
following operations:
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter
wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved closer to the double-slit plane;
(e) the width of the source slit is increased;
(f ) the monochromatic source is replaced by a source of white light?
( In each operation, take all parameters, other than the one specified, to remain unchanged.) [NCERT]
4] In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4
m away. The distance between the central bright fringe and the fourth bright fringe is measured to be
1.2 cm. Determine the wavelength of light used in the experiment. [NCERT] Ans: 600nm
5] A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference
fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the
screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the
central maximum where the bright fringes due to both the wavelengths coincide?
[NCERT] Ans: 1950 (D/d)nm, 2600(D/d) nm
6] In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m
away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire
experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
[NCERT] Ans: 0.15
0

35. DUSHYANTHA RAO BB MOB:[phone] Page 2

7] In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on
a distant screen is 0.1
0
. What is the spacing between the two slits? [NCERT] Ans: 0.344mm
8] A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction
pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5
mm from the centre of the screen. Find the width of the slit. [NCERT] Ans: 0.2mm
9] In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light
at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point
where path difference is λ/3 ? [NCERT] Ans: K/4
10] In a young’s double slit experiment the distance between the slits is 1mm. the fringe width is
found to be 0.6mm. When the screen is moved through a distance of 0.25m away from the plane of the
slit, the fringe width becomes 0.75mm. Find the wavelength of the light used. [M-15]
11] A beam of light consisting of two wavelengths 420 nm and 560 nm is used to obtain interference
fringes in Young’s double slit experiment. The distance between the slits is 0.3 mm and the distance
between the slits and the screen is 1.5 m. Compute the least distance of the point from the central
maximum, where the bright fringes due to both the wavelengths coincide. [J-15]
12] In young’s double slit experiment, fringes of certain width are produced on the screen kept at a
distance from the slits. When the screen is moved away from the slits by 0.1m, fringes width increases
by 6 x 10
-5
m. The separation between the slits is 1mm. calculate the wavelength of the light used.
[M-16]
13] In the young’s double slit experiment by using a source of light of wavelength 4500 Å, the fringe
width is 5mm. If the distance between the screen and plane of the slits is reduced to half, what should
be the wavelength of light to get the fringe width of 4mm? [J-16]
14] Light of wavelength 6000Å is used to obtain interference fringes of width 6mm in Young’s double
slit experiment. Calculate the wavelength of light required to obtain fringe width of 4mm when the
distance between the screen and slits is reduced to half of its initial value. [M-17]
15] In young’s double slit experiment distance between the slits is 0.5mm, when the screen is kept at a
distance of 100cm from the slits, the distance of the 9th bright fringe from the centre of the fringe
system is 8.835mm. Find the wavelength of light used. [J-17]
16] In young’s double slit experiment the slits are separated by 0.28mm and the screen is placed at a
distance of 1.4m away from the slits. A distance between the central bright fringe and 5th dark fringe
is measure to be 1.35cm. Calculate the wavelength of light used. Also find the fringe width if the screen
is moved 0.4m towards the slit, for the same experimental setup. [M-18]
17] In a Young’s double slit experiment wave length of light used in 5000 Å and distance between the
slits is 2mm, distance of screen from the slits is 1m. Find fringe width and also calculate the distance
of 7th dark fringe from central bright fringe. [M-19]
18] A rigid beam of light consisting of two wavelengths 500nm and 400nm is used to obtain
interference fringes in young’s double slit experiment. The distance between the slits is 0.3mm and
the distance between the slits and screen is 1.5m. Compute the least distance of the point from the
central maximum, where the bright fringes due to both the wavelengths coincide.

36. DUSHYANTHA RAO BB MOB:[phone] Page 3

19] In young’s double slit experiment, the distance between the slits is 1.2mm and the screen is 0.75m
from the slits. If the distance of the 5th fringe from the central fringe on the screen is 1.8mm. Calculate
the wavelength of light used. What will be the distance of the 5th dark fringe from the centre of the
screen?
20] In Young’s double slit experiment the distance of the screen from the slits is 0.5m and the distance
between the slits is 1.5mm. If the distance of the fourth bright fringe from the center of the screen is
0.8 mm. Calculate the wavelength of light used. What will be the distance of the fifth dark fringe from
the central point on the fringe?
21] In a double slit experiment, the distance of the screen from the slits is 1 m and the distance
between the slits is 1mm. If the third dark fringe is formed at a distance 1.5 mm from the central point
on the screen, calculate the wavelength of light used.
22] In a Young’s double slit experiment light of wavelength 620nm is used to Illuminate slits of width
0.3mm. A screen is placed at a distance of 0.9m. Calculate fringe width (b) distance between 5th and
9th bright fringe on screen.
23] In Young’s double slit experiment the screen is at distance of 1.25m from the slit. When the slits
are illuminated by a light of wavelength 546nm, the width of 20 fringes is 8mm. Find the separation
between the slits. Find also the width of 20 fringes if yellow light of wavelength 594nm is used.
24] In Young’s double -slit experiment the slit separation is 0.3mm and wavelength of light used is
6500 A
0
. A screen is placed 1m away from the slits. Calculate (a) Distance of the 3rd bright fringe and
(b) Distance of the 2nd dark fringe from the central bright fringe.
Problems on diffraction of light
25] Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a
telescope whose objective has a diameter of 100 inch? [NCERT] Ans: 2.9 X10
-7
rad
26] Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and
wavelength 400 nm. [NCERT] Ans: 40m
27] Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill
halfway between the towers. what is the longest wavelength of radio waves, which can be sent
between the towers Without appreciable diffraction effects? [NCERT] Ans: 12.5cm
Problems on polarization of light
28] What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
[NCERT] Ans: 56.31
0

37. Dushyantha Rao mob:[phone] Page 1

DUAL NATURE OF RADIATION AND MATTER

Electron emission: On supplying energy to metal surface, the free electrons in the metal surface leave the
metal surface .This phenomenon of electron leaving the metal surface is called emission.
Types of electron emission (M-14, M-19)
1. Thermionic emission: When a metal is heated to a suited temperature it liberates electrons. The
phenomenon of librating electrons by supplying heat energy is called thermionic emission.
2. Photo electric emission: The phenomenon of emission electrons from the surface of a metal with
absorption of light or photon having suitable frequency is called photo electric effect.
3. Secondary emission:The phenomenon of emission of electrons from the surface of a metal when it is
bombarded with high-energy electrons is called secondary emission.
4. Field emission: The phenomenon of emission of electron from the surface of a metal under the influence of
electric field is called field emission.

Hertz’s Observation: While studying the production of EM Waves by means of spark discharge, he found
that the high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated
by uv light from an arc lamp. when uv light is incident on the metal surface, some electrons near the surface
absorb energy and overcome the attraction of positive ions in the material, they escape from the surface of
the metal.

Hallwach’s and Lenard’s observations. (J-15)
Lenard observed that when UV radiation are made to strike the emitter plate, current was recorded in the
external circuit. When UV radiation were stopped, the current flow also stopped. Thus indicate that the
negatively charged particles emitted when UV rays strike the metal. These attracted towards the anode
caused electric current in the circuit
Hallwach observed the following when zinc plate was illuminated by UV rays
(i) Negatively charged zinc plate becoming neutral.
(ii) Neutral zinc plate turning into positively charged plate.
(iii) Positively charged plate enhancing its positive charge.
Hallwach and Lenard together observed that these was no photoemission if the frequency of is less than a
frequency called threshold frequency. The value of threshold frequency depends on the nature of the
material of the emitter.

Photoelectric effect
The phenomenon of emission of electrons from a metal surface when light of suitable frequency is made to
incident on it is known as photoelectric effect

38. Dushyantha Rao mob:[phone] Page 2

Radiation
G
electrons
C A o
o
o
o
o
o
o
o
o
K
Rh
h
Ba
V
Photo metals: The metals which exhibit photo electric effect are called photo metals.
Eg. Na, K, Mg (alkali metal)

Experimental study of photoelectric effect
Experimental set up consists of
G evacuated glass tube with a side window
Cathode C is connected to negative terminal of a battery
Anode A is connected to negative terminal of a battery A
micro ammeter connected in series
Rh rheostat
V  Voltmeter
K reversing key
Ba Battery

When light of suitable frequency is made to incident on the cathode, photoelectrons are liberated. These
electrons accelerated towards the anode. The electrons flow in the external circuit results in electric current.
The micro-ammeter records the photoelectric current.
The experiment is performed by varying the quantities like frequency, intensity of incident radiation, applied
potential and using different photo emissive material.

Experimental observations: (M-16, M-17, J-18, M-19, J-20)

Effect of intensity of light on photocurrent: Above threshold frequency, the
intensity of radiation is varied and the photoelectric current is recorded each
time. The strength of photoelectric current is directly proportional to intensity
of radiation and it is independent of frequency of radiation.

Effect of potential on photoelectric current: With the increase in positive
anode potential photoelectric current increases and becomes maximum. It
remains constant (or saturates) for the further increase in potential.
When anode is made negative and slowly its negative potential is increased,
the photoelectric current decreases and becomes zero at a fixed potential
called stopping potential (V0). Stopping potential is the measure of
maximum kinetic energy of emitted electrons.
It is directly dependent on frequency of radiation and is independent of intensity of radiation.
current
intensity
I1
I2
current
Potential
Vo
-V
0
I2>I1

39. Dushyantha Rao mob:[phone] Page 3

KEmax
frequency
νo
KEmax
frequency
νo
W
Slope = h
Vo
frequency
νo
W/e
Slope = h/e

Effect of frequency of incident radiation on stopping potential:
Incident radiation of same frequency and different intensities are
plotted against potential. The stopping potential is greater for higher
frequency radiation



Important terms associated with Photoelectric effect (J-15, J-16, J-17, M-20)
1. Threshold frequency (o
 ): It is the minimum frequency of incident radiation below which there is no
photo emission takes place.
2. Threshold wavelength (λo): It is the maximum wavelength of the incident radiation above which no
photoemission takes place.
3. Stopping potential (Vs): It is the minimum negative potential of anode for which the emitted electrons are
prevented from reaching the anode. It is also called retarding potential.
4. Work function (W): The minimum energy required just to liberate a free electron from the metal surface is
called photoelectric work function.
The relationship between threshold frequency νo and work function w is given by W = hνo

Other related graphs of photoelectric effect





Note:
1.The slope of the graph of KEmax versus frequency gives Planck’s constant h and the intercept gives work
function.
2. The slope of the graph of stopping potential versus frequency gives the ratio of Planck’s constant to the
elementary charge (h/e).



O

-V02

Photoelectric
current

-V03

Potantial(V)

-V01

-V

ν3>ν2>ν1

ν3

ν2

ν1

40. Dushyantha Rao mob:[phone] Page 4

Laws of photoelectric emission
1. For every photo emissive material, there is a minimum frequency below which no photoemission
takes place. This minimum frequency is called threshold frequency.
2. Greater than the threshold frequency, the strength of the photoelectric current is directly proportional
to the intensity of incident radiation.
3. The maximum kinetic energy of the photoelectrons is directly proportional to the frequency of
incident radiation.
4. The photoemission is instantaneous. The time lag between the instant of light incident and current
recorded is about 10
-9
s.

Einstein’s explanation of photoelectric effect
According to Einstein the photoelectric effect is due to the collision between the incident photon and the free
electron inside the metal. During collision the free electron absorbs the photon and gains an energy. Certain
Minimum energy needed just to liberate a free electron from the metal surface is called photoelectric work
function. The excess energy of the photon appears as the kinetic energy of the emitted photoelectron.
Energy of incident photon = Work function + Maximum kinetic energy of photoelectron
2
1
2o max
E W KE
h h mv

   
This equation is called Einstein’s photoelectric equation. (M-20)
Other forms of Einstein’s photoelectric equation

hν = w + ½ mv
2
max

hν = hνo+ ½ mv
2
max
 hc
λ =o
hc
λ + ½ mv
2
max
 hc
λ =o
hc
λ + eVs

Einstein’s explanation of laws of photoelectric emission (M-15)
1. From Einstein’s photoelectric equation o max
h h KE   
or KEmax = h (ν - νo )
If ν  νo, kinetic energy will be negative or the velocity will become imaginary. Therefore the
photoemission is not possible. Hence a minimum frequency called threshold frequency is needed for the
photoemission.

2. Greater the intensity of incident radiation, the greater will be the number of photons in it. Hence, the
number of photons undergoing collision with the electrons increases resulting in larger photoelectric current.

41. Dushyantha Rao mob:[phone] Page 5

3. From Einstein’s photoelectric equation hν = W + ½ mv
2
max, it is clear that with the increase in frequency
of incident radiation the maximum kinetic energy of photoelectrons increases as the work function w is a
constant for a given material. Hence the maximum kinetic energy of the photoelectrons is directly
proportional to the frequency of incident radiation.
4. According to Einstein, photoemission is due to the elastic collision between photon and the free electron.
In this process all the energy of photon is transferred to the free electron at once. Hence the photo emission
is instantaneous.

Properties of photon: (J-16)
(1) Each photon has energy, E = hν and momentum, h
P=



where h is planck’s constant, ν and λ are the frequency and wavelength of radiation.
(2) Photon energy is independent of intensity of incident radiation.
(3) All photons travel with same speed in free space that is 3×10
8
ms
-1

(4) When photons travel from one medium to another the frequency remains constant.
(5) Wavelength & velocity of photons change when they travel from one medium to another.
(6) Rest mass of photon is zero.
(7) Photons are not affected by electric and magnetic fields.

Dual nature of matter and radiation

Dual nature of radiation:
Radiation behaves sometimes as a wave, sometimes as a particle that is radiation has dual nature.
(i) The phenomena like interference, diffraction & polarisation can be explained by wave nature of radiation
(ii) The phenomena like photoelectric effect, Compton Effect can be explained by quantum nature of
radiation.
(iii) Phenomena like rectilinear propagation of light, reflection, refraction each can be explained on basis of
either wave nature (or) particle nature of radiation.

Wave nature of matter
de-Broglie’s matter wave hypothesis: (J-17)
de Broglie proposed hypothesis that like radiation, matter also exhibits dual nature i.e., the material particles
such as electrons behave like waves.
According to de Broglie, every moving particle is associated with a wave which controls the particle.
Matter waves: The waves associated with the material particles in motion are called matter waves.
de Broglie wave length: The wave length of matter waves is called de Broglie wave length.

42. Dushyantha Rao mob:[phone] Page 6

Expression for de Broglie wave length
Consider a photon of energy, E = h ν ……(1)
Where  is the frequency of radiation
h is the Planck’s constant.
According to Einstein’s mass-energy relation, E = mc
2
…….(2) 2
2
From (1) & (2) we have,


h = mc
hc
mc
λ
h
mc


= mc
2

The above equation is the expression for wavelength of a photon.
The de Broglie wave length (the wavelength of the material particle) is given by (M-17, J-17)
h
λ=
mv
Where m & v are the mass & velocity of the material particle respectively.
Note : The different forms of de Broglie wavelengths are
1) h
λ=
p Where p  is the momentum of the electron
2) h
λ=
2meV Where V is the accelerating potential. For an electron, 0
12.27A
λ=
V (M-19, M-20)
3) h
λ=
2mE Where E  is the kinetic energy of the particle.
4) For a thermal neutron, h
λ=
3mKT
Where K is the Boltzmann’s constant
T is the absolute temperature of neutron.
Experimental demonstration of wave nature of electron (Davisson & Germer experiment):
Experimental arrangement used by Davisson and Germer in as shown. Electrons emitted from the hot
filament F is accelerated towards anode A which is connected to a cylinder with two slits. A fine collimated
beam of electrons is made to fall on the nickel target. The diffracted electrons are collected by the electron
collector which can be moved on a circular scale and it is connected to a sensitive galvanometer.

43. Dushyantha Rao mob:[phone] Page 7

The experiment is performed by varying the
accelerating potential between 44V to 68V. At the
scattering angle of 50
0
and accelerating potential of
54V, the intensity of electrons reaching the collector
was maximum. This is due to the constructive
interference of electrons scattered from different
layers of atoms of the crystal.

According to Bragg’s law of x-ray diffraction
2d sinθ = nλ
where d is the inter-planar spacing = 0.91A
0
for nickel, angle between atomic plane and the scattering
direction, θ = 65
o
and n =1 
2dsinθ
λ=
n


10 0
= 2 0.9 10 sin65

  
λ =1.65A
o

According to de-Broglie’s hypothesis wavelength associated with an electron is given by

0
12.27A
λ=
V

10
12.27 10
=
54


λ =1.67A
o

Conclusion: The close agreement with experimental value by Davisson & Germer and theoritical value by
de-Broglie proves the wave nature of electrons. (M-15, M-17)
*************************

44. DUSHYANTHA RAO MOB: [phone] Page 1

NUMERICAL PROBLEMS ON DUAL NATURE OF MATTER AND RADIATION
Problems on incident light (energy, frequency, wavelength)
1] Monochromatic light of frequency 6.0 ×10
14
Hz is produced by a laser. The power emitted is 2.0
×10
–3
W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on
an average, are emitted by the source?
[NCERT] Ans: 3.98 × 10
–19
J, 5.0 ×10
15
photons per second
2] The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm
(average wavelength) for yellow-green colour and about 760 nm for red colour.(a) What are the
energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and
(iii) red end of the visible spectrum? (Take h = 6.63×10
–34
J s and 1 eV = 1.6×10
–19
J.)
[NCERT] Ans: 3.19 eV, 2.26 eV, 1.64 eV
Problems on photoelectric effect
3] The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the
wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60
V.
[NCERT] Ans: (a) 5.16X10
14
Hz (b) 454nm
4] The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10
14
Hz is incident on
the metal surface, photoemission of electrons occurs. What is the (a) energy of the incident photons
(b) maximum kinetic energy of the emitted electrons. (c) Stopping potential, and (d) maximum speed
of the emitted photoelectrons? Given h = 6.63 X 10
-34
Js, e = 1.6 x 10
-19
C, me = 9.1 × 10
-31
kg [J-14]
[NCERT] Ans: (a) 0.34 eV (b) 0.34 V (c) 344 km/s
5] The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic
energy of photoelectrons emitted?
[NCERT] Ans: 1.5 eV = 2.4 × 10
–19
J
6] In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of
incident light is found to be 4.12 × 10
–15
V s. Calculate the value of Planck’s constant.
[NCERT] Ans: 6.59 × 10
–34
J s
7] The threshold frequency for a certain metal is 3.3 × 10
14
Hz. If light of frequency 8.2 × 10
14
Hz is
incident on the metal, predict the cut-off voltage for the photoelectric emission.
[NCERT] Ans: 2.0 V
8] The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for
incident radiation of wavelength 330 nm?
[NCERT] Ans: No, because ν < νo
9] Light of frequency 7.21 × 10
14
Hz is incident on a metal surface. Electrons with a maximum speed of
6.0 × 10
5
m/s are ejected from the surface. What is the threshold frequency for photoemission of
electrons? [NCERT] Ans: 4.73 × 10
14
Hz

10] Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric

45. DUSHYANTHA RAO MOB: [phone] Page 2

effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of
photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
[NCERT] Ans: 2.16 eV = 3.46 × 10
–19
J
11] Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made
of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How
would the photo-cell respond to a high intensity (∼105 W m
–2
) red light of wavelength 6328 Å
produced by a He-Ne laser ? [NCERT] Ans: 4.2 eV, The photo-cell will not respond
howsoever high be the intensity of laser light.
12] Calculate the maximum velocity of photoelectron emitted when light of frequency 3 x 10
12
Hz is
incident on a metal surface of threshold frequency 2 x 10
12
Hz.
Given h = 6.625 x 10
-34
Js, mass of the electron = 9.1 x 10
-31
kg. [J-09]
13] For a metal the maximum wavelength required for photoelectric emission is 210 nm. Find the
work function. If radiation of wavelength 150nm falls on the surface of the given metal, find the
maximum kinetic energy of the emitted photoelectrons. [A-09]
14] A photon of frequency 1.5 x 10
15
Hz is incident on a metal surface of work function 1.672 eV.
Calculate the stopping potential. h = 6.625 x 10
-34
Js. [M-10]
15] Light of frequency 8.47x10
14
Hz is incident on a metal surface. Electrons with their maximum
speed of 7.5x10
5
ms
-1
are ejected from the surface. Calculate the threshold frequency for the photo
emission of electrons. Also find the work function of the metal in eV. Given h= 6.625 x 10
-34
Js and
mass of the electron = 9.1x 10
-31
kg. [M-18]
16] Light of wavelength 430 nm is incident on a) nickel surface of work function 5 eV and b)
potassium surface of work function 2.3 eV. Find out from which metal electrons area emitted. Also
calculate the maximum velocity of electrons emitted from this metal.
17] Find the maximum velocity of photoelectrons emitted by radiation of frequency 3X10
15
Hz from a
photoelectric surface having a work function of 4.0 eV
18] Light of frequency 8×10
15
Hz is incident on a substance of photo electric work function 6.125 ev.
Calculate the max velocity of the emitted photoelectrons. Given: the mass of the electron = 9.1×10
-31

kg, Planck’s constant = 6.625×10
-34
Js
Problems on de-Broglie wavelength
19] What is the de Broglie wavelength associated with (a) an electron moving with a speed of
5.4×10
6
m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?
[NCERT] Ans: 0.135 nm, 1.47 ×10
–34
m
20] A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of
the particle to that of the electron is 1.813 × 10
–4
. Calculate the particle’s mass and identify the
particle.
[NCERT] Ans: 1.675 × 10
–27
kg. Thus, the particle, with this mass could be a proton or a neutron.

46. DUSHYANTHA RAO MOB: [phone] Page 3

21] What is the de Broglie wavelength associated with an electron, accelerated through a potential
differnece of 100 volts? [NCERT] Ans: 0.123 nm
22] Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV
electrons. [NCERT] Ans: (a) 7.24 × 10
18
Hz (b) 0.041 nm
23] Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power
emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How
many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the
beam to have uniform cross-section which is less than the target area ), and (c) How fast does a
hydrogen atom have to travel in order to have the same momentum as that of the photon? [NCERT]
Ans: (a) 3.14 × 10
–19
J, 1.05 × 10
–27
kg m/s (b) 3 × 10
16
photons/s (c) 0.63 m/s
24] A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre
of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the
sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At
what rate are the photons delivered to the sphere?
[NCERT] Ans: (a) 3.38 × 10
–19
J = 2.11 eV (b) 3.0 × 10
20
photons/s
25] Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a
potential difference of 56 V.
[NCERT] Ans: (a) 4.04 × 10
–24
kg m s
–1
(b) 0.164 nm
26] What is the (a) momentum,(b) speed, and (c) de Broglie wavelength of an electron with kinetic
energy of 120 eV. [NCERT] Ans: (a) 5.92 ×
10
–24
kg m s
–1
(b) 6.5 × 10
6
m s
–1
(c) 0.112 nm
27] The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic
energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.
[NCERT] Ans: (a) 6.95 × 10
–25
J = 4.34 μeV (b) 3.78 × 10
–28
J = 0.236 neV
28] What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0
km/s,(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10
–9
kg drifting with a speed of 2.2 m/s ?
[NCERT] Ans: (a) 1.7 × 10
–35
m (b) 1.1 × 10
–32
m (c) 3.0 × 10
–23
m
29] An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta,(b) the
energy of the photon, and (c) the kinetic energy of electron.
[NCERT] Ans: (a) 6.63 × 10–25 kg m/s (for both) (b) 1.24kev (c) 1.51 eV
30] (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10
–10

m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an
average kinetic energy of (3/2) kT at 300 K.
[NCERT] Ans: (a) 6.686 × 10
–21
J = 4.174 × 10
–2
eV (b) 0.145 nm

47. Dushyantha Rao mob:[phone] Page 1

ATOMS
INTRODUCTION
Dalton was the first to postulate that matter is made of atoms which are indivisible.
The first atom model(plum-pudding model) was proposed by J.J. Thomson.
Thomson’s model failed to explain the origin of spectral series of hydrogen & other atoms. It also failed to
explain large angle scattering of α particles from thin metal foils.

ALPHA-PARTICLE SCATTERING EXPERIMENT :
Alpha particles (of energy 5.5MeV) emitted
by a radioactive source (83Bi
214
) are
collimated into a narrow beam with the help
of a lead slit. The collimated alpha
particles is allowed to fall on a thin gold foil
of thickness 2.1x10
-7
m.The alpha particles
scattered in different directions are observed
through a rotatable detector consisting of a
zinc sulphide (ZnS) screen and a
microscope.The alpha particles produce bright flashes or scintillations on the ZnS screen. These are
observed in the microscope and counted at different angles from the direction of the incident beam. The
angle of deviation θ of the alpha particle from its initial direction is called scattering angle.

Observation: The graph obtained by plotting the number of
alpha particles scattered in a given time as a function of
scattering angle is as shown in figure. It is observed from the
graph that
(i) most of the alpha particles pass straight through the gold
foil. That means they do not suffer any collision with gold
atoms.
(ii) Only about 0.14% of the incident alpha particles are
scattered by more than 1
0
.
(iii) about 1 in 8000 of the incident alpha particles is deflected by more than 90
0
.

Conclusion: Rutherford said that, to deflect alpha particles backwards (by more than 90
0
) , it must
experience a large repulsive force. This force could be provided if the greater part of mass of the atom and
its positive charge were concentrated tightly at its centre in a nucleus. If the alpha particles could get very

48. Dushyantha Rao mob:[phone] Page 2

close to the positive charge at the centre without penetrating it , then it would result in large deflection of the
alpha particle. Thus Rutherford’s assumption could explain the large angle scattering of alpha particles.

Distance of closest approach: It is the distance of alpha particle from the nucleus in which its kinetic energy
becomes zero.
2
0
12
4



ze
d
k
Where z Atomicnumber
k kineticenergy

Impact parameter: It is defined as the perpendicular distance of the initial velocity of the alpha particle
from the central line of the nucleus.
22



d
b cot
Where d distance of closest approach
scattering angle



RUTHERFORD’S ATOM MODEL: (Postulates)
1. Every atom consists of a central core, called the atomic nucleus, in which the entire positive charge and
almost entire mass of the atom are concentrated.
2. The size of nucleus is of the order of 10
-15
m, which is very small as compared to the size of the atom
which is of the order of 10
-10
m.
3. The atomic nucleus is surrounded by certain number of electrons which revolve around the nucleus in
various circular orbits as do the planets around the sun. The necessary centripetal force is provided by the
electrostatic force of attraction between the electrons and the nucleus.
4. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus
is equal to the total positive charge on the nucleus.

Limitations of Rutherford atom model
Failed to explain stability of atom , and the spectrum f radiation emitted by atom

BOHR MODEL OF HYDROGEN ATOM
Postulates of Bohr atom model:
First postulate : An electron in an atom could revolve in certain stable orbits without the emission of
radiant energy. Such orbits are called stationary orbits.

49. Dushyantha Rao mob:[phone] Page 3

Second postulate: Electron revolves around the nucleus only in those orbits for which the angular
momentum of the electron is an integral multiple of 2
h
 where h is planck’s constant.
That is 2
h
L mvr n

 where n = 1 , 2 , 3…….
Third postulate: When an electron makes a transition from a higher energy orbit to lower energy orbit , the
difference in the energies emitted as a photon of frequency fi
EE
h



Expression for radius of stationary orbits:
Consider an atom of atomic number Z
+Ze  the charge on its nucleus
m mass of an electron
-e charge on electron
r the radius of the circular orbit
v the linear velocity.
The necessary centripetal for the electron is provided by the electro static force of attraction between the
nucleus & the electron is
ceF = F
22
2
0
1
4

mv Ze
rr

22
0
1
1
4
  mv r Ze

From Bohr’s quantization rule, we have 2
nh
mvr



squaring on both sides we get

22
222
2
2
4

nh
m v r

22
2222
2
2
4
Dividing (2) by (1) we get
1
4







nh
m v r
mv r
ze
c


22
0
2

n
nh
r
mze


For H-atom Z = 1 , 22
0
2n
nh
r
me




50. Dushyantha Rao mob:[phone] Page 4

Note :
1] From the equation for radius, it follows that 2
n
rn
Then r1 : r2 : r3 : = ??????
1
2
: ??????
2
2
: ??????
3
2
= 1
2
: 2
2
: 3
2
= 1 : 4 : 9
This means electron orbits of an atom are not equally spaced.
2] The radius of the first orbit (n=1) of hydrogen atom is known as the Bohr’s radius, for hydrogen atom Z = 1
∴ Bohr’s radius, 2
0
1 2
h
r
me


 Substituting the values of ??????
0, h, m and e
We get 0
1
r 0.53A


Expression for Orbital Velocity of electron:
Consider an atom of atomic number Z
+Ze  the charge on its nucleus
m mass of an electron
-e charge on electron
r the radius of the circular orbit
v the linear velocity
The necessary centripetal for the electron is provided by the electro static force of attraction between the
nucleus & the electron is
ceF = F
22
2
0
1
4

mv Ze
rr 
22
0
1
1
4
  mv r Ze


From Bohr’s quantization rule, we have
2
2

nh
mvr

2
2
0
22
2
1
4
Dividing (1) and (2) we get
4







ze
mv r
mvr nh



2
0
2

n
ze
v
nh

For H-atom Z =1, 2
0
2
n
e
v
nh


51. Dushyantha Rao mob:[phone] Page 5


Note : From the equation for radius, it follows that 1
n
v
n

For the 1
st
orbit of a hydrogen atom n = 1, 2
61
1
0
2 10
2
e
v ms
h

  
This shows that the electron revolves around the nucleus with a very high speed

Expression for total Energy of electron:
Consider an atom of atomic number Z
+Ze  the charge on its nucleus
m mass of an electron
-e charge on electron
r the radius of the circular orbit
v the linear velocity.
The electrons revolving around the nucleus possess
Potential energy 
2
0
1
1
4
  
ze
U
r 2
2
22
00
1
Kinetic energy of the electron
2
1 1 1
2 4 4

   
 
   
K mv
ze
K mv r Ze
r 


Total energy of electron E = K + U
22
00
1 1 ze 1 ze
E
2 4 r 4 r
   
     
 
   
2
0
11
24
ze
E
r

Substituting 22
0
2n
nh
r
mze


 in the above equation, we get
24
222
0
8
n
mz e
E
nh
 This is the equation for Total energy of electron
For H-atom Z =1, 4
222
0
8
n
me
E
nh

Where n is principal quantum number. i.e. n = 1,2,3,4,5,…………∞

52. Dushyantha Rao mob:[phone] Page 6

Note:
1. The –ve sign indicates that, the electron is bound to the nucleus of an atom. Therefore energy must be
supplied (or) work must be done to remove electron.
2. As the orbit of electron increases (n), the energy of electron (En) also increases. This means electron in the
outer orbit has greater energy than the electron in the inner orbit.
3. Neglecting the –ve sign, give the binding energy of electron, which decreases with the increase in the orbit
of electron.
Expression for wave number
From Bohr’s postulate, the energy of photon emitted when electron jumps from higher energy state ‘n2’ to
lower energy state ‘n1’is 21
24
222
0
2 4 2 4
2 2 2 2 2 2
0 2 0 1
24
2 2 2 2
0 1 2
24
2 2 2 2
0 1 2
For the electron in the nth state, energy is,
8
88
11
8
11
8
nn
n
h E E
mz e
E
nh
mz e mz e
h
n h n h
mz e
h
h n n
hc mz e
h n n










     


 


 

24
2 3 2 2
0 1 2
4
2 7 1
2 2 2 3
1 2 0
22
12
1 1 1
8
1 1 1
1.097 10
8
1,
11


 


     





mz e
ch n n
me
Rz where R m called Rydberg constant
n n ch
for H atom z
then R
nn






Bohr’s explanation of spectral series of hydrogen atom
When hydrogen gas is heated, the electrons go to different excited states depending upon the amount of
thermal energy received by each electron. When they return to the lower energy states, they emit energy in the
form of electromagnetic radiation giving rise to several spectral series.
22
12
11
R
nn





53. Dushyantha Rao mob:[phone] Page 7

Following are the spectral lines formed in hydrogen
1. Lyman series: This series obtained due to the transition of an electron from higher orbits to the first
orbit. i.e from n2 = 2,3,4,……..to n1=1. These spectral lines in the ultraviolet region of Electromagnetic
spectrum.
2. Balmer series: this series is obtained due to the transition an electron from higher orbits to the second
orbit. i.e from n2 = 3,4,6…….. to n1=2. There spectral lines lie in the visible region. These spectral
lines are called H??????,??????
&#3627409149;,??????
&#3627409150;
3. Paschen series: This series is obtained due to transition of an electron from higher orbits to the third
orbit. i.e from n2 = 4,5,6,7……to n1 =3. These spectral lines lie in the infrared region.
4. Brackett series: This series is obtained due to the transition of an electron from higher orbits to the
fourth orbit. i.e from n2 = 5,6,7…..to n1 =4 . These spectral lines lie in the infrared region.
5. Pfund series: This series is obtained due to the transition of an electron from higher orbits to the fifth
orbit. i.e from n2 = 6,7,8… to n1 =5. There spectral lines lie in infrared region.
Note :
1] In a given series, first spectral lines has the maximum wavelength and last spectral line has shortest
wavelength.
2] By substituting n2 =∞ in each of these spectral series, the series limits can be obtained. The series
limit con be obtained. The series limit corresponds to the last spectral line of the series. This spectral
line has lowest wavelength or highest frequency.
Energy level Diagram:
Transition of electron from higher orbit to lower orbits are represented by diagrams know as energy
level diagrams. The energy of the electron in the n
th
orbit of a hydrogenation atom is given

54. Dushyantha Rao mob:[phone] Page 8


Excitation and Excitation Potential
Excitation: The process of transferring an electron from its orbit to a higher orbit is called excitation.
Excitation energy: The energy required to transfer an electron from its orbit to a higher orbit is known as
excitation energy.
Excitation potential: The minimum accelerating potential which provided an electron energy sufficient to
jump from inner orbit to one of the outer orbits is known as excitation potential.

Eg. For hydrogen atom, the energy required to raise the electron from n =1 to n = 2 state is
E = E2-E1
E = -3.4 – (-13.6) = +10.2ev
∴ First excitation potential is +10.2V
Ionosation and Ionization Potential
Ionisation: of an atom is the process of removing electron from the atom.
Ionization energy: The energy required to remove an electron from the atom is called ionization energy.
Ionization potential: The minimum accelerating potential which provides an electron, energy sufficient just to
remove it from the atom is known as ionization potential.

Eg. Therefore Ionization potential of H2 atom=13.6 volt.
Ionisation energy of H-atom is =E∞ - E1 = 0-(-13.6) = 13.6ev the energy required to remove an electron out
of an atom is known as Ionisation energy.

Demerits of Bohr’s theory:
1. Bohr’s theory is applicable only to hydrogen atom and hydrogen like atoms
2. Theory is not applicable to atoms having more than one electron
3. Bohr’s theory fails to explain the fine structure of spectral lines even in hydrogen atom.
4. It does not explain the relative intensity of spectral lines.
5. Bohr’s theory does not take into account wave nature of electrons
6. This model couldn’t explain stark effect and Zeeman effect.

de-Broglie’s explanation of Bohr’s second postulate:
According to de-Broglie , the electron in its circular orbit must be seen as
a matter wave. We know that when a stretched string fixed at two ends is
plucked, the waves which survive are those which travel a total distance
down the string and back equal to integral number of wavelengths .
For an electron moving in the n
th
circular orbit of radius r with speed v

55. Dushyantha Rao mob:[phone] Page 9

Total distance travelled = circumference of the orbit = 2πrn.
For a stationary orbit , 2πr = nλ where n = 1 , 2 , 3 …. For a de-Broglie wave , the wavelength is
2
2



h
mv
h
rn
mv
h
mvr n
Whichis Bohr second postulate




56. DUSHYANTHA RAO MOB:[phone] Page 1

NUMERICAL PROBLEMS ON ATOMS
1. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7
MeV α-particle before it comes momentarily to rest and reverses its direction?
[NCERT] Ans: 3.0 × 10
–14
m
2. It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton
and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
[NCERT] Ans: 5.3 × 10
–11
m, 2.2 ×10
6
m/s
3. Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman
series of the hydrogen spectrum.
[NCERT] Ans: λ21 = 1218 Å, λ31 = 1028 Å, λ41 = 974.3 Å, and λ51 = 951.4 Å.
4. What is the shortest wavelength present in the Paschen series of spectral lines?
[NCERT] Ans: 820 nm.
5. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation
emitted when the atom make a transition from the upper level to the lower level?
[NCERT] Ans: 5.6 × 10
14
Hz
6. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies
of the electron in this state? [NCERT] Ans: 13.6 eV; –27.2 eV
7. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level.
Determine the wavelength and frequency of photon. [NCERT] Ans: 9.7 × 10
–8
m; 3.1 × 10
15
Hz
8. (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2,
and 3 levels. (b) Calculate the orbital period in each of these levels.
[NCERT] Ans: (a) 2.18 × 10
6
m/s; 1.09 × 10
6
m/s; 7.27 × 10
5
m/s
(b) 1.52 × 10
–16
s; 1.22 × 10
–15
s; 4.11 × 10
–15
s.
9. The radius of the innermost electron orbit of a hydrogen atom is 5.3×10
–11
m. What are the radii of
the n = 2 and n =3 orbits? [NCERT] Ans: 2.12×10
–10
m; 4.77 × 10
–10
m
10. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series
of wavelengths will be emitted? [NCERT] Ans: Lyman series: 103 nm and 122 nm
11. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s
revolution around the sun in an orbit of radius 1.5 × 10
11
m with orbital speed 3 × 10
4
m/s. (Mass of
earth = 6.0 × 10
24
kg.) [NCERT] Ans: 2.6 × 10
74
12. The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
[NCERT] Ans: + 3.4 eV, – 6.8 eV

57. DUSHYANTHA RAO MOB:[phone] Page 2

13. Calculate the shortest and longest wavelength of balmer series of hydrogen atom. Given R = 1.097
x 10
7
m
-1
. [M-16]
14. The first member of the Balmer series of hydrogen atom has wavelength of 6563Å, calculate the
wavelength and frequency of the second member of the same series. Given: C=3x10
8
ms
-1
.[M-17]
15. An electron transmission occurs from n=4 and n=1 energy level in hydrogen atom. Find the
wavelength of the emitted radiation if the energy of the electron in the ground state is -13.6 eV. To
which series does the spectral line belong?
16. Calculate the wave number, wavelength and frequency of spectral line of hydrogen for transition
n2 = 3 n1 = 2, R = 1.097 x 10
7
m
-1
. Ans: (1.524 x 10
6
m
-1
, 6563Å, 4.57 x 10
14
Hz)
17. The wavelength of first member of Lyman series is 1215 Å. Calculate the wavelength of the third
member of Balmer series. Ans: (4543 Å)
18. The series limit of hydrogen spectrum in the visible region is 364.6 nm. Calculate the longest
wavelength of the region and the frequency of the corresponding line.
Ans: (656.3 nm, 4.571 x 10
14
Hz)
19. Calculate the wavelength of the first member and the series limit of Lyman series of hydrogen
spectrum (Rydberg constant = 1.096 x 10
7
m
-1
). Ans: (1217Å, 912Å)
20. Calculate the wave number, wavelength and frequency of Hα , Hβ , and Hγ line of hydrogen
spectrum. Given R = 1.097 x 10
7
m
-1
. Ans: (2.057 x 10
6
m
-1
, 4861 Å, 6.171 x 10
14
Hz, 6563 Å)

58. Dushyantha rao mob: [phone] Page 1

Nuclei
Nuclear physics deals with the study of properties and behavior of nucleus and its constituents.
Nucleus of an atom made up of positively charged protons and electrically neutral neutrons.
Both protons and neutrons are collectively called nucleons
Atomic number(Z): The total number of protons present in the nucleus.
Mass number(A): The total number of both protons and neutrons.
(A-Z) The number of neutrons.
Total number of nucleons = protons+ neutrons
A = Z+N
A typical nucleus is represented as  
A proton nuetron
z proton
XX

Classification of nuclei
1) Isotopes : Isotopes of an element are the atoms of the element which have same atomic number (Z)
but different mass number (A)
Eg: 1 2 3
1 1 1
H , H & H are the isotopes of hydrogen
10 11 12 13 14
6 6 6 6 6
C , C , C , C & C are the isotopes carbon.
2) Isobars :are atoms of different elements which have the same mass number (A) but different atomic
number (Z)
Eg : 22 22
11 10
Na & Ne are isobars
3) Isotones: isotones are the nuclides which contain the same number of neutrons .
Eg: 34
12
H & He are isotones.
4) Mirror nuclei: are those nuclei in which the number of protons of one is equal to the number of
neutrons of the other.
Eg.: 77
43
Be & Li are mirror nuclei

General properties of nucleus
1.Nuclear size :∝-Scattering experiments showed that the mean radius of an atomic nucleus is about 10
-15
m.
Note: The volume of a nucleus is proportional to its mass number
1
3
1
3
3
3
0
00
VA
4
RA
3
RA
RA
R R A
Where R constant Value of R 1.3fermi







59. Dushyantha rao mob: [phone] Page 2

2. Nuclear density( ): It is the mass per unit volume of the nucleus.
Note: mA is the nuclear mass
Where m average mass the nucleons
A mass number
Volume of the nucleus 34
VR
3

33
0
3
0
17
3
massof nucleus
Nucleardensity
volumeof nucleus
mA mA
44
R R A
33
m
4
R
3
kg
2.3 10
m

  



  
Nuclear density does not depend on the mass number (A) hence nuclear density is same for all nuclei
.
?????? Nuclear charge: Nucleus contains both protons & neutrons. Protons are positively charged and neutrons
are electrically neutral. The nucleus is (+ve).
If Z is the number of protons then the charge one nucleus is (+Ze)
4. Nuclear mass: Mass of the nucleus is always equal to sum of the masses of nucleons present in it
Nuclear mass = total mass of protons + total mass of neutrons
M = [Z mp + (A-Z) mn]
Where mp  mass of each proton, Z  atomic no.
mn  mass of each neutron, (A - Z)  no. of neutrons.
5. Nuclear spin: The nucleons present in the nucleus have orbital as well as spin angular momentum. The
total angular momentum of the nucleus is the resultant of spin and orbital angular momentum of the
nucleons. The resultant angular momentum of the nucleus is called nuclear spin.

Properties of nuclear forces
1. Nuclear forces are strong attractive forces: The magnitude of nuclear forces is 100 times that of
electrostatic force and 10
38
times that of gravitational force between nucleons.
2. Nuclear forces are short range forces: These act for a very short distances. When the distance between
nucleons is greater than 10 fermi, force become negligible.
3. Nuclear forces are non – central forces:the force existing between two nucleons doesnot act along the line
joining their centres.
4. Nuclear forces exhibit saturation property:each nucleon interacts only with a limited number of nucleons
nearest to it.

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5. Nuclear forces do not obey inverse square low
6. Nuclear forces are charge independent: The force of attraction between a proton and proton is the same
as the force of attraction b/n proton and neutron or a neutron and neutron.
7. Nuclear force depends on the spin of the nuclei: the force of attraction between two nucleons having
parallel spin is stronger than nucleons having anti-parallel spin.
8. Nuclear forces are exchange forces: the nuclear forces are brought into existence due to the exchange of
particles called ?????? meson.
9. Nuclear forces have a repulsive core: at too close distance (0.5 fermi) of approach between two nucleons
force becomes repulsive.

H. Yukawa s theory: There is a continuous exchange of ?????? meson between the nucleons. This results in
exchange force and keeps them bound together. A proton inside the nucleus emits ??????
+
meson and becomes
neutron. A neutron absorbs ??????
+
meson and becomes proton. A neutron emits ??????
−
meson and becomes proton.
A proton absorbs ??????
−
meson and becomes neutron.
The attractive force between P-P and n-n develops due to the exchange of ??????
0
meson.

Einstein’s mass – energy relation: Einstein showed that energy and mass are inter convertible i.e. mass can
be converted into energy and energy can be converted into mass.
The inter – conversion of mass and energy is governed by the mass – energy relation.
E = mc
2

E  energy equivalent to mass m
C speed of light

Eg. 1) when an electron and a positron come close to each other they pair annihilate and destroy each other.
Their mass converted into energy and released in the form of gamma rays.
2) During pair production energy is converted into mass. When a gamma ray of photon approaching a
nucleus, converted into a pair of particles- an electron and a positron.
3) During nuclear fission of heavy nucleus into smaller nuclei, there is a decrease in total mass. This
decrease in mass is converted into energy.

Atomic mass unit (amu):Atomic masses and nuclear masses are measures in an unit called atomic mass unit
Define 1 atomic mass unit: One atomic mass unit is defined as (1/12)
th
the mass of an atom of C -12 isotope.

Show that 1 amu = 1.66 x 10
-27
kg
From Avogadro hypothesis
6.023 x 10
23
atoms of C-12 weigh →12 x 10
-3
kg

61. Dushyantha rao mob: [phone] Page 4

∴ mass of 1 atom of C – 12 weigh → 12 x 10
-3
/ 6.023 x 10
23
kg
By the definition
1amu = 1/12 (mass of 1 atom of C-12 isotope)
⇒ 1 amu = 1/12 [12 x 10
-3
/ 6.023 x 10
23
]
 1 amu = 1.66 x 10
-27
kg

Electron – volt (ev) – It is the unit of energy.
Define 1 electron – volt: It is the energy gained by an electron, when accelerated through a p.d of one volt.
Work done = charge x potential
∴ 1 ev = 1.6 x 10
-19
c × 1v
19
1eV 1.6 10 J



Relation between amu and eV (or) To show 1amu = 931Mev
According to Einstein mass – energy relation
E = mc
2
Take m = 1 amu = 1.66 x 10
-27
kg
E = 1.66 x 10
-27
x (3 x 10
8
)
2

E = 14.944 x 10
-11
J
⇒ E = 14.944 x 10
-11
/ 1.6 x 10
-19
eV
⇒ E = 931 x 10
6
eV
⇒ E = 931 M eV
Mass defectm : It is the difference between rest mass of a nucleus and sum of the mass of nucleons,
which forms nucleus.
Mass defect  
Pn
m Zm A Z m M      

Where Z atomic number
A Mass number
P
m  Mass of each proton
n
m  Mass of neutron
M Mass of nucleon
This mass defect will be converted into energy, which is used to bound the nucleons in the nucleus called
Binding energy.

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Packing fraction (pf): It is the ratio of mass defect to the mass number of the given isotope
 MAMass defect
P.F
Mass number A


Binding energy: It is the energy required to bind (or) to separate the nucleons in the nucleus.
Specific binding energy (Binding energy per nucleon):
It is the ratio of the binding energy of the nucleus to the Mass number
It is also defined as average energy required to remove each nucleon from the nucleus.
Binding energy BE
Specific BE
Mass number A

Binding energy curve: The curve obtained by plotting specific binding energy v/s mass number is called
B.E curve.

Some of the results obtained from the above graph are as follows
1) Average B.E/nucleon for light nuclei is very small.
2) Specific B.E increases with increase in mass number & becomes maximum (8.8 MeV) at A=56
3) Between A=56 & A=120 curve remains almost constant & all the nuclei lying in this region possess
high specific B.E &hence they are stable. In this region specific BE decreases slowly from 8.8 MeV
to 7.6 MeV
4) Above A=120 specific BE decreases rapidly & thus nuclei lying in this region are unstable.
5) Towards lighter nuclei side, nuclei like ??????&#3627408466;,
2
4
&#3627408438;
6
12
,??????,
8
16
etc…. which are having even & equal number
of proton & neutron are found to be highly stable & nuclei like ????????????
3
6
, &#3627408437;,
5
10
having add & equal number
of proton & neutron are found to be unstable

63. Dushyantha rao mob: [phone] Page 6

Importance of B.E curve
It gives the information about how much energy is needed to fuse two nuclei
Or how much energy is liberated when a nuclei is broken .

RADIOACTIVITY
Radioactivity: The spontaneous dis-integration of heavy nucleus with the emission of certain radiations is
called radioactivity.
The radiations emitted during radioactivity phenomenon are called radioactive radiations.
Elements which under goes radioactive phenomenon are called radioactive elements.
Example: Uranium, Thorium, Neptunium, Radium, Plutonium etc……..
The different radioactive radiations are &#3627409148;,&#3627409149; &#3627408436;??????&#3627408465; &#3627409150; radiations.
Note: Elements with atomic number ≥83 and mass number A > 200 exhibit the phenomenon of
radioactivity naturally.
Radioactivity was discovered by Henry Becquerel
Radioactive decay law:
Statement: The rate of disintegration of a radioactive nuclei is directly proportional to the total number of
nuclei present at that instant of time. dN
Rateof disin tegration N
dt
dN
N
dt
Where N Number of nuclei present
Decay constant

 




Derive the relation t
o
N N e


Let o
N be the number of nuclei present initially (t = 0)
N  be the number of nuclei present after time t dN
Then the rateof disin tegration N
dt
dN
N
dt
Where Decay constant
negative sign indicate that the number of atoms will be decreasing with respect to time.

 


64. Dushyantha rao mob: [phone] Page 7


o
Nt
N0
o
0
t
0
t
o
dN
dt
N
Integrating on both sides with limits
dN
dt
N
ln N ln N t
N
ln t
N
N
e
N
N N e


 
 
  

 




Graph of number of atoms with time is as shown.





Half life: The time required to disintegrate half the initial nuclei is called half life period.
Derive an expression for half life:
From decay law the number of undis-integrated nuclei 
t
o
N N e 1





1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
o
T
o
o
T
N
When t T N
2
N
1 becomes N e
2
1
e
2
1
ln T
2
ln 2 T
2.303log 2 T
0.693 T
0.693
T






 







Note: At 1
2
o
n
N
t nT N
2

Thus after n half lives the number of atoms that remain unchanged is
1
2
?????? times the initial number of atoms.

65. Dushyantha rao mob: [phone] Page 8

Decay constant: It is defined as the reciprocal of the time during which 37 % of nuclei will be left
undisintegrated. (63%disintegrated)
Mean life: It is the ratio of the sum of the lives of all the individual atoms to the total number of atoms
present in the sample. 1
2
1
2
m
m
m
Total life time of all the atoms
Mean life
Total number of atoms present in the sample
1
T
T
T
0.693
TT






Activity (A): The number of disintegration per second is called activity.
dN
N
dt
AN


Note: Activity t
o
A A e



Where A  Activity after time t
A0  Initial activity of the sample
SI unit of activity is Becquerel.
Note: 1 Becquerel = 1 disintegration/second.
1 curie = 3.7 x 10
10
disintegration/second.
1 Rutherfort = 10
6
disintegration/second.
Note :1) Radioactivity is a natural phenomenon. It is not affected by external factors such as temperature,
pressure, electric and magnetic fields, chemical reactions etc.
2) The activity depends only on the radioactive substance and the number of atoms taken. 
- decay: Alpha decay is the process in which a nucleus decays spontaneously emitting alpha particles. The
alpha particle emitted have discrete values of energy.
When a radioactive substance emits ∝ - particle, its atomic number decreases by two and mass number
decreases by four. The new element so formed falls in a group two to the left of the parent element in the
periodic table.

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Generally decayA A 4 4
z z 2 2
X Y He
 

 

decay238 234 4
92 90 2
Exmaple U Th He

 

This spontaneous decay is possible only when the total mass of the decay products is less than the mass of
the initial nucleus. The decrease in mass during the decay appear as KE of the products.
The disintegration energy or the Q-value of the of the reaction: is the difference between the initial mass
energy and the total mass energy of the decay products.
β-decay: When a radioactive substance emits &#3627409149; – particle, its atomic number increases by one and mass
number remains same. During beta decay electron or positron are emmited. Along with beta particle one
more particle called neutrino (ν) or antineutrino(??????̅)
The beta particle emitted have continuous range of energies upto maximum value called end point energy.
i. During electron emission (
 decay): Antineutrino (??????̅) is also emitted. And the atomic no of product
nucleus increased by one but mass no remains the same.
Example: decay234 234 0
90 91 1
Th Pa e



   
decay32 32 0
15 16 1
P S e



   
The electron emited as beta particle is not atomic electron but emitted due to decay of the nucleus. In the
nucleus, a netron in converted into a proton, an electron and antineutrino.
decay1 1 0
0 1 1
N H e



   
ii. During electron emission (
 decay): Neutrino is also emitted. In this process, the atomic no of product
nucleus decreased by one but mass no remains the same.
Example: decay22 22 0
11 10 1
Na Ne e



   
In the nucleus, a proton is converted into a neutron, positron and neutrino.
decay1 1 0
1 0 1
H N e



   
An alternate process to positron emission is K-electron capture. In an excited nucleus, a proton may turn
into a neutron by absorbing an electron in the K-shell instead of emitting a positron.
63 0 63
30 1 29
Zn e Cu

   

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Gamma decay: Gamma decay is the process in which an excited nucleus makes a transition to a state of
lower energy by emitting a photon. When a radioactive substance emits ray neither the atomic number
nor the mass number changes.
The emitted gamma rays have discrete values of energy.
When a nucleus of a radioactive substance undergoes alpha or beta
decay, the product nucleus may be in one of the excited states. This
excited nucleus returns to a lower energy state by emitting a photon.

Example: Co-60 decays by emitting a electron, the product nucleus formed is Ni-60 which may be in an
excited state. The successive transition from the excited state to the ground state leads to the emission of
gamma ray of energies 1.17MeV and 1.33MeV
Nuclear fission and Nuclear fusion
Nuclear Fission: It is the process in which a heavier nucleus breaks into two lighter nuclei of comparable
masses releasing large amount of energy .

Example: When U
235
is bombarded with a neutron, it captures the neutron and becomes unstable compound
nucleus U
236
. The compound nucleus then splits into two fragments such as B
141
& Kr
92
. In addition three
neutrons are produced together with large amount of heat energy.
Example: 1.  
*
235 1 236 141 92 1
92 0 92 56 36 0
U n U Ba Kr 3 n Energy     

2.  
*
235 1 236 140 94 1
92 0 92 54 38 0
U n U Xe Sr 2 n Energy     
Chain reaction: When an atom of
235
U undergoes fission, along with the fission fragments, about 2 to 3
neutrons are also produced, under suitable conditions, these neutrons can cause fission in the other atoms of
the sample & release more number of neutron. This process continues till all the atoms of the sample
undergo fission. This process is known as nuclear chain reaction.
There are two types of chain reaction
1) Controlled chain reaction In the controlled chain reaction the neutron number is controlled using
cadmium rods
Example: Nuclear power reactor
2) Uncontrolled chain reaction  In this reaction the neutron number is uncontrolled
Example: Atom bomb

68. Dushyantha rao mob: [phone] Page 11

Critical size: The minimum size of the sample for which the sustained chain reaction takes place is called
critical size.
Note: For the chain reaction to be taken place the released neutron should not escape from sample. The
leakage of neutrons could be minimized by taking suitable shape of the fissile material in the form of sphere.

Parts of power reactor
1) Nuclear fuel: Normally, heavy nuclei Th–232, U–235, Pu–239, are taken as nuclear fuel. They are
embedded in graphite blocks.
2) Moderators: Moderators are used to slow down the fast moving neutrons. Moderators should not
absorb neutrons. Heavy water, graphite are used as moderators.
3) Control rods: the function of control rods is to control the neutron no. cadmium or boron are used as
control rods.
4) Coolant: During the fission reaction large energy will be released. This energy will be absorbed by
the coolant such as D2O (Heavy water), CO2, etc. in turn coolant transfers the energy to water and
water will be converted into steam, which is used for different forms of energy.
5) Safety wall: To avoid the leakage of harmful radiations like &#3627409148;,&#3627409149;,&#3627409150;, the enter chamber is surrounded
by a concrete wall of thickness 10 – 15 feet.








Uses of power reactor:
1) Power reactors are used to generate electric power.
2) Used to run submarines.
Nuclear fusion: It is a process in which two or more lighter nuclei combine a heavier nucleus with the
release of large energy.
Example: 4??????
1
1
??????&#3627408466;
2
4
+ 2 &#3627408466;
1
0
+ Energy
??????
1
2
+ ??????
1
2
??????&#3627408466;
2
3
+ ??????
0
1
+ Energy
??????
1
2
+ ??????
1
3
??????&#3627408466;
2
4
+ ??????
0
1
+ Energy

69. Dushyantha rao mob: [phone] Page 12

Stellar energy: The nuclear fusion reaction are main responsible for large amount of energy produced in
stellar bodies like sun.
The types of thermonuclear reactions have been postulated in the stars for fusion hydrogen into
helium along with emission of enormous amount of energy.
Proton – Proton cycle 1 1 2 0
1 1 1 1
1 2 3
1 1 2
3 4 1
2 2 1
2 H 2 H 2 H 2 e Energy
2 H 2 H 2 He Energy
2 He He 2 H Energy
   
  
  

The same cycle can be summed up as follows.
1 4 0
1 2 1
4 H He 2 e Energy  
Difference b/n fusion and fission reaction
Nuclear fission Nuclear fusion
1. It is a process of heavier nucleus splits into
two lighter nuclei of comparable masses with
the release of large energy
1. It is process in which two lighter nuclei
combined to form a stable heavier nucleus with
the release of energy
2. The energy released/ nucleon is less 2. The energy released/nucleon is more
3. It takes place at normal temp. 3. It takes place at very high temp.
4. It can be conducted artificially 4. It cannot be conducted artificially but take
place at stars
5. It can be controlled 5. It cannot be controlled

70. DUSHYANTHA RAO MOB: [phone] Page 1

NUMERICAL PROBLEMS ON NUCLEI
1. Obtain approximately the ratio of the nuclear radii of the gold isotope 197
79Au and the silver isotope 107
47Ag
[NCERT] Ans: 1.23
2. Given the mass of iron nucleus as 55.85u and A=56, find the nuclear density?
[NCERT] Ans: 2.29 × 10
17
kg m
–3

3. Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this,
express the mass defect of 8O
16
in MeV/c
2
. [NCERT] Ans: 127.5 MeV/c
2
4. Calculate the binding energy and binding energy per nucleon (in MeV) of nitrogen nucleus (7N
14
)
from the following data: Mass of proton=1.00783u, Mass of neutron=1.00867u and Mass of nitrogen
nucleus=14.00307u. [M-14]
5. Calculate the binding energy and binding energy per nucleon of oxygen nucleus (8O
16
) using the
following data in Mev. Mass of proton = 1.007825u, mass of neutron = 1.008665 u and mass of
oxygen nucleus =15.995u. [J-17]
6. A copper coin has a mass of 63.0 g. Calculate the nuclear energy that would be required to separate
all the neutrons and protons form each other. The coin is entirely made of 63
29Cu atoms.
Mass of 63
29Cu atom = 62.92960 u
Mass of proton = 1.00727 u
Mass of neutron = 1.00866 u
Avogadro’s number = 6.022×10
23
[M-20] Ans: 535.94MeV
7. We are given the following atomic masses: 238
92U
= 238.05079 u 4
2He = 4.00260 u 234
90Th
= 234.04363 u 1
1H = 1.00783 u 237
91Pa
= 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
(a) Calculate the energy released during the alpha decay of 238
92U
b) Show that 238
92U can not spontaneously emit a proton.
[NCERT] Ans: 4.25 MeV, – 7.68 MeV, the Q of the process is negative and therefore it cannot
proceed spontaneously.

71. DUSHYANTHA RAO MOB: [phone] Page 2

8. Obtain the binding energy (in MeV) of a nitrogen nucleus 14
7N , given m 14
7N =14.00307 u
[NCERT] Ans: 104.7 MeV
9. Obtain the binding energy of the nuclei 56
26Fe and 209
83Bi in units of MeV from the following data:
Mass of 56
26Fe = 55.934939 u Mass of 209
83Bi = 208.980388 u
[NCERT] Ans: 8.79 MeV, 7.84 MeV
10. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate
all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 63
29Cu
atoms (of mass 62.92960 u). [NCERT] Ans: 1.584 × 10
25
MeV or 2.535×10
12
J
11. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of
(a) 226
88Ra and (b)220
86Rn
Given mass of 226
88Ra = 226.02540 u, mass of 222
86Rn = 222.01750 u,
Mass of 220
86Rn = 220.01137 u, 216
84Po = 216.00189 u, mass of 4
2He = 4.00260 u
[NCERT] Ans: (a) Q = 4.93 MeV, Eα = 4.85 MeV (b) Q = 6.41 MeV, Eα = 6.29 MeV
12. The radionuclide
11
C decays according to
11 11
156
2
e T 20.3minCB

    
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values: mass of 11
6C =11.011434 u and 11
5B =11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.
[NCERT] Ans: 0.9671MeV
13. The nucleus 23
10Ne ecays by β
–
emission. Write down the β-decay equation and determine the
maximum kinetic energy of the electrons emitted. Given that: 23
10Ne
=22.994466 u 23
11Na =22.089770 u [NCERT] Ans: 4.37MeV
14. The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA + mb – mC – md]c
2
where the masses refer to the respective nuclei. Determine from the given data the Q-value of the
following reactions and state whether the reactions are exothermic or endothermic. 1 3 2 2
1 1 1 1
12 12 20 4
6 6 10 2
H H H H
C C Ne He
  
  

Atomic masses are given to be

72. DUSHYANTHA RAO MOB: [phone] Page 3
2
1H
= 2.014102 u, 3
1H = 3.016049 u, 12
6C = 12.000000 u, 20
10Ne = 19.992439 u, 1
1H
=1.007825u, 4
2He = 4.002603u
[NCERT] Ans: 4.62Mev
15. Suppose, we think of fission of a 56
26Fe nucleus into two equal fragments, 28
13Al . Is the fission
energetically possible? Argue by working out Q of the process. Given m 
56
26Fe = 55.93494 u and m 
28
13Al
=27.98191 u. [NCERT] Ans: -26.88728 MeV, the energy is negative so, fission is not
possible
16. The fission properties of 239
94Pu are very similar to those of 235
92U . The average energy released per
fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 239
94Pu undergo
fission? [NCERT] Ans: 4.53X10
26
MW
17. The neutron separation energy is defined as the energy required to remove a neutron from the
nucleus. Obtain the neutron separation energies of the nuclei 41
20Ca and 27
13Al from the following data:
m 
27
13Al =26.981541 u, m 
26
13Al = 25.986895 u, m 
41
20Ca = 40.962278 u,
m 
40
20Ca = 39.962591 u, m
1
0n =1.008665 [NCERT] Ans: 8.362MeV, 13.06MeV
18. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-
particle. Consider the following decay processes: 223 209 16
88 82 4
223 4 219
8688 2
Ra Pb C
Ra HeRn



Calculate the Q-values for these decays and determine that both are energetically allowed.
m 
223
88Ra = 223.01850u, m 
209
82Pb =208.98107u, m
16
4C =14.00324u,
m 
219
86Rn =219.948u, m 
4
2He = 4.00260u.
[NCERT] Ans: 31.83MeV, 5.98MeV
19. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a)
3.125%, b) 1% of its original value? [NCERT] Ans: (a) 5 T years, (b) 6.65 T years
20. The normal activity of living carbon-containing matter is found to be about 15 decays per minute
for every gram of carbon. This activity arises from the small proportion of radioactive 14
6C present
with the stable carbon isotope 12
6C .When the organism is dead, its interaction with the atmosphere
(which maintains the above equilibrium activity) ceases and its activity begins to drop. From the
known half-life (5730 years) of 14
6C , and the measured activity, the age of the specimen can be
approximately estimated. This is the principle of 14
6C dating used in archaeology. Suppose a specimen

73. DUSHYANTHA RAO MOB: [phone] Page 4

from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the
approximate age of the Indus-Valley civilisation. [NCERT] Ans: 4224 years
21. Obtain the amount of 60
27Co necessary to provide a radioactive source of 8.0 mCi strength. The half-
life of 60
27Co is 5.3 years. [NCERT] Ans: 7.126 ×10
–6
g
22. The half-life of 38
90Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?
[NCERT] Ans: 7.877 ×10
10
Bq (or) 2.13 Ci
23. The half-life of 92U
238
undergoing α-decay is 4.5 × 10
9
years. What is the activity of 1g sample of
92U
238
? [NCERT] Ans: 1.23 × 10
4
Bq
24. Determine the mass of Na
23
which has an activity of 5mCi. Half life of Na
23
is 2.6 years. Avogadro
number = 6.023 x 10
23
atoms. [M-15]
25. Calculate the half life and mean life of radium – 226 of activity 1Ci. Given mass of radium 226 is 1g
and 226g of radium consist of 6.023 x 10
23
atoms. [J-15]
26. The activity of radioactive substance 4700 per minute. 5 minutes later the activity is 2700 per
minute. Find decay constant and half of radioactive substance. [J-16]
27. The half life of a radioactive sample 38????????????
90
is 28 years. Calculate the rate of disintegration of 15 mg
of this isotope. Given Avogadro’s number = 6.023X10
23
[J-18]
28. Half life of U-238 undergoing α- decay is 4.5X10
9
years. What is the activity of one gram of U-238
sample? [M-19]

74. DUSHYANTHA RAO MOB:[phone] Page 1

SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE
CIRCUITS
INTRODUCTION: Semiconductor electronics is a branch of Electronics which deals with the study of
manufacturing and operation of semiconductor devices.
Following are the advantages of semiconductor devices over their predecessor ’valves’ or vacuum tubes.
1. In a semiconductor device, the supply and flow of charge carriers are within the solid (Semiconductor)
itself. Whereas in a valve, charge carriers are supplied by heated cathode and they are made to move in an
evacuated space.
2. Semiconductor devices are small in size, consume low power, operate at low voltages and have long life
and high reliability. Whereas, valves are bulky, consume high power, operate generally at high voltages,
have limited life and low reliability.
On the basis of relative values of electrical resistivity, solids are broadly classified as –
(i) Metals (Or Conductors): Which possess very low resistivity of the order of 10
-2
– 10
-8
Ω m
(ii) Insulators: Which possess very high resistivity of the order of 10
11
– 10
19
Ω m and
(iii) Semiconductors: Which possess resistivity whose value is intermediate to that of conductors and
insulators (about 10
-5
– 10
6
Ω m).
Band theory of solids :
In a solid, the atoms are closely arranged. Due to interaction with electrons of neighboring atoms, each
discrete energy level of an electron splits into number of separate energy levels. Since these energy levels lie
very close to one another, they form an energy band.
The energy bands which are completely filled at 0K (zero kelvin) are known as valence bands. These are
occupied by valence electrons.
The energy bands with energies higher than valence bands are known as conduction bands. These may be
partially filled or empty.
The energy difference between the highest level of valence band and the lowest level of conduction band is
known as 'Energy gap'.
Classification of solids based on band theory of solids:
1. Conductors : In conductors, the valence and conduction bands
overlap on each other. Hence there is no energy gap between
them. Hence electrons can move easily from valence band to
conduction band even in the presence of a small applied field.
The energy band structure of a conductor is as shown.
2. Insulators : In insulators, the conduction band and the valence band are
separated by a large energy gap. Therefore electrons in the valence band
cannot go to the conduction band and hence insulators do not conduct
electric current.
The energy band structure of an insulator is as shown.
In an insulator, the valence band is completely filled while the conduction band is empty. BandBand
BandBand

75. DUSHYANTHA RAO MOB:[phone] Page 2

3. Semiconductors : In semiconductors, the conduction band and the
valence band are separated by a small energy gap. At 0 K, the
semiconductor behaves as a insulator. But at room temperature, some
electrons in valence band acquire thermal energy and jump to the
conduction band. Hence semiconductors conduct electricity at room
temperature. But the conductivity is less than that of good conductors.
The energy band structure of a semiconductor is as shown.
SEMICONDUCTORS
Semiconductors are materials whose conductivity lies between that of conductors and insulators. Silicon
and Germanium are the commonly used semiconductors.
INTRINSIC SEMICONDUCTORS:
A pure semiconductor is known as an intrinsic semiconductor.
Ge or Si are tetravalent elements. The atoms share their
valence electrons and form covalent bonds. No bond will
be broken at lower temperatures and hence there will be
no free electrons for conduction. Therefore, it behaves as
an insulator. As the temperature increases, due to
increased thermal energy, some bonds break creating
free electrons. The positions of free electrons in broken
covalent bonds behave as particles with a positive charge
known as ‘holes’. Free electrons and holes will be equal
in number (??????
&#3627408466;=??????
ℎ). Hence an intrinsic semiconductor
is electrically neutral. As the charge carriers are small in
number, its conductivity is low.
Under the action of an external field, both free electrons and holes
move in opposite directions constituting electron current (??????
&#3627408466;) and hole
current (??????
ℎ) respectively. The total current in the solid, ??????=??????
&#3627408466;+??????
ℎ.
In an intrinsic semiconductor, at 0K, the valence band is occupied and
conduction band is empty. At temperatures greater than 0K, some
electrons jump into conduction band due to their thermal energies leaving
behind an equal number of holes in the valence band. The energy – band
diagram for an intrinsic semiconductor is as shown.

EXTRINSIC SEMICONDUCTORS:
A doped or an impure semiconductor is known as extrinsic semiconductor.
Doping: Doping is a process of adding impure atoms to an intrinsic semiconductor.
Dopant: The impure atoms added to an intrinsic semiconductor are known as dopants.
Types of extrinsic semiconductors:
1) p-type semiconductor
2) n-type semiconductor

76. DUSHYANTHA RAO MOB:[phone] Page 3

p – type semiconductor:
It is a type of extrinsic semiconductor in which holes are the majority carriers.
It is formed when an intrinsic semiconductor is doped with a trivalent atom like Boron, Indium or Gallium.
When Silicon is doped with Boron impurity, the crystal structure is as follows :
Boron provides electrons to fill only three covalent bonds and the vacancy in the
fourth bond constitutes a hole. The hole behaves like a positively charged
particle and attracts electrons from the neighboring covalent bond. A new hole is
created in the place of the displaced electron and in this way the hole moves in a
crystal.
Since the hole can accept one electron from the neighboring atom, the impurity
atom which contributes this hole is known as acceptor impurity and the
semiconductor is known as acceptor type semiconductor.
At any instant, the number of holes is greater than the number of free electrons.
Hence holes are majority carriers and electrons are minority carriers in a p – type
semiconductor.
The energy band diagram (or structure) of a p – type semiconductor is as shown.
n – type semiconductors :
It is a type of extrinsic semiconductor in which electrons are the majority carriers.
It is formed when an intrinsic semiconductor is doped with a pentavalent atom like Phosphorus, Arsenic
or Antimony.
When silicon is doped with Phosphorus impurity, the crystal structure is as follows :
Of the five valence electrons of a phosphorus atom, four electrons involve in
covalent bonding with four neighboring silicon atoms. The fifth electron is
loosely bound to its parent nucleus and can become a free electron even at room
temperature. In this way, a large number of free electrons can be generated by the
doping of small quantity of impure atoms.
Since each impurity atom donates a free electron to the semiconductor, the
impurity atom is known as donor atom and the semiconductor is known as donor-
type semiconductor.
At any instant, the number of electrons is greater than the number of holes.
Hence electrons are majority carriers and holes are minority carriers in a n – type
semiconductor.
The energy band diagram of a n – type semiconductor is as shown.

77. DUSHYANTHA RAO MOB:[phone] Page 4

Difference between n-type and p-type semiconductor

n-type semiconductor p-type semiconductor
1. It is obtained when a pure semiconductor
is doped with pentavalent impurities.
1. It is obtained when a pure
semiconductor is doped with trivalent
impurities.
2. Electrons are the majority charge
carriers and holes are the minority charge
carriers.
2. Holes are the majority charge carriers
and electrons are the minority charge
carriers.
3. The impurity added is called donor. 3. The impurity added is called acceptor.
4. Majority charge carriers are in the
conduction band.
4. Majority charge carriers are in the
valence band.
5. Donor impurity level lies just below the
conduction band.
5. Acceptor impurity level lies just above
the valence band.

P-N JUNCTION (SEMICONDUCTOR DIODE)
A p-n junction is formed by adding pentavalent impurity on
one half of the pure semiconductor and trivalent impurity on
the other half of it.
Holes are the majority carriers in p-type region and electrons
are majority carriers in n-type. Due to concentration gradient
of majority carriers, few electrons diffuse from n-region to p-
region while holes diffuse from p-region to n-region
producing diffusion current. Hence a positive charge is
developed on n-side of the junction and a negative charge on
the p-side. The minority carriers are swept across the junction
plane, forming a drift current. When these two currents are
equal in magnitude, a potential difference is established across
the junction and is called junction potential difference or barrier potential difference (0.3V for Ge &
0.7V for Si) which prevents further diffusion of majority charge carriers. This potential is called junction
potential difference or barrier potential difference. The region at the vicinity of the junction which is
completely depleted of mobile charge carriers is called depletion region.
Biasing of p-n junction :
A p-n junction is said to be biased when an external voltage is applied across the
junction.
Forward Bias : A p – n junction is said to be forward biased, when positive
terminal of a battery is connected to the p – region and negative terminal of the
battery is connected to n – region of the p – n junction diode.
When the applied voltage is greater than the barrier potential, the majority

78. DUSHYANTHA RAO MOB:[phone] Page 5

carriers in both the regions acquire enough energy to cross over the barrier potential across the junction.
Hence a flow of electrons from n – region to p – region and holes from p – region to n – region begins. The
diode offers low resistance for the flow of charges. Thus a continuous current flows through the circuit.
Reverse Bias : A p – n junction is said to be reverse biased if the positive terminal of a battery is connected
to the n – region and negative terminal is connected to p – region of the p – n junction diode.
The negative potential at p – type region attracts the holes while the positive
potential at n – type region attracts the electrons. Due to this the junction widens
and the barrier potential increases. As there is no flow of majority charge
carriers across the junction, the diode does not conduct. But there will be a small
current through the circuit due to thermally generated minority charge carriers.
This current is known as 'Saturation current' (IS).
As the applied voltage is increased, the current through the circuit practically remains a constant. But when
the voltage becomes greater than 'break down voltage', the diode starts to conduct heavily due to Avalanche
effect.
Definitions :
1. Knee voltage : It is the applied voltage at which the forward current through a forward biased diode
starts increasing rapidly.
2. Saturation current : It is the small current through a reverse biased diode due to thermally generated
minority charge carriers.
3. Break down voltage : It is the applied voltage at which the current through a reverse biased diode
abruptly increases due to Avalanche effect.
4. Dynamic Resistance : It is the ratio of small change in voltage ∆?????? to a small change in current ∆?????? i.e.
??????
&#3627408465;=
∆??????
∆??????

V – I characteristics of a diode:
The circuit arrangement for studying V – I characteristics of a diode is as shown.
A battery is connected to a diode through a rheostat so that the applied voltage can be varied. For different
values of voltage, the corresponding value of current is noted. For forward bias, a milliammeter is used
while for reverse bias, a microammeter is used in the circuit. A plot of values of V against I is obtained

79. DUSHYANTHA RAO MOB:[phone] Page 6

using which the values of knee voltage, breakdown voltage and dynamic resistance of the diode can be
obtained.

Rectifiers :
RECTIFIER: A device which converts alternating current into direct current is known as a 'rectifier'.
The process of converting an alternating voltage (or current) into a direct voltage (or current) is known as
'rectification'.
A diode can be used for rectification process because it conducts easily when forward biased and does not
conducts when reverse biased i.e., it has the property of unidirectional conducting.
Rectifiers are of two types namely — (i) half wave rectifier and (ii) full wave rectifier
Half wave rectifier (HWR) :
The rectifier which converts only one half cycle of input
AC into rectified output (DC) is known as a half wave
rectifier.
Circuit: The circuit consists of a diode D connected in
series with a load resistance RL across which the rectified
output is obtained. The AC to be rectified is fed to the
primary of a transformer and the transformer output from secondary S is applied between A and B to the
diode.
Working: When the voltage at A is positive i.e., during positive half cycle of transformer output, the diode
D is forward biased and hence it conducts. There will be current through RL and hence a voltage across it.
During negative half-cycle, i.e., when voltage at A is negative, the diode is reverse biased and it does not
conducts. The voltage across RL is practically zero. Hence there is rectified output voltage only during
positive half cycles of input AC to the diode. The rectified output will be unidirectional but varying in
nature.
The input and output waveforms of a half wave rectifier are as shown :





Full wave rectifier (FWR) :
The rectifier which converts the complete cycle of input
AC into rectified output (DC) is known as a full wave
rectifier.
Circuit: The circuit consists of two diodes D1 and D2
whose positive ends are connected to secondary of the
transformer. The negative ends are connected to the
centre-tap of the secondary through a load resistance

80. DUSHYANTHA RAO MOB:[phone] Page 7

RL. The AC to be rectified is applied across the primary of the transformer. The centre-tap divides the
transformer output into two equal halves.
Working: During positive half cycles of transformer output, A is positive and B is negative with respect to
C. Diode D1 is forward biased and D2 is reverse biased. Hence D1 conducts and there will be current
through RL from M to N and hence an output voltage.
During negative half cycles, A is negative, B is positive Diode D2 is forward biased and D1 is reverse biased.
Hence D2 conducts and there will be current through RL from M to N and hence an output voltage.
Therefore, during both half cycles, there is an output voltage across RL. Hence the circuit is known as full
wave rectifier.
The input and output waveforms of full wave rectifier are as shown :





Note: Filter circuits are used to convert pulsating dc into steady dc.
SPECIAL PURPOSE DIODES
Zener Diode:
It is a special purpose semiconductor diode designed to operate under reverse bias in the breakdown region
Zener diode used as a voltage regulator.
The symbol for Zener diode is as shown.
Zener diode as voltage regulator:
The circuit diagram of a voltage regulator using Zener diode is as shown.
The Zener diode is reverse connected with the unregulated supply through a series resistance Rs. The
regulated output is taken across the load RL.
If the input voltage increases, the current through Rs
and Zener diode also increases. This increases the voltage
drop across Rs without any change in the voltage across the
Zener diode.
Similarly, if the input voltage decreases, the current through
Rs and Zener diode also decreases. The voltage drop across Rs
decreases without any change in the voltage across the Zener
diode. Thus any change in the input voltage results in the change in the voltage drop across Rs without any
change in voltage across the Zener diode. Hence the Zener diode acts as a voltage regulator.

81. DUSHYANTHA RAO MOB:[phone] Page 8

D
R
Ba
Optoelectronic junction devices
(i) Photodiode: The photodiode is operated in the reverse bias mode. When
light is made to incident on the depletion region of the diode, there is an
increase in the concentration of minority charge carriers. As a result the
reverse current increases which is directly proportional to the intensity of
light.
Applications
1. detection of both visible and invisible radiations
2. switching circuits
3. measurement of very low intensity of light.
4. In radio and TV receivers.

(ii) Light Emitting Diode (LED)
A heavily doped p-n junction in forward biased condition, constitutes a light emitting diode. In such an
arrangement, a large number of electron–hole recombination takes place due to the transition of electrons
from conduction band to the valence band.
During this process the energy is released as
photons whose wavelengths are in the visible
region. Depending upon the level of doping and the doping material,
different colours are obtained.
The wavelength of the emitted light is given by where Eg is the energy
gap.
Eg: Gallium arsenide, Gallium phosphide etc.
Note:
 To emit visible light, the energy gap must be between 1.8eV to 2.8eV.
 In pure silicon and germanium semiconductors when the electrons from the conduction band fall into
the valence band, the energy released is in the form of heat.

Applications:
1. In fancy lights for decorative purpose.
2. In calculators as seven segment displays.
3. In traffic signals
4. Infra red LED’s in remote controls
5. Message displays at airports and railway stations.



D
R
Ba

82. DUSHYANTHA RAO MOB:[phone] Page 9

(iii) Solar cell
A solar cell is basically a p-n junction which generates emf when solar
radiation falls on the p-n junction. It works on the same principle as
the photodiode, except that no external bias is applied. The junction
area is kept much larger for solar radiation so that more power can be
generated.
When light is incident on the p-n junction, the electron- hole pairs are
created at the junction. Due to the junction potential difference,
electrons move to n-side and holes to p-side. An emf is developed on
two sides of the semiconductor. A current flows in the circuit when an
external resistor R is connected to the cell.
Applications:
1. Solar cells are used as power sources in artificial satellites
2. To supply power in calculators
3. Solar cells are used in water heaters, street lights in solar cookers etc.
DIGITAL ELECTRONICS
Digital electronics is a branch of electronics in which the data is represented as binary numbers 0 and 1
using digital signals.
A signal which has only two values i.e. high or low is known as a digital signal.
The response of a digital circuit to input signal is known as its ‘logic’.
A ‘logic gate’ is an electronic device or circuit which makes logical decisions.
A logic gate has two or more inputs and one output.
A table which represents all inputs and corresponding output values of a logic gate is known as ‘truth table’.
An equation which represents the logic function performed by a logic gate is known as ‘Boolean equation’.

OR Gate: It is a logic gate which gives an output when one or both the inputs are present.

Boolean equation: Y = A + B Circuit symbol:
Truth table:







A B Y
0 0 0
0 1 1
1 0 1
1 1 1

83. DUSHYANTHA RAO MOB:[phone] Page 10

AND Gate: It is a logic gate which gives an output when both the inputs are present.
Boolean equation: ??????=&#3627408436; ∙&#3627408437; Circuit symbol:
Truth table:





NOT Gate (Inverter): It is a logic gate in which the output is the compliment of input.
Boolean equation: ??????= &#3627408436;̅ Circuit symbol:
Truth table:



Note: NOT Gate is the only logic gate to have one input and one output.
NOR Gate: It is a logic gate whose output is compliment of output of OR gate.
Boolean equation: ??????= &#3627408436;+&#3627408437;̅̅̅̅̅̅̅̅ Construction: Circuit symbol:
Truth table:



NAND Gate: It is a logic gate whose output is compliment of output of AND
gate.
Boolean equation: ??????= &#3627408436; ∙&#3627408437;̅̅̅̅̅̅̅ Construction:
Circuit symbol:
Truth table:





Note: 1. OR gate, AND gate and NOT gate are known as ‘basic gates’.
2. NOR gate and NAND gate are known as ‘Universal gates’, as all the basic gates can be constructed using
them.
3. The Boolean equations for NOR and NAND gates can also be written as ??????= &#3627408436;̅ ∙ &#3627408437;̅ and ??????= &#3627408436;̅+ &#3627408437;̅
respectively.
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
A Y
0 1
1 0
A B Y
0 0 1
0 1 0
1 0 0
1 1 0
A B Y
0 0 1
0 1 1
1 0 1
1 1 0

84. DUSHYANTHA RAO MOB:[phone] Page 11

4. Construction of AND gate using NAND gate:

5. Construction of OR gate using NAND gate:





Exclusive OR gate (EOR Gate or XOR Gate): It is a logic gate which gives an output when ONLY ONE
input is present. (Or It is a logic gate which gives an output when the inputs are different).
Boolean equation: ??????=&#3627408436; ⨁ &#3627408437; Circuit symbol:
Truth table:





Note: The Boolean equation for XOR gate can also be written as ??????=&#3627408436; &#3627408437;̅+ &#3627408436;̅ &#3627408437;

*******

A B Y
0 0 0
0 1 1
1 0 1
1 1 0