optics- REFRACTION SPHERICAL SURFACE.ppt

refraction

1. RAY OPTICS
Refraction of Light

2. Spherical Refracting Surfaces:
A spherical refracting surface is a part of a sphere of refracting material.
A refracting surface which is convex towards the rarer medium is called
convex refracting surface.
A refracting surface which is concave towards the rarer medium is
called concave refracting surface.
••
C C
P P
R R
A
B AB
APCB – Principal Axis
C – Centre of Curvature
P – Pole
R – Radius of Curvature
••
Denser MediumDenser Medium Rarer MediumRarer Medium

3. Refraction at Convex Surface:
(From Rarer Medium to Denser Medium - Real Image)
•
CP
R
O
•
Denser MediumRarer Medium
• •
IM
μ
2
μ
1
α
βγ
i
r
i = α + γ
γ = r + β or r = γ - β
A
tan α =
MA
MO
tan β =
MA
MI
tan γ =
MA
MC
or α =
MA
MO
or β =
MA
MI
or γ =
MA
MC
According to Snell’s law,
μ
2 sin i
sin rμ
1
= or

i
r μ
1
=
μ
2
or

μ
1
i = μ
2
r
Substituting for i, r, α, β and γ, replacing M by P and rearranging,
μ
1
PO
μ
2
PI
μ
2
-

μ
1
PC
+
=
Applying sign conventions with values,
PO = - u, PI = + v and PC = + R
vu
μ
2
V
μ
1
U
μ
2
-

μ
1
R
-
=
N

4. Refraction at Convex Surface:
(From Rarer Medium to Denser Medium - Virtual
Image)
Refraction at Concave Surface:
(From Rarer Medium to Denser Medium - Virtual
Image)
•
CP
R
•
Denser MediumRarer Medium
• •
I M
μ
2
μ
1
αβ γ
i r
A
v
u
O
N
O
•
C
PR
•
Denser MediumRarer Medium
•
I M
μ
2
μ
1
α
β
γ
r
A
v
u
•
i
N
μ
2
V
μ
1
U
μ
2
-

μ
1
R
-
=
μ
2
V
μ
1
U
μ
2
-

μ
1
R
-
=

5. Refraction at Convex Surface:
(From Denser Medium to Rarer Medium -
Real Image)
•
C P
R
O
•
Denser Medium Rarer Medium
• •
IM
μ
1
μ
2
α
βγ
r
A
vu
N
i
μ
2
V
μ
1
U
μ
2
-

μ
1
R
- =
Refraction at Convex Surface:
(From Denser Medium to Rarer Medium - Virtual
Image)
Refraction at Concave Surface:
(From Denser Medium to Rarer Medium -
Virtual Image)
μ
2
V
μ
1
U
μ
2 -
μ
1
R
- =
μ
2
V
μ
1
U
μ
2
-

μ
1
R
- =

6. Note:
1.Expression for ‘object in rarer medium’ is same for whether it is real or
virtual image or convex or concave surface.
2. Expression for ‘object in denser medium’ is same for whether it is real or
virtual image or convex or concave surface.
3.However the values of u, v, R, etc. must be taken with proper sign
conventions while solving the numerical problems.
4. The refractive indices μ
1
and μ
2
get interchanged in the expressions.
μ
2
V
μ
1
U
μ
2 -
μ
1
R
-
=
μ
2
V
μ
1
U
μ
2
-

μ
1
R
- =

7. LATERAL MAGNIFICATION
The lateral magnification m is the ratio of the image height to the object
height
























u
v
h
h
m
i
2
1
0

μ
2 sin i
sin r
=
μ
1

8. Light from α point source in air falls on a convex spherical glass surface (μ =1.5,
radius of curvature =20 cm). The distance of light source from the glass surface is
100 cm. At what position is the image formed ?

9. A glass dumbbell of length 30 cm and refractive index 1.5 has ends of 3 cm
radius of curvature. Find the position of the image formed due to refraction at
one end only, when the object is situated in air at a distance of 12 cm from the
end of the dumbbell along the axis.

10. A globe of 30 cm diameter is made of glass of refractive index μ = 1.5 A ray
enters the globe parallel to the axis. Find the position from the centre of the
sphere where the ray crosses the axis.

11. A small air bubble in a glass sphere of radius 2 cm appears to be 1 cm from
the surface when looked at, along a diameter. If the refractive index of glass
is 1.5, find the true position of the air bubble.

12. The diameter of a glass sphere is 15 cm. A beam of light strikes the sphere,
which converges at point 30 cm behind the pole of the spherical surface. Find the
position of the image if μ = 1.5.
Solution. In the absence of glass sphere,
the light rays will converge at point O. So O
acts as virtual object for the image I for
refraction at the first surface.

13. Light from α point source in air falls on a convex spherical glass surface (μ =1.5,
radius of curvature =20 cm). The distance of light source from the glass surface is
100 cm. At what position is the image formed ?
A glass dumbbell of length 30 cm and refractive index 1.5 has ends of 3 cm
radius of curvature. Find the position of the image formed due to refraction at
one end only, when the object is situated in air at a distance of 12 cm from the
end of the dumbbell along the axis.

14. A globe of 30 cm diameter is made of glass of refractive index μ = 15. A ray
enters the globe parallel to the axis. Find the position from the centre of the
sphere where the ray crosses the axis.
As refraction takes place from rarer to
denser medium, we use the relation
For refraction at surface BP
2. The ray
AB (before meeting point I
1) suffers
another refraction at surface BP
2
. The
real image I
1 acts as virtual object for
refraction at surface BP
2
and I is the
real image.
As refraction occurs from denser to rarer medium, so we use the relation

15. What curvature must be given to the bounding surface of μ =1.5 for virtual
image of an object in the medium of μ = 1 at 10 cm to be formed at a
distance of 40 cm. Also calculate power of the surface and two principal focal
lengths of the surface.

16. SPHERICAL LENSES

17. DEFINITIONS
(i) Centre of curvature (C).
The centre of curvature of the surface of a
lens is the centre of the sphere of which it
forms a part. Because a lens has two
surfaces, so it has two centres of curvature.
(ii) Radius of curvature (R).
The radius of curvature of the surface of a
lens is the radius of the sphere of which the
surface forms a part.
(iii) Principal axis (C
1
C
2
)
It is the line passing through the two
centres of curvature of the lens.
(iv) Optical centre.
If a ray of light is incident on a lens such
that after refraction through the lens the
emergent ray is parallel to the incident ray,
then the point at which the refracted ray
intersects the principal axis is called the
optical centre of the lens.

18. Lens Maker’s Formula:
R
1
P
1
•
O
•
μ
2
μ
1
i
A
vu
N
1
R
2
C
1
• •
I
1
N
2
L
C
N
P
2
•
C
2
•
I
•
μ
1
For refraction at
LP
1N,
1
V
(μ
2
-

μ
1
)
R
1
- =
1
1
U R
2
- )
1
(
μ
1
μ
2
V
1
μ
1
U
μ
2
-

μ
1
R
1
-
=
μ
1
V
μ
2
V
1
μ
1
-

μ
2
R
2
-
=
v
1
μ
1
V
(μ
2
-

μ
1
)(

R
1
-
=
1
μ
1
U R
2
- )
1

19. Since μ
2
/ μ
1
= μ
1
V

μ
2
R
1
- =
1
1
U R
2
- )
1
(
μ
1
- 1)(
or
1
V
(μ – 1)
R
1
- =
1
1
U R
2
- )
1
(
When the object is kept at infinity, the image is formed at the principal focus.
i.e. u = - ∞, v = + f.
So, (μ – 1)
R
1
=
1
1
f R
2
- )
1
(
This equation is called ‘Lens Maker’s Formula’.
Also, from the above equations we get,
1
v f
-
=
1
1
u

20. First Principal Focus:
First Principal Focus is the point on the principal axis of the lens at which if
an object is placed, the image would be formed at infinity.
F
1
f
1
F
2
f
2
Second Principal Focus:
Second Principal Focus is the point on the principal axis of the lens at
which the image is formed when the object is kept at infinity.
F
2
f
2
F
1
f
1

21. Linear Magnification:
Linear magnification produced by a lens is defined as the ratio of the size of
the image to the size of the object.
m=
I
O
A’B’
AB
=
OB’
OB
- I
+ O
=
+ v
- u
According to new Cartesian sign conventions,
A’B’ = - I, AB = + O, OB’ = + v and OB = - u.
m
I
O
=
v
u
=

22. Linear Magnification:
m
I
O
=
v
u
=
Magnification in terms of v and f:
m=
f - v
f
Magnification in terms of u and f:
m=
f
f + u
Power of a Lens:
Power of a lens is its ability to bend a ray of light falling on it and is reciprocal
of its focal length. When f is in metre, power is measured in Dioptre (D).
P=
1
f

23. Longitudinal Magnification:
m
I
O
=
v
2
-v
1
u
2-u
1
=
222

























f
vf
uf
f
u
v
du
dv
m
Areal magnification
2
2










uf
f
m
A
A
m
o
i
s
(A
i
= Area of image, A
o
= Area of
object)
Relation between object and image speed
If an object moves with constant speed Vo towards a convex lens from
infinity to focus, the image will move slower in the beginning and then
faster.
oi
V
uf
f
V .
2











2
.
i o
V m V

24. Formation of images by convex lens :
(a) Object beyond 2F. The image is
(i)between F and 2F
(ii)real
(iii) inverted
(iv) smaller
(b) Object at 2F. The image is
(i)at 2F
(ii)real
(iii) inverted
(iv) same size

25. (c) Object between 2F and F. The image is
(i)beyond 2F
(ii)real
(iii) inverted
(iv) larger
(d) Object between F and O. The image is
(i)behind object (on the same side)
(ii)virtual
(iii) erect
(iv) larger

26. (e) Object in any position. The image is
(i)in front of object
(ii)virtual
(iii) erect
(iv) smaller
Formation of images by concave lens :

27. To find the focal length of convex lens
a) Distant object method.
b) u-v method.
1 1 1
f v u
 

28. To find the focal length of convex lens
c) graphical method.

29. LENS DISPLACEMENT METHOD:
U
2
•
X
U
1
C
A
B
A’
B’
D
••••
V
2
V
1
U V D 
1 1 1
v u f
 
1 1 1
D u u f
 
 
( )Df D u u 
2
0u Du Df  
2
4
2
D D Df
u
 

2
4 0
4
D Df
D f
 


30. CASE-I : D=4f
2
D
U V 
CASE-II : 4D f
We get two different position of lens for which the image of object on the screen
Is distinct and clear:
2
1
4
2
D D Df
u
 

2
2
4
2
D D Df
u
 

2
1
4
2
D D Df
v
 

2
2
4
2
D D Df
v
 

2
1 2
4x v v D Df   
2 2
4x D Df 
2 2
4
D x
f
D


1
1
ID x
m
O D x

 

2
2
ID x
m
O D x

 

1 2
1mm
1 2
1
I I
O O

1 2
O I I

31. The radius of curvature of each face of biconcave lens, made of glass of
refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air.
The radii of curvature of the faces of a double convex lens are 10 cm and
15 cm. If focal length is 12 cm, what is the refractive index of glass ?

32. A convex lens of focal length 0.2 m and made of glass (μ=1.50) is immersed
in water (μ=133). Find the change in the focal length of the lens.

33. A convergent beam of light passes through a diverging lens of focal length
0.2 m and comes to focus at distance 0.3 m behind the lens. Find the
position of the point at which the beam would converge in the absence of the
lens.

34. A double convex lens made of glass of refractive index 1.5 has its both
surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height
is placed at a distance of 10 cm from the lens. Find the position, nature and
size of the image.

35. In the Fig, the direct image formed by the lens (f = 10 cm) of an object
placed at O and that formed after reflection from the spherical mirror are
formed at the same point O'. What is the radius of curvature of the mirror ?

36. In the following ray diagram are given the positions of an object O, image I
and two lenses L
1
and L
2
. The focal length of L
1
is also given. Find the focal
length of L
2

37. A convex lens of focal length 10 cm is placed coaxially 5 cm away from a
concave lens of focal length 10 cm. If an object is placed 30 cm in front of
the convex lens, find the position of the final image formed by the combined
system.

38. A lens forms α real image of an object. The distance of the object to the lens
is u cm and the distance of the image from the lens is v cm. The given graph
shows the variation of v with u. (i) What is the nature of the lens ? (ii) Using
this graph, find the focal length of this lens.

39. Lens in contact
vu
v
1
1 1
1 1 1
F v u
 
2 1
1 1 1
F v v
 
1 2
1 1 1 1 1
F F v u F
   
21
111
ffF

21
PPP 
(1) In case when two thin lens are in contact : Combination will behave
as a lens, which have more power or lesser focal length.

40. Combination of Lens
(2) For a system of lenses, the net power, net focal length and
magnification are given as follows :
..........
321 PPPP 
...........
1111
321

fffF
............
321  mmmm
(3) If two lens of equal focal length but of opposite nature are in contact
then
combination will behave as a plane glass plate and

ncombinatioF
(4) When two lenses are placed co-axially at a distance d from each other
then equivalent focal length (F).
2121
111
ff
d
ffF

2121 PdPPPP 

41. Cutting of Lens
(1) A symmetric lens is cut along optical axis in two equal parts.
Intensity of image formed by each part will be same as that of complete
lens. Focal length is double the original for each part.
(2) A symmetric lens is cut along principle axis in two equal parts.
Intensity of image formed by each part will be less compared as that of
complete lens.(aperture of each part is times that of complete lens).
Focal length remains same for each part.
2
1

42. SILVERING OF LENS
On silvering the surface of the lens it behaves as a mirror whose effective
power is given by
1 2
1 1 1
( 1)
L
L
P
f R R

 
     
 
2
net L m
P P P 
1 2
m
m m
P
f R
 
So the combination acts like a mirror having net focal length given by
1
net
net
f
P

1 2 1
net
net L m
P
F f f
  
1.If F
net
is –ve , it is a concave mirror
2.If F
net
is +ve , it is a convex mirror
+

43. Silvering of Lens
(1) Plano convex is
silvered
,
2
m
R
f


( 1)
L
R
f



2
R
F


,
mf
)1(


R
f
l
2( 1)
R
F



1 2 1
net L m
F f f
  
1 2( 1) 2
net
F R R

  
Concave
mirror
Concave
mirror

44. (ii) Double convex lens is silvered
,
2( 1) 2
l m
R R
f f


 

2(2 1)
R
F




46. 11v cm

49. Thin Lens Formula (Gaussian Form of Lens Equation):
For Convex Lens:
f
•
R
u
C
A
B
A’
B’
M
Triangles ABC and A’B’C are similar.
A’B’
AB
=
CB’
CB
Triangles MCF
2 and A’B’F
2 are similar.
A’B’
MC
=
B’F
2
CF
2
v
A’B’
AB
=
B’F
2
CF
2
or
•
2F
2
•
F
2
•
F
1
•
2F
1
CB’
CB
=
B’F
2
CF
2
CB’
CB
=
CB’ - CF
2
CF
2
According to new Cartesian sign
conventions,
CB = - u, CB’ = + v and CF
2
= + f.
1
v f
- =
1
1
u

50. (v) Principal foci and focal length :
First principal focus.
It is a fixed point on the principal axis such
that rays starting from this point (in convex
lens)
or appearing to go towards this point (in
concave lens), after refraction through the
lens, become parallel to the principal axis.
Second principal focus.
It is a fixed point on the principal axis such
that the light rays incident parallel to the
principal axis, after refraction through the
lens, either converge to this point (in
convex lens) or appear to diverge from this
point (in concave lens).
(vi) Aperture.
It is the diameter of the circular boundary of the lens.

51. Lens Displacement Method
Consider an object and a screen placed at a distance D (> 4f) apart. Let a
lens of focal length f be placed between the object and the screen.
(1) For two different positions of lens two images (I
1
and I
2
) of an object
are formed at the screen.
(2) Focal length of the lens
21
22
4 mm
x
D
xD
f




O
I
m
1
1

O
I
m
2
2 .1
21
mm
(3) Size of object 21.IIO