wave optics all derivations.pdf. for class 12

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1. WAVE OPTICS – ALL DERIVATIONS
Proof of laws of reflection
Consider a beam of light LM, whose wave front AP reaches A’P’ in time t, hence for any
point Q on the AP wave front must also reach A’P’ in time t.

Let speed of light in the medium be c, then
PP’ = ct and AA’ = ct
o
In AP'P and AA'P'
PP' AA' [Proved above]
AP' AP' [common]
APP' AA'P' [both 90 ]
AP'P AA'P' [by RHS congruence rule]
i r [by cpct]
 


  
  
  

Proof of law of refraction (Snell’s law) using wave theory of light
Let speed of light in medium 1 be c1 and the speed of light in medium 2 be c2.
Let time taken by light to travel from AP to A’P’ be t, then

1
PP' c t,
2
AA' c t

2. In ABD
BD
sini
AD
In ADC
AC
sinr
AD





11
2 2
1 1
2
2
1
2
BD
csini BD c tAD
ACsinr AC c t c
AD
c
Also
c
sini
sinr
    
 
  

Which is Snell’s law.
Condition for constructive and destructive interference
We can represent the displacement of the waves from the sources S1 and S2 (see figure) at
point P on the screen at any time t by


1 1
y a sinωt
and
2 2
y a sin( t )   
where  is the constant phase difference between the two waves.
The resultant at P, y
1 2
 y y

3. 1 2
1 2 2
1 2 2
1 2
2
2 2 2 2 2 2
1 2
2
1 2
a sinωt a sin(ωt )
a sinωt a sinωtcos a cosωtsin
(a a cos )sinωt a sin cosωt
Taking (a a cos ) Rcosθ
and a sin Rcosθ
y Rcosθsinωt Rsinθcosωt Rsin(ωt θ)
For R cos θ R sin θ (a a cos ) (sin )
a a
  
  
  
 

    
   
 

 
 


 
2 2 2
1 2
(cos sin ) 2a a cos   


2 2 2
1 2 1 2
R a a 2a a cos   
But intensity of light
2
I R
2 2
1 2 1 2
1 2 1 2
I a a 2a a cos
I I I 2 II cos
   
   



Constructive interference
For intensity of light to be maximum at P,
cos 1
phase difference 2nπ, where n 0,1,2,3,4........   
If x be the path difference between S1P and S2P, then respective phase difference between
two waves will be

2π
x
λ
2π
2nπ x
λ
x nλ


 


Destructive interference
For intensity of light to be minimum at P,
cos 1
(2n 1)π
2π
x (2n 1)π
λ
λ
x (2n 1)
2
 
 
 
 



4. Ratio of intensity of light at Maxima and minima
2 2 2
max 1 2 1 2 1 2
2 2 2
min 1 2 1 2 1 2
2
max 1 2
2
min 1 2
I a a 2a a (a a )
I a a 2a a (a a )
I (a a )
I (a a )
    
    




Ratio of intensity of light due to two sources
Let I1 and I2 and a1 and a2 be the intensities and amplitudes of light from slits S1 and S2
respectively. Then,
2
1 1
2
2 2
I a
I a

Relation between slit width (), I and a
2
1 1 1
2
2 2 2
ω I a
ω I a
 

Angular width in central maxima in diffraction pattern
The angular width of the central maxima is the angular separation between the directions of
the first minima on the two sides of the central maximum.

As condition for nth minima is
n
dsinθ nλ, so first minima is formed where n = 1. Therefore,
1
λ
sinθ
d
.

5. As we know, for small angles, sinθ θso
λ
θ
d
…..(i)
Now from figure
1
x
tanθ
D
 and for small angles tanθ θ, therefore
1
x
θ
D
…….(ii)
So, from (i) and (ii), we get
1
1
xλ
D d
λD
x
d

 

Therefore, width of central maxima is
1
2λD
2x
d
 .
Angular width of central maxima is
λ
2θ 2
d

With of nth secondary maxima
Nth maxima lies between nth minima and (n+1)th minima
As, direction of nth minima,
n
nλ
θ
d

Direction of (n+1)th minima,
n 1
(n 1)λ
θ
d



Therefore, angular width of nth secondary minima
n 1 n
(n 1)λ nλ λ
θ θ
d d d


   
So, linear width of nth secondary maxima = Angular width X D =
λ
D
d