Question

explain superposition of waves and interference of light and find condition for maximum and minimum intensity

Answer 1

Explanation:

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for the maximum and zero intensity.

Answer 2

Answer:

Explanation:

Let y  

1

​

 and y  

2

​

 be displacement of two waves having same amplitude a and phase difference ϕ between them.

y  

1

​

=asinωt

y  

2

​

=asin(ωt+ϕ)

Resultant displacement is: y=y  

1

​

+y  

2

​

 

y=asinωt+asin(ωt+ϕ)=asinωt(1+cosϕ)+cosωt(asinϕ)

Rcosθ=a(1+cosϕ)

Rsinθ=asinϕ

y=Rsin(ωt+θ)

Where, R is resultant amplitude at P, I is intensity, squaring the equations we get,

I=R  

2

=a  

2

(1+cosϕ)  

2

+a  

2

(sinϕ)  

2

=2a  

2

(1+cosϕ)=4a  

2

cos  

2

 

2

ϕ

​

 

Maximum intensity:

cos  

2

 

2

ϕ

​

=1

ϕ=2nπ where n=0,1,2,3,....

Therefore, I  

max

​

=4a  

2

 

Minmum intensity:

cos  

2

 

2

ϕ

​

=0

ϕ=(2n+1)π where n=0,1,2,3,....

Therefore, I  

max

​

=0