Question

The width of one of the two slits in a Young's double slit experiment is double of the other slit.
Assuming that the amplitude of the light coming from a slit is proportional to the slit-width,
find the ratio of the maximum to the minimum intensity in the interference pattern.​

Answer 1

Answer:

Let the amplitude of light wave coming from the narrower slit be a then the amplitude of light wave from the wider slit =2a

The maximum intensity occurs where the constructive interference takes place and the minimum intensity where the destructive interference takes place.

⇒a  

max

​  

=2a+a=3a

⇒a  

min

​  

=2a−a=a

Ratio of maximum to minimum intensity,  

I  

min

​  

 

I  

max

​  

 

​  

=  

a  

min

2

​  

 

a  

max

2

​  

 

​  

=  

a  

2

 

3  

2

a  

2

 

​  

=9

Explanation: