Question

In Young’s double-slit experiment, the width of one of the slits is three times the other slit. 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.

a. 4: 1

b. 2: 1

c. 3: 1

d. 1: 4

Answer 1

The correct answer is  a) 4:1  

Explaination ::

A₂=3A₁  

\( Amplitude ∝ width \space of \space slit\)

I_max = (A₁ + 3A₁)²     ..............1

and     I_min = (A₁ – 3A₁)²           .................2

\({I_{max} \over I_{min }}= {(A_1+3A_1)^2\over (A_1-3A_1)^2}\)  \(={16A^2_1\over 4A^2_1}={4\over1}\)  

 

4:1