Question

The width of one of the two slits in 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

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.

$\Rightarrow a_{max}=2a+a=3a\Rightarrow a_{min}=2a-a=a$

Ratio of maximum to minimum intensity,

$\frac{I_{max}}{I_{min}}=\frac{a_{max}^2}{a_{min}^2}=\frac{3^2{a}^2}{a^2}=9$