Q 12-10-108JEE MainJEE Main 2021 (24 Feb, Shift 1)Easy
In a Young's double slit experiment, the width of the one of the slit 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.
Answer: (D) $4 : 1$
$a_1 : a_2 = 3 : 1$.
$$\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{4^2}{2^2} = 4 : 1$$
Solution by Sreeraj P, M.Sc Physics