Q 12-10-055JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is
Answer: (C) $9:1$
Intensity is proportional to slit width: $I_1 : I_2 = 4:1$, so amplitudes are $2:1$.
$$\frac{I_{max}}{I_{min}} = \frac{(2 + 1)^2}{(2 - 1)^2} = 9:1$$
Solution by Sreeraj P, M.Sc Physics