Q 12-10-160JEE MainJEE Main 2025 (3 Apr, Shift 2)Medium
The width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:
Answer: (B) $(3 + 2\sqrt2) : (3 - 2\sqrt2)$
The intensity from a slit is proportional to its width, so $I_2 = 2I_1$.
$$\frac{I_{\max}}{I_{\min}} = \frac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1\right)^2} = \frac{3 + 2\sqrt2}{3 - 2\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics