In Young’s double slit experiment, the ratio of slit widths is 4 : 1. The intensity ratio in the interference pattern would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 : 1
The light passing through a slit is proportional to its width, so the intensity from a slit is \( I\propto w \) and the amplitude is \( a\propto\sqrt{I} \). In the pattern the maximum intensity comes from \( (a_1 + a_2)^{2} \) and the minimum from \( (a_1 - a_2)^{2} \).
\( \displaystyle \frac{I_1}{I_2} = \frac{w_1}{w_2} = \frac{4}{1} \Rightarrow \frac{a_1}{a_2} = 2 \)
\( \displaystyle \frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^{2}}{(a_1 - a_2)^{2}} = \frac{(2 + 1)^{2}}{(2 - 1)^{2}} = \frac{9}{1} \)
Note: A similar question was also asked in CUET (UG) 2026 on 24-05-2026 (Slot 2), in CUET (UG) 2025 on 22-05-2025 (Slot 1), and in CUET (UG) 2023 on 08-06-2023 (Slot 3).