If the two slits in a Young's double slit experiment have their widths in the ratio 4:1, then the ratio of intensities at maxima and minima in the interference pattern will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9 : 1
The light through a slit is proportional to its width, so the intensities are in the ratio of the widths, and the amplitude is proportional to the square root of the intensity.
\( \displaystyle \frac{I_1}{I_2} = \frac{4}{1} \)
\( \displaystyle \frac{a_1}{a_2} = \sqrt{\frac{4}{1}} = \frac{2}{1} \)
At a maximum the amplitudes add, and at a minimum they subtract:
\( \displaystyle \frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^{2}}{(a_1 - a_2)^{2}} = \frac{(2 + 1)^{2}}{(2 - 1)^{2}} = \frac{9}{1} \)
Note: A similar question was also asked in CUET (UG) 2025 on 22-05-2025 (Slot 1), and in CUET (UG) 2023 on 08-06-2023 (Slot 3) and on 07-06-2023 (Slot 3).