The two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beams are:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9I and I
For coherent beams the amplitudes add, and the intensity is proportional to the square of the amplitude. So \( \sqrt{I_1} \) and \( \sqrt{I_2} \) add for constructive interference and subtract for destructive interference.
\( I_{\max} = \left(\sqrt{I} + \sqrt{4I}\right)^{2} = (1 + 2)^{2}I = 9I \)
\( I_{\min} = \left(\sqrt{4I} - \sqrt{I}\right)^{2} = (2 - 1)^{2}I = I \)
Shortcut: Interference only redistributes energy, so the average of the maximum and minimum is the sum of the two intensities: \( I_{\max} + I_{\min} = 2(I_1 + I_2) = 10I \). Only Option (2) satisfies this (9I + I = 10I); the others add up to 6I, 8I and 12I.
Note: The same question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1), and in CUET (UG) 2023 on 22-05-2023 (Slot 3), and a similar question in CUET (UG) 2026 on 18-05-2026 (Slot 2).