Q 12-10-060JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
Two coherent monochromatic light beams of intensities $I$ and $4I$ are superimposed. The difference between the maximum and minimum possible intensities in the resulting beam is $xI$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 8
$$I_{max} = (\sqrt I + \sqrt{4I})^2 = 9I,\qquad I_{min} = (\sqrt{4I} - \sqrt I)^2 = I$$
Difference $= 8I$, so $x = 8$.
Solution by Sreeraj P, M.Sc Physics