JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 23)

Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is $x \mathrm{I}$. The value of $x$ is___________ .
Answer
24

Explanation

Calculate Maximum Intensity ($I_{\text{max}}$):

$ I_{\max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 $

Substituting the values, we have:

$ I_{\max} = \left(\sqrt{4I} + \sqrt{9I}\right)^2 $

$ I_{\max} = \left(2\sqrt{I} + 3\sqrt{I}\right)^2 = \left(5\sqrt{I}\right)^2 = 25I $

Calculate Minimum Intensity ($I_{\text{min}}$):

$ I_{\min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 $

Substituting the values, we have:

$ I_{\min} = \left(\sqrt{4I} - \sqrt{9I}\right)^2 $

$ I_{\min} = \left(2\sqrt{I} - 3\sqrt{I}\right)^2 = \left(-1\sqrt{I}\right)^2 = I $

Calculate the Difference Between Maximum and Minimum Intensities:

$ I_{\max} - I_{\min} = 25I - I = 24I $

Thus, the value of $x$, which represents the difference between the maximum and minimum intensities, is $24$.

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