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