JEE Main 2025 — Wave Optics Question with Solution

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 $24$.