JEE Main 2025 — Wave Optics Question with Solution

To find the ratio of the intensities of the maximum to the minimum in an interference pattern formed by two monochromatic light beams with intensity ratios of 1:9, you use the formula:

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

Given that the intensity ratio is 1:9, we have $I_1=1$ and $I_2=9$.

Calculate $\sqrt{I_1}$ and $\sqrt{I_2}$:

$\sqrt{I_1}=\sqrt{1}=1$

$\sqrt{I_2}=\sqrt{9}=3$

Substitute these values into the formula:

$\frac{I_{\max }}{I_{\min }}=\frac{(1+3{)}^2}{(1-3{)}^2}=\frac{4^2}{2^2}=\frac{16}{4}=4$

Thus, the ratio of the intensities of the maximum to the minimum is 4:1.