JEE Physics 2025 Question with Solution

Given: The two monochromatic light beams have intensities in the ratio $1:9$.

Find: The ratio of maximum intensity to minimum intensity in the interference pattern.

For two interfering beams of intensities $I_1$ and $I_2$,

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

and

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

Given

$$\frac{I_1}{I_2} = \frac{1}{9}$$

Let

$$I_1 = I, \qquad I_2 = 9I$$

Then,

$$I_{\max} = (\sqrt{I} + \sqrt{9I})^2 = (\sqrt{I} + 3\sqrt{I})^2 = (4\sqrt{I})^2 = 16I$$

Similarly,

$$I_{\min} = (\sqrt{I} - \sqrt{9I})^2 = (\sqrt{I} - 3\sqrt{I})^2 = (-2\sqrt{I})^2 = 4I$$

Therefore,

$$\frac{I_{\max}}{I_{\min}} = \frac{16I}{4I} = 4$$

So, the ratio of maximum to minimum intensity is $4:1$. Therefore, the correct option is D.

Given: The intensity ratio of the two beams is $1:9$.

Find: The ratio $I_{\max}:I_{\min}$.

The intensity at any point in an interference pattern is

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$

Maximum intensity occurs for constructive interference, that is, when $\phi = 0$:

$$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

Minimum intensity occurs for destructive interference, that is, when $\phi = \pi$:

$$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$

Now take

$$I_1 = I, \qquad I_2 = 9I$$

Then

$$I_{\max} = I + 9I + 2\sqrt{I \cdot 9I} = 10I + 6I = 16I$$

and

$$I_{\min} = I + 9I - 2\sqrt{I \cdot 9I} = 10I - 6I = 4I$$

Hence,

$$I_{\max}:I_{\min} = 16I:4I = 4:1$$

Therefore, the ratio of the intensities of maximum to minimum is $4:1$, so the correct option is D.