JEE Physics 2025 Question with Solution

Given:

• Refractive index of liquid = $1.44$
• Slit separation = $d = 1.5 \, \text{mm}$
• Wavelength in air = $\lambda = 690 \, \text{nm}$
• Distance of screen from slits = $D = 0.72 \, \text{m}$

Find: The fringe width $\beta$.

In a Young's double slit experiment, the fringe width is given by

$$\beta = \frac{\lambda D}{d}$$

where $\lambda$ is the wavelength in the medium.

The wavelength in the liquid is

$$\lambda' = \frac{\lambda}{n}$$

Substituting $n = 1.44$,

$$\lambda' = \frac{690 \times 10^{-9}}{1.44} = 479.17 \, \text{nm}$$

Now,

$$\beta = \frac{479.17 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3}} = 0.33 \, \text{mm}$$

Therefore, the fringe width is $0.33 \, \text{mm}$ and the correct option is A.

Given: A Young's double slit interference apparatus is immersed in a liquid of refractive index $1.44$. The slit separation is $d = 1.5 \, \text{mm}$, the wavelength in air is $\lambda_{\text{air}} = 690 \, \text{nm}$, and the screen is at $L = 0.72 \, \text{m}$.

Find: The fringe width on the screen.

Step 1: Determine the effective wavelength in the liquid.

$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n}$$

So,

$$\lambda_{\text{liquid}} = \frac{690 \, \text{nm}}{1.44} = 479.17 \, \text{nm}$$

Step 2: Use the formula for fringe width.

$$\beta = \frac{\lambda_{\text{liquid}} L}{d}$$

Substituting the values,

$$\beta = \frac{479.17 \times 10^{-9} \, \text{m} \times 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}} = 0.33 \, \text{mm}$$

Therefore, the fringe width is $0.33 \, \text{mm}$. Hence, the correct option is A.