QJEEM25Ph4901: JEE Main 2025 Physics - Class 12 JEEM25Ph4901 Solution: Young's double slit inteference apparatus is immersed in a…

In Young's double-slit experiment, the fringe width ( $\beta$ ) can be calculated using the formula:

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

where:

$\lambda'$ is the wavelength of light in the medium,

$D$ is the distance between the slits and the screen,

$d$ is the separation between the slits.

The wavelength in the medium ( $\lambda'$ ) is given by:

$\lambda' = \frac{\lambda_0}{n}$

where:

$\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air,

$n = 1.44$ is the refractive index of the liquid.

Substituting the values,

$\lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm}$

Now, convert $\lambda'$ to meters:

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

Next, use the values for $D$ and $d$ :

$D = 0.72 \, \text{m}$

$d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$

Calculating the fringe width:

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

$\beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}}$

$\beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm}$

The closest answer choice is:

Option D: 0.23 mm