JEE Main 2025 — Wave Optics Question with Solution

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

$$\beta =\frac{{\lambda }^{\prime }\cdot D}{d}$$

where:

${\lambda }^{\prime }$ 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 (${\lambda }^{\prime }$) is given by:

$${\lambda }^{\prime }=\frac{{\lambda }_0}{n}$$

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,

$${\lambda }^{\prime }=\frac{690\text{nm}}{1.44}\approx 479.17\text{nm}$$

Now, convert ${\lambda }^{\prime }$ to meters:

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

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