JEE Main 2025 — Wave Optics Question with Solution

In Young's double slit experiment, the fringe width $$\beta$$ is given by:

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

where:

$$\lambda$$ is the wavelength of light,

$$D$$ is the distance from the slits to the screen,

$$d$$ is the separation between the slits.

Here's how the change affects the fringe width:

Initial Situation:

When $$d=0.2\text{ mm}$$, the fringe width is:

$${\beta }_{\text{initial}}=\frac{\lambda D}{0.2\text{ mm}}$$

After Increasing Slit Separation:

When $$d$$ is increased to $$0.4\text{ mm}$$:

$${\beta }_{\text{new}}=\frac{\lambda D}{0.4\text{ mm}}$$

Comparing the Two Fringe Widths:

Notice that:

$${\beta }_{\text{new}}=\frac{1}{2}\times \frac{\lambda D}{0.2\text{ mm}}=\frac{1}{2}{\beta }_{\text{initial}}$$

This means the fringe width is halved, which is a reduction of $$50\%$$.

Thus, the fringe width decreases by $$50\%$$ when the slit separation is increased from $$0.2\text{ mm}$$ to $$0.4\text{ mm}$$.

The correct answer is Option B: $$50\%$$.