JEE Physics 2026 Question with Solution

Given: slit separation $d = 0.1 \, \text{cm}$, screen distance $D = 50 \, \text{cm}$, refractive index $n = 1.5$, fringe shift $S = 0.2 \, \text{cm}$.

Find: thickness $t$ of the transparent sheet.

A transparent sheet introduced in front of one slit produces an additional optical path difference

$$\Delta x = (n-1)t$$

The corresponding shift of the fringe pattern is

$$S = \frac{D}{d}(n-1)t$$

So,

$$t = \frac{S d}{D(n-1)}$$

Substituting the given values,

$$t = \frac{(0.2)(0.1)}{50(1.5-1)} \, \text{cm}$$ $$t = \frac{0.02}{50 \times 0.5}$$ $$t = \frac{0.02}{25} \, \text{cm}$$ $$t = 8 \times 10^{-4} \, \text{cm}$$

Therefore, the thickness of the transparent sheet is $8 \times 10^{-4} \, \text{cm}$ and the correct option is A.

Given: the central fringe shifts by $S = 0.2 \, \text{cm}$ when a sheet of thickness $t$ and refractive index $n=1.5$ is placed before one slit.

Find: the value of $t$.

The fringe shift can also be written in terms of the extra phase difference introduced by the sheet. The additional phase difference is

$$\Delta \phi = \frac{2\pi}{\lambda}(n-1)t$$

If $\beta = \frac{\lambda D}{d}$ is the fringe width, then the shift is

$$S = \frac{\beta}{2\pi}\Delta \phi$$

Substituting for $\beta$ and $\Delta\phi$,

$$S = \frac{\lambda D}{d} \cdot \frac{1}{2\pi} \cdot \frac{2\pi}{\lambda}(n-1)t$$ $$S = \frac{D}{d}(n-1)t$$

Hence,

$$t = \frac{S d}{D(n-1)} = \frac{(0.2)(0.1)}{50(0.5)} \, \text{cm} = 8 \times 10^{-4} \, \text{cm}$$

Therefore, the required thickness is $8 \times 10^{-4} \, \text{cm}$.