JEE Physics 2026 Question with Solution

Given: slit separation $d = 0.4 \, \text{mm} = 4 \times 10^{-4} \, \text{m}$, screen distance $D = 1 \, \text{m}$, sheet thickness $t = 20 \, \mu\text{m} = 2 \times 10^{-5} \, \text{m}$, fringe shift $\Delta y = 20 \, \text{mm} = 2 \times 10^{-2} \, \text{m}$.

Find: $\alpha$ if refractive index $\mu = \frac{\alpha}{10}$.

When a transparent sheet is placed in front of one slit in YDSE, it introduces an additional optical path length and shifts the entire fringe pattern.

Using the shift formula,

$$\Delta y = \frac{D}{d}(\mu - 1)t$$

Rearranging,

$$\mu - 1 = \frac{\Delta y \cdot d}{D \cdot t}$$

Substitute the values,

$$\mu - 1 = \frac{(2 \times 10^{-2})(4 \times 10^{-4})}{1 \times (2 \times 10^{-5})}$$ $$\mu - 1 = \frac{8 \times 10^{-6}}{2 \times 10^{-5}} = 4 \times 10^{-1} = 0.4$$

Therefore,

$$\mu = 1 + 0.4 = 1.4$$

Given that

$$\frac{\alpha}{10} = 1.4$$

so,

$$\alpha = 14$$

Therefore, the value of $\alpha$ is $14$.

Given: a transparent sheet of thickness $t$ is inserted before one slit.

Find: the value of $\alpha$.

The extra path difference introduced by the sheet is

$$(\mu - 1)t$$

In YDSE, converting path difference to linear shift on the screen gives

$$\Delta y = \frac{D}{d}(\mu - 1)t$$

Now substitute directly:

$$2 \times 10^{-2} = \frac{1}{4 \times 10^{-4}}(\mu - 1)(2 \times 10^{-5})$$

This gives

$$\mu - 1 = 0.4$$

Hence,

$$\mu = 1.4 = \frac{\alpha}{10}$$

So,

$$\alpha = 14$$

Thus, the required value is $14$.