JEE Main 2026 — Wave Optics Question with Solution

In Young's Double Slit Experiment, when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is introduced in front of one slit, the central bright fringe shifts by a distance $\Delta y$.
The formula for the shift is given by $\Delta y=\frac{D}{d}(\mu -1)t$.
Given values:
Distance between slits $d=0.4\text{ mm}=0.4\times 10^{-3}\text{ m}$
Distance to screen $D=1\text{ m}$
Thickness of sheet $t=20\mu \text{m}=20\times 10^{-6}\text{ m}$
Shift $\Delta y=20\text{ mm}=20\times 10^{-3}\text{ m}$
Substituting the values into the formula:
$20\times 10^{-3}=\frac{1}{0.4\times 10^{-3}}\times (\mu -1)\times 20\times 10^{-6}$
$20\times 10^{-3}=\frac{20\times 10^{-6}}{0.4\times 10^{-3}}\times (\mu -1)$
$20\times 10^{-3}=50\times 10^{-3}\times (\mu -1)$
$\mu -1=\frac{20}{50}=0.4$
$\mu =1.4$
The refractive index is given as $\frac{\alpha }{10}$.
So, $1.4=\frac{\alpha }{10}$
$\alpha =14$