In a Young's double slit experiment set up, the two slits are kept...

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, $1.4 = \frac{\alpha}{10}$
$\alpha = 14$