In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed

Question

In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness $20 \mu \mathrm{~m}$ is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by $\frac{\alpha}{10}$, where $\alpha$ is $\_\_\_\_$。

Professor PC Thomas Classes · Answer

The shift in the central bright fringe due to the introduction of a transparent sheet is given by $\Delta y = \frac{D}{d}(\mu - 1)t$. Given $\Delta y = 20	ext{ mm} = 20 	imes 10^{-3}	ext{ m}$, $D = 1	ext{ m}$, $d = 0.4	ext{ mm} = 4 	imes 10^{-4}	ext{ m}$, and $t = 20\mu	ext{m} = 20 	imes 10^{-6}	ext{ m}$. Substituting the values: $20 	imes 10^{-3} = \frac{1}{4 	imes 10^{-4}} (\mu - 1) (20 	imes 10^{-6})$. This simplifies to $20 	imes 10^{-3} = (\mu - 1) 	imes 0.05$. Therefore, $\mu - 1 = \frac{0.02}{0.05} = 0.4 \Rightarrow \mu = 1.4 = \frac{14}{10}$. Hence, $\alpha = 14$.