Question 183: Wave Optics

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 $\_\_\_\_$.

Accepted Answer

When a transparent sheet of thickness t and refractive index $\mu$ is placed in front of one slit, the optical path length of the light from that slit increases.
Extra path length introduced is $\Delta \mathrm{x}=(\mu-1) \mathrm{t} \ldots$ (i)
In YDSE, the path length at a point is given as $\Delta \mathrm{x}=\mathrm{d} \sin 	heta \Rightarrow \sin 	heta=\frac{\Delta \mathrm{x}}{\mathrm{d}} \ldots$ (ii)
For small angle, $\sin 	heta \approx 	an 	heta \approx 	heta$ (in radians).
Also, $	an 	heta=\frac{\Delta y}{D} \ldots$ (iii)
From (i), (ii) and (iii);
$$ \frac{\Delta y}{D}=\frac{\Delta x}{d} $$
$$\Rightarrow \Delta y=\frac{D \Delta x}{d}=\frac{D(\mu-1) t}{d} $$
$$\Rightarrow $$ $\Delta y=\frac{D(\mu-1) t}{d} \Rightarrow(\mu-1)=\frac{\Delta y \cdot d}{D \cdot t}$
$$ \Rightarrow \mu=\frac{\Delta y \cdot d}{D \cdot t}+1 $$
Where :
$\mathrm{D}=$ distance between the slits and the screen $=1 \mathrm{~m}$.
$\mathrm{d}=$ distance between the two slits $=0.4 \mathrm{~mm}=0.4 	imes 10^{-3} \mathrm{~m}$.
Sheet thickness is $\mathrm{t}=20 \mu \mathrm{~m}=20 	imes 10^{-6} \mathrm{~m}$
Fringe shift is $\Delta y=20 \mathrm{~mm}=20 	imes 10^{-3} \mathrm{~m}$
Substituting these values,
$$ \mu=\frac{\left(20 	imes 10^{-3}ight) 	imes\left(0.4 	imes 10^{-3}ight)}{1 	imes\left(20 	imes 10^{-6}ight)}+1 $$
$$\Rightarrow $$ $\mu=\frac{8 	imes 10^{-6}}{20 	imes 10^{-6}}+1$
$$\Rightarrow $$ $\mu=0.4+1=1.4=\frac{14}{10}$
$$ \Rightarrow \frac{14}{10}=\frac{\alpha}{10} \Rightarrow \alpha=14 $$
Therefore, the value of $\alpha$ is 14 .

Wave Optics

Integer / Numeric Answer Type