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Refraction Of Light Through Glass Slab - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Refraction Through A Glass Slab is considered one the most difficult concept.
10 Questions around this concept.

Solve by difficulty

A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :

14 cm below flat surface

30 cm below flat surface

20 cm below flat surface

16 cm below flat surface

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A ray of light is incident on surface of glass slab at an angle $45^{\circ}$ . If the lateral shift produced per unit thickness is $\mathrm{\frac{1}{\sqrt{3}}m,}$ , the angle of refraction produced is:

$\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\tan ^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$

$\sin ^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$

$\tan ^{-1}\left(\sqrt{\frac{2}{\sqrt{3}-1}}\right)$

View Solution

Concepts Covered - 2

Refraction Through A Glass Slab

Consider an object O placed at distance d in front of a glass slab of thickness "t" and refractive index $\mu$ . The observer is on
the other side of the slab. A ray of light from the object first refracts at the surface 1 and then refracts at the surface 2 before reaching the observer as shown in the above figure.

So for the refraction at the surface (1)

$\text { Apparent depth, } d_1'=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d}{\left(\frac{n_{\text {incident }}}{n_{\text {reftaction }}}\right)}= \frac{d}{\frac{1}{\mu }}=d\mu$

Similarly for the refraction at the surface (2)

$\text { Apparent depth, } d_2'=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d_1'+t}{\left(\frac{n_{\text {incident }}}{n_{\text {reftaction }}}\right)}= \frac{d_1'+t}{\frac{\mu }{1 }}= \frac{d_1\mu +t}{\mu }$

As you observe, The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction.

i.e. the ray undergoes no deviation ( $\delta =0$ ).

the object appears to be shifted towards the slab by the distance known as apparent shift or Normal shift.

And the apparent shift= OA-I₂A

I.e $\begin{array}{l}{\text { Apparant shift }=t\left\{1-\frac{1}{\mu}\right\}} \\ \end{array}$

If the slab is placed in the medium of the refractive index $\mu _{sur}$

then $\begin{array}{l}{\text { Apparant shift }=t\left\{1-\frac{\mu _{sur}}{\mu}\right\}} \\ \end{array}$

Lateral Displacement Of Emergent Ray Through A Glass Slab

In the above figure Incident, ray AO is an incident on the EF surface of the slab at an angle of incident i, and PB is the emergent ray emerging out of the HG surface of the slab.

for the surface EF

Applying Snell's law at the surface EF and HG

$\begin{array}{l}{\mu_{a} \sin i=\mu \sin r \quad \text { and } \quad \mu \sin r^{\prime}=\mu_{a} \sin e} \\ Using \ \ {r^{\prime}=r \quad \text { and } \quad \mu_{a}=1, \text { we get }} \\ {\sin i=\sin e \text { or } e=i}\end{array}$