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Refraction Of Light Through Glass Slab MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Refraction Through A Glass Slab is considered one the most difficult concept.

  • 14 Questions around this concept.

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A ray of light enters a rectangular slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{\circ}$. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between incident and emergent rays is:

Concepts Covered - 2

Refraction Through A Glass Slab

 

Consider an object O placed a 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) 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 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}

i.e the emergent ray is parallel to the incident ray.

If  PQ is the perpendicular dropped from P on the incident ray produced.

Then PQ=d is known as lateral displacement which is given as

\begin{aligned} d &=P Q=O P \sin (i-r)=\frac{O M}{\cos r} \sin (i-r) =\frac{t \sin (i-r)}{\cos r} \end{aligned}

  • \text { If } i \text { is very small, } r \text { is also very small, then } \ \ d=\left(1-\frac{1}{\mu}\right) t i 

 

 

Study it with Videos

Refraction Through A Glass Slab
Lateral Displacement Of Emergent Ray Through A Glass Slab

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