Refraction Of Light Through Glass Slab

Refraction of light through a glass slab is a fundamental optical phenomenon that occurs when light passes through a transparent medium, such as glass, and changes direction due to a change in speed. As light enters the glass slab from air, it bends towards the normal because glass has a higher refractive index than air. Upon exiting the glass slab back into the air, the light bends away from the normal. This change in direction is governed by Snell's Law and is crucial in understanding how lenses and optical devices work. In real life, this principle is evident in everyday objects like eyeglasses, which correct vision by refracting light to focus correctly on the retina. Additionally, it is also observed in photographic lenses and microscopes, where precise light bending enables clear and detailed imaging. In this article, we will study refraction through the glass slab along with lateral displacement of emergent rays and some solved examples based on this concept.

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This Story also Contains

1. Refraction Through a Glass Slab
2. Solved Examples Based On Refraction of Light Through Glass Slab
3. Summary

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Refraction Through a Glass Slab

Refraction through a glass slab occurs when light travels from one medium (air) into another medium (glass) with a different refractive index. As light enters the glass slab, it slows down and bends towards the normal due to the higher refractive index of glass compared to air. When light exits the slab, it speeds up and bends away from the normal as it re-enters the air.

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 surface 1 and then refracts at surface 2 before reaching the observer as shown in the above figure.

So for the refraction at the surface (1)

Apparent depth, $d_1^{\prime}=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d}{\left(\frac{n_{\text {incidew }}}{n_{\text {refluction }}}\right)}=\frac{d}{\frac{1}{\mu}}=d \mu$

Similarly for the refraction at the surface (2)

Apparent depth, $d_2^{\prime}=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d_1^{\prime}+t}{\left(\frac{n_{\text {incident }}}{n_{\text {reflaction }}}\right)}=\frac{d_1^{\prime}+t}{\frac{\mu}{1}}=\frac{d_1 \mu+t}{\mu}$

Lateral Displacement of Emergent Ray Through a Glass Slab

Lateral displacement of the emergent ray through a glass slab refers to the horizontal shift between the original incident ray and the ray that emerges from the slab. This phenomenon occurs because light bends at both the entry and exit surfaces of the glass slab, causing a deviation from its original path.

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 Apparent shift $=t\left\{1-\frac{1}{\mu}\right\}$

If the slab is placed in the medium of the refractive index $\mu_{\text {sur }}$

then Apparent shift $=t\left\{1-\frac{\mu_{\max }}{\mu}\right\}$

In the above figure Incident, ray AO is 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

$\mu_a \sin i=\mu \sin r \quad$ and $\quad \mu \sin r^{\prime}=\mu_a \sin e$ Using $r^{\prime}=r$ and $\mu_a=1$, we get $\sin i=\sin e$ or $e=i$

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

$
d=P Q=O P \sin (i-r)=\frac{O M}{\cos r} \sin (i-r)=\frac{t \sin (i-r)}{\cos r}
$

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

Solved Examples Based On Refraction of Light Through Glass Slab

Example 1: An object is placed on the principle axis of a concave mirror of focal length 10 cm at a distance of 21 cm from it. A glass slab is placed between the mirror and the object

The distance of the final image formed by the mirror is:

1) 10cm

2) 20cm

3) 30cm

4) 21cm

Solution:

As we learn

Refraction through the parallel slab

$
s=t\left(1-\frac{1}{\mu}\right)
$
- wherein
$s=$ shifting of an object from slab
$t=$ thickness of slab
$\mu=$ Refractive Index of the slab.
$
\text { shift }=3\left(1-\frac{1}{3}\right)=1 \mathrm{~cm}
$

Therefore mirror object distance =(21-1)cm = 20cm

Therefore object is at the centre of the curvature of the mirror. Hence light rays will retrace and an image will form of the object itself.

Example 2: A ray of light is incident from the air on a glass plate having thickness $\sqrt{3} \mathrm{~cm}$ and refractive index $\sqrt{2}$ The angle of incidence of a ray is equal to the critical angle for the glass-air interface. The lateral displacement of the ray when it passes through the plate is $\qquad$ $\times 10^{-2} \mathrm{~cm}$. (given $\sin 15^{\circ}=0.26$ )

1) 52

2) 54

3) 56

4) 58

Solution:

$
\sin c=\frac{1}{\sqrt{2}} \Rightarrow c=45^{\circ}
$

Using Snell's law on 1st surface, sinc $=\sqrt{2} \sin r$
$
\begin{gathered}
\Rightarrow \sin r=\frac{1}{2} \Rightarrow r=30^{\circ} \\
d=t \operatorname{secr} \times \sin (c-r)=\sqrt{3} \times \frac{2}{\sqrt{3}} \times 0.26=0.52 \mathrm{~cm}=52 \times 10^{-2} \mathrm{~cm}
\end{gathered}
$

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

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

$
\begin{aligned}
2) & \tan ^{-1}\left(1-\sqrt{\frac{2}{3}}\right) \\
3) & \text { 2) } \sin ^{-1}\left(1-\sqrt{\frac{2}{3}}\right) \\
4) & \tan ^{-1}\left(\sqrt{\frac{2}{\sqrt{3}-1}}\right)
\end{aligned}
$

Solution:

Here, the angle of incidence $\mathrm{i}=45^{\circ}$
$
\frac{\text { Lateralshift }(\mathrm{d})}{\text { Thickness of glass } \operatorname{slab}(\mathrm{t})}=\frac{1}{\sqrt{3}}
$

Lateral shift, $d=\frac{\mathrm{t} \sin \delta}{\cos \mathrm{r}}=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
$
\begin{aligned}
& \Rightarrow \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}} \\
& \quad \text { or } \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin \mathrm{i} \cos \mathrm{r}-\cos \sin r}{\cos \mathrm{r}} \text { or } \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin 45^{\circ} \cos r-\cos 45^{\circ} \sin r}{\cos \mathrm{r}}=
\end{aligned}
$
or $\quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r}) \quad$ or $\quad \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan r) \quad$ or $\quad \tan \mathrm{r}=1-\frac{\sqrt{2}}{\sqrt{3}}$ or Angle of refraction, $r=\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$

Hence, the answer is option (2).

Example 4: On which factors lateral displacement of emergent ray from glass slab depends on

1) Thickness of glass slab and the refractive index of the material

2) Only on the refractive index of the material

3) Only on the thickness of the glass slab

4) Depends on the molecular structure of the material only

Solution:

The lateral displacement of the emergent ray from a glass slab depends on both the thickness of the glass slab and the refractive index of the material. The thickness determines the path length within the slab, while the refractive index affects the bending of light as it enters and exits the slab. Together, these factors influence the overall shift of the emergent ray.

Hence, the answer is option (1).

Summary

The lateral displacement of an emergent ray through a glass slab is influenced by both the thickness of the slab and its refractive index. The thickness dictates the path length of the light within the slab, while the refractive index determines the extent of light bending at the entry and exit points. This combined effect results in the overall lateral shift of the emergent ray.