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# Refractive index of a rectangular glass slab is $m = \sqrt 3$. A light ray incident at an angle ${60^0}$ is displaced laterally through $2.5\,cm$ . Distance travelled by light in the slab isA. ${\text{ }}4cm$B. ${\text{ }}5cm$C. ${\text{ }}2.5\sqrt 3 cm$D. ${\text{ }}3cm$

Last updated date: 14th Aug 2024
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Hint- In order to find the distance by light in the slab first we will assume the travelled distance as a variable. Then we will proceed further by using the lateral shift formula as Snell's law equation which is mentioned in the solution. We will make a clear ray diagram, and then we will proceed further by finding the angle between incident ray and emergent ray.

Formula used- $d = t\dfrac{{\sin \left( {{\theta _a} - {{\theta '}_b}} \right)}}{{\cos {{\theta '}_b}}},m = \dfrac{{\sin {\theta _a}}}{{\sin {{\theta '}_b}}}$
Given that- $m = \sqrt 3$ , incident angle= ${60^0}$
We will use the following figure to solve this problem.

Let the length traveled in glass be l.
We know that Lateral shift is given as
$d = t\dfrac{{\sin \left( {{\theta _a} - {{\theta '}_b}} \right)}}{{\cos {{\theta '}_b}}}............(1)$
From figure by using the geometry
$t = l\cos {\theta '_b}..............(2)$
Substitute the value of t in above formula we have
$\because d = t\dfrac{{\sin \left( {{\theta _a} - {{\theta '}_b}} \right)}}{{\cos {{\theta '}_b}}}$
$t = l\cos {{\theta '}_b} \\$
$\Rightarrow d = \left( {l\cos {{\theta '}_b}} \right) \times \dfrac{{\sin \left( {{\theta _a} - {{\theta '}_b}} \right)}}{{\cos {{\theta '}_b}}} \\$
$\Rightarrow d = l\sin \left( {{\theta _a} - {{\theta '}_b}} \right)..............(3) \\$
As we know that Snell's law equation is given as
$\Rightarrow m = \dfrac{{\sin {\theta _a}}}{{\sin {{\theta '}_b}}}$
Put the value of m and incident angle in order to find the value of ${\theta '_b}$ , we have
$\because m = \dfrac{{\sin {\theta _a}}}{{\sin {{\theta '}_b}}} \\$
$\Rightarrow \sqrt 3 = \dfrac{{\sin {{60}^0}}}{{\sin {{\theta '}_b}}} \\$
$\Rightarrow \sqrt 3 = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{\sin {{\theta '}_b}}} \\$
$\Rightarrow \sin {{\theta '}_b} = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{\sqrt 3 }} = \dfrac{1}{2} \\$
$\Rightarrow {{\theta '}_b} = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\$
$\Rightarrow {{\theta '}_b} = {30^0} \\$
Now, substitute this value in above equation (3) to obtain the distance travelled by light in the slab, we obtain
$\because d = l\sin \left( {{\theta _a} - {{\theta '}_b}} \right) \\$
$\Rightarrow 2.5 = l\sin \left( {{{60}^0} - {{30}^0}} \right) \\$
$\Rightarrow 2.5 = l\sin \left( {{{30}^0}} \right) \\$
$\Rightarrow 2.5 = l \times \dfrac{1}{2} \\$
$\Rightarrow l = 2.5 \times 2 \\$
$\therefore l = 5\,cm$
Hence, the distance traveled by light in the slab is 5 cm.
Therefore, the correct answer is option B.

Note- Refraction is the transition of a wave's direction from one medium to another, or from a subtle change in the medium. Light refraction is the most frequently observed phenomenon but other waves like sound waves and water waves also experience refraction. Students must remember the nature of light when it moves from rarer to denser medium and vice-versa. Students must remember Snell’s law to solve such problems.