Question

Parallel side glass slab of thickness $8\text{cm}$ is made of a material of refractive index $\sqrt{3}$. When light is incident on one of the parallel faces at an angle of ${60}^{\circ }$ with the normal to the surface and emerges from other parallel face, find the lateral displacement of the emergent ray.

• A. $\frac{4}{\sqrt{3}}\text{cm}$
• B. $\sqrt{3}\text{cm}$
• C. $\frac{8}{\sqrt{3}}\text{cm}$
• D. $8\sqrt{3}\text{cm}$

Answer 1

The correct option is C $\frac{8}{\sqrt{3}}\text{cm}$



Given that,

thickness $\left(t\right)=8\text{cm},\mu =\sqrt{3},i={60}^{\circ }$

From Snell's law,

$\frac{\sin i}{\sin r}=\mu$

Substituting the given data we get,

$\frac{\sin {60}^{\circ }}{\sin r}=\sqrt{3}\Rightarrow \sin r=\frac{\sin {60}^{\circ }}{\sqrt{3}}$

$\Rightarrow \sin r=\frac{\sqrt{3}}{2\sqrt{3}}=\frac{1}{2}$

$\Rightarrow r={30}^{\circ }$

For lateral displacement,
$\delta =\frac{t\sin (i-r)}{\cos r}$

Substituting the given data we get,

$\delta =\frac{8\times \sin ({60}^{\circ }-{30}^{\circ })}{\cos {30}^{\circ }}=\frac{8\sin {30}^{\circ }}{\cos {30}^{\circ }}$

$\Rightarrow \delta =\frac{8\times \frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{8}{\sqrt{3}}\text{cm}$

Hence, option (c) is the correct answer.

Why this question? This question helps student in application part of the formula of the lateral displacement due to a glass slab.