Question

A monochromatic ray of light is incident on the surface of a glass slab of thickness $8cm$, making and angle ${30}^{\circ }$ with the surface. The lateral displacement of the ray of light will be $({\mu }_g=\sqrt{3})$.

• A. $5.62cm$
• B. $4.62cm$
• C. $3.62cm$
• D. $8cm$

Answer 1

Answer: (B)

$\begin{array}{c}{\mu }_1\sin i={\mu }_2\sin \gamma \\ \text{(1)}\sin 60=\sqrt{3}\sin \gamma \\ \begin{array}{cc}\sin \gamma =\frac{1}{2}& \text{so consider}△OHA\\ \gamma ={30}^{\circ }& \end{array}\end{array}$

$HA=$ lateral displacement

$\tan 30=\frac{HA}{OH}$

$\frac{1}{\sqrt{3}}=\frac{HA}{8}$

$HA=\frac{8}{\sqrt{3}}=4.62cm$

so lateral displacement $=4.62cm$

Hence option $A$ is correct.