Question

A ray of light enters a rectangular glass 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 slab. The perpendicular distance between the incident ray and the emergent ray is

• A. $5/\sqrt{3}cm$
• B. $\sqrt{3}/5cm$
• C. $5/3cm$
• D. $5/2cm$

Answer 1

The correct option is D $5/2cm$

Using snell's law:

$\frac{sin{60}^{\circ }}{sin{r}_1}=\sqrt{3}\Rightarrow sin{r}_1=\frac{sin{60}^{\circ }}{\sqrt{3}}\Rightarrow sin{r}_1=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=\frac{1}{2}\Rightarrow r_1={30}^{\circ }\text{Now, we know that,}sin(i_1-r_1)=\frac{d}{5}\Rightarrow d=5sin(i_1-r_1)\Rightarrow d=5sin({60}^{\circ }-{30}^{\circ })\Rightarrow d=5sin{30}^{\circ }=\frac{5}{2}cm$