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 $5cm$ inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

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

Answer 1

The correct option is B $\frac{5}{2}cm$

Using Snell's Law:

$\frac{sin{60}^o}{sin{r}_1}=\sqrt{3}$

$⟹sin{r}_1=\frac{sin{60}^o}{\sqrt{3}}$

$⟹sin{r}_1=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=\frac{1}{2}$

$⟹r_1={30}^o$

Now, again Applying Snell's Law :

$sin(i_1-r_1)=\frac{d}{5}$

$⟹d=5sin(i_1-r_1)$

$⟹d=5sin({60}^o-{30}^o)$

$⟹d=5sin{30}^o=\frac{5}{2}cm$