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

Answer: (B)

Snell's law gives
$1\sin 60^{\circ }=\sqrt{3}\cdot \sin r$
$\sin r=\frac{\sin 60^{\circ }}{\sqrt{3}}=\frac{1}{2}$
$\Rightarrow r=30^{\circ }$
$x=5\sin (i-r)=5\sin \left(60^{\circ }-30^{\circ }\right)$
$\Rightarrow x=5\sin 30^{\circ }=\frac{5}{2}cm$