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 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. $5\, cm$

Answer 1

Answer: (B)

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