Question

A slab of glass of thickness of $3$ units and refractive index $\sqrt{3}$ is surrounded by vacuum and is placed as shown. A light ray is incident at the point $(2,5\sqrt{3})$ as shown. The point where the ray will cut the $X-$ axis is:

• A. $(6,0)$
• B. $(5+\sqrt{3},0)$
• C. $(9,0)$
• D. $(8,0)$

Answer 1

The correct option is C $(9,0)$

Given thickness of slab $BC=3$ units
coordinate of B is $(2,5\sqrt{3})$
$\therefore CE=5\sqrt{3}$

Applying Snell's law at B,
$1\text{sin i}=\mu \text{sin r}$
$1\text{sin}{60}^{\circ }=\sqrt{3}\text{sin r}\frac{\sqrt{3}}{2}=\sqrt{3}\text{sin r}$
$\therefore r={30}^{\circ }$

From $\Delta BCD$
$CD=BC\tan r=3\text{tan}{30}^{\circ }=\sqrt{3}$
$\therefore DE=CE-CD=[5\sqrt{3}-\sqrt{3}]=4\sqrt{3}$
From triangle $DEF$
$\therefore EF=DE\cot {60}^{\circ }=4\sqrt{3}\text{cot}{60}^{\circ }=4$
$\therefore$ Coordinate of point $E=(3+2+4,0)=(9,0)$