Question

Find the lateral shift $\left(\Delta \right)$ of a ray passing through a slab having refrective index $\sqrt{3}$ incident at ${60}^{\circ }$.

• A. $3\sqrt{2}$
• B. $2\sqrt{3}$
• C. $\sqrt{2}$
• D. $4\sqrt{3}$

Answer 1

The correct option is B $2\sqrt{3}$

Given :
$\mu =\sqrt{3};i={60}^{\circ }$


From geometry of diagram,

$\sin (i-r)=\frac{\Delta }{AB}$

And,

$\cos r=\frac{6\text{cm}}{AB}$

$\therefore \Delta =\frac{\sin (i-r)}{\cos r}\times 6.....\left(1\right)$

Using Snell's Law, we get

$1\cdot \sin {60}^{\circ }=\sqrt{3}\cdot \sin r$

$\Rightarrow \sin r=\frac{1}{2}$

$\Rightarrow r={30}^{\circ }$

Putting the value in eq. $\left(1\right)$

$\therefore \Delta =6\times \frac{\sin ({60}^{\circ }-{30}^{\circ })}{\cos {30}^{\circ }}$

$\Rightarrow \Delta =\frac{6\times \frac{1}{2}}{\frac{\sqrt{3}}{2}}=2\sqrt{3}$

Hence, option $\left(b\right)$ is correct.