Question

In the given figure , find the lateral shift produced by the glass slab.

• A. $\sqrt{3}\text{m}$
• B. $\frac{1}{\sqrt{2}}\text{m}$
• C. $\frac{1}{\sqrt{3}}\text{m}$
• D. $\sqrt{2}\text{m}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}\text{m}$

We know,

Lateral shift, $d$ $=\frac{t\sin (i-r)}{\cos r}$

From Snells law,

${\mu }_{air}\sin {60}^{\circ }={\mu }_{glass}\sin r$

Substituting the given data we get,

$1\times \sin {60}^{\circ }=\sqrt{3}\times \sin r$
$\Rightarrow \sin r=\frac{1}{2}\Rightarrow r={30}^{\circ }$

Substituting the values of $i$ and $r$ in lateral shift formula we get,

$d$ $=\frac{1\times \sin ({60}^{\circ }-{30}^{\circ })}{\cos {30}^{\circ }}$

$\therefore d=\frac{1}{\sqrt{3}}\text{m}$

Hence, option (c) is the correct answer.