Question

A plane wavefront is travelling from air to glass. It is incident at an angle of incidence of ${45}^{\circ }$ on the glass surface. If the ratio of width of wavefront in air to that in glass is $\frac{\sqrt{2}}{\sqrt{3}}$, find the refractive index of glass.

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\sqrt{6}$
• D. $\frac{\sqrt{3}}{\sqrt{2}}$

Answer 1

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

Consider a plane wavefront travelling from air to glass, as shown below.

Here, $\frac{PQ}{P_1{Q}_1}=\frac{\sqrt{2}}{\sqrt{3}}$ and $i={45}^{\circ }$

Now, in $\Delta PQ{Q}_1$
$\sin (90-i)=\cos i=\frac{PQ}{P{Q}_1}$

And, in $\Delta P{P}_1{Q}_1$
$\sin (90-i)=\cos r=\frac{P_1{Q}_1}{P{Q}_1}$

$\therefore \frac{\cos i}{\cos r}=\frac{PQ}{P_1{Q}_1}=\frac{\sqrt{2}}{\sqrt{3}}$

$\cos r=\frac{\sqrt{3}\cos {45}^{\circ }}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{2}$
$\therefore r={30}^{\circ }$
Thus, refractive index of glass is given as
$\mu =\frac{\sin i}{\sin r}=\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{2}$

Hence, $\left(B\right)$ is the correct answer.