1(iv)
From the figure given below, the refractive index of medium B with respect to medium A (AμB) is:
- (a)$$\begin{aligned}\frac{\text {sin }45^{\circ}}{\text {sin }30^{\circ}} \\[1em]\end{aligned}$$
- (b)$$\begin{aligned}\frac{\text {sin }30^{\circ}}{\text {sin }45^{\circ}} \\[1em]\end{aligned}$$
- (c)$$\begin{aligned}\frac{\text {sin }45^{\circ}}{\text {sin }60^{\circ}} \\[1em]\end{aligned}$$
- (d)$$\begin{aligned}\frac{\text {sin }60^{\circ}}{\text {sin }45^{\circ}} \\[1em]\end{aligned}$$
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$$\begin{aligned}\frac{\text {sin }60^{\circ}}{\text {sin }45^{\circ}} \\[1em]\end{aligned}$$
Reason — The angle given in medium A is 30° with the surface, so the angle with the normal is:
$i = 90^{\circ} - 30^{\circ} = 60^{\circ}$
The angle given in medium B is 45° with the surface, so the angle with the normal is:
$r = 90^{\circ} - 45^{\circ} = 45^{\circ}$
Now, refractive index of medium B with respect to medium A is:
$$\begin{aligned}{}_A\mu_B=\frac{\sin i}{\sin r} \\[1em] \Rightarrow {}_A\mu_B=\frac{\sin 60^\circ}{\sin 45^\circ}\end{aligned}$$