From the figure given below, the refractive index of medium B with respect to medium A (_Aμ_B) is:

1. $\dfrac{\text {sin }45\degree}{\text {sin }30\degree} \\[1em]$
2. $\dfrac{\text {sin }30\degree}{\text {sin }45\degree} \\[1em]$
3. $\dfrac{\text {sin }45\degree}{\text {sin }60\degree} \\[1em]$
4. $\dfrac{\text {sin }60\degree}{\text {sin }45\degree} \\[1em]$

$\dfrac{\text {sin }60\degree}{\text {sin }45\degree} \\[1em]$

Reason — The angle given in medium A is 30° with the surface, so the angle with the normal is:

$i = 90\degree - 30\degree = 60\degree$

The angle given in medium B is 45° with the surface, so the angle with the normal is:

$r = 90\degree - 45\degree = 45\degree$

Now, refractive index of medium B with respect to medium A is:

${}$A\muB=\frac{\sin i}{\sin r} \\[1em] \Rightarrow {}A\muB=\frac{\sin 60^\circ}{\sin 45^\circ}