The reflected and the refracted rays are perpendicular to each other when light is incident on a transparent denser medium having refractive index $\sqrt{3}$. The angle of incidence is

60°

The correct answer is Option (1) → 60°

Given:

Refractive index of medium, $\mu = \sqrt{3}$

Condition: Reflected ray and refracted ray are perpendicular to each other.

Hence, angle between them = $90^\circ$.

By Snell’s law:

$\mu = \frac{\sin i}{\sin r}$

Also, from geometry of the condition:

$i + r = 90^\circ$

Substitute $r = 90^\circ - i$ in Snell’s law:

$\mu = \frac{\sin i}{\sin (90^\circ - i)} = \frac{\sin i}{\cos i} = \tan i$

Therefore, $\tan i = \sqrt{3}$

$i = 60^\circ$

Final Answer: $i = 60^\circ$