The reflected and the refracted rays are perpendicular to each other…

The correct answer is Option (1) → \( 60^\circ \)

When the reflected and refracted rays are perpendicular, light is incident at the Brewster angle \( i_B \), and Brewster's law gives it.

    \( \displaystyle \tan i_B = \mu \)

    \( \displaystyle \tan i_B = \sqrt{3} \;\Rightarrow\; i_B = 60^\circ \)

Check: the angle of refraction is \( r = 90^\circ - 60^\circ = 30^\circ \), and \( \sin 60^\circ/\sin 30^\circ = \sqrt{3} \), as required by Snell's law.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Polarisation by Reflection – “… Thus we have seen that when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarised wave. The angle of incidence in this case is called Brewster’s angle and is denoted by i_B. We can see that i_B is related to the refractive index of the denser medium. Since we have i_B + r = π/2, we get from Snell’s law

    \( \displaystyle \mu = \frac{\sin i_B}{\sin r} = \frac{\sin i_B}{\sin(\pi/2 - i_B)} = \frac{\sin i_B}{\cos i_B} = \tan i_B \qquad (10.36) \)

This is known as Brewster’s law.”