A ray of light strikes a glass plate at an angle of 60°. If the refle…

The correct answer is Option (2) → 1.732

When the reflected and refracted rays are at right angles, the angle of incidence is the Brewster angle \( i_B \) and the refractive index of the glass is \( \mu = \tan i_B \) (Brewster's law). The angle of incidence here is \( 60^\circ \).

    \( \mu = \tan 60^\circ = \sqrt{3} = 1.732 \)

The same follows from Snell's law: if the reflected and refracted rays are perpendicular, the angle of refraction is \( r = 90^\circ - 60^\circ = 30^\circ \), so

    \( \displaystyle \mu = \frac{\sin60^\circ}{\sin30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} = 1.732 \)

Note: A similar question was also asked in CUET (UG) 2023 on 12-06-2023 (Slot 2).

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 – “… When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. … The angle of incidence in this case is called Brewster’s angle and is denoted by i_B. … 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.”