When a ray of light is incident at an angle \( \theta=60^{\circ} \) on an air-glass interface, the reflected ray is 100 percent plane polarised. The value of the refractive index of the glass with respect to air \( \left({}_{a}n_{g}\right) \) is given by:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.732
The reflected light is completely plane polarised only when the angle of incidence is the Brewster angle \( \theta_B \), and then Brewster’s law gives the refractive index: \( n = \tan\theta_B \).
\( {}_{a}n_{g} = \tan60^{\circ} = \sqrt{3} = 1.732 \)
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 iB. …
\( \displaystyle \mu = \frac{\sin i_B}{\cos i_B} = \tan i_B \qquad (10.36) \)
This is known as Brewster’s law.”