Unpolarised light is incident on a medium at an angle of incidence 60°. The reflected ray is perpendicular to the refracted ray. Refractive index (μ) of the medium is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.73
When the reflected and refracted rays are at right angles, the angle of incidence is the Brewster angle \( i_B \), and Brewster’s law gives \( \mu = \tan i_B \). Here \( i_B = 60^\circ \).
\( \mu = \tan60^\circ = \sqrt{3} = 1.73 \)
Check with Snell’s law: the angle of refraction is \( r = 90^\circ - 60^\circ = 30^\circ \), so \( \mu = \displaystyle\frac{\sin60^\circ}{\sin30^\circ} = \frac{0.866}{0.5} = 1.73 \).
Note: A similar question was also asked in CUET (UG) 2023 on 06-06-2023 (Slot 1).
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. We can see that iB is related to the refractive index of the denser medium. Since we have iB + 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.”