Which of the following option shows the Brewster's law in polarisatio…

The correct answer is Option (3) → \( \mu\cot i_B = 1 \)

Brewster’s law gives the angle of incidence \( i_B \) at which the reflected light is completely plane polarised: \( \mu = \tan i_B \).

Since \( \cot i_B = 1/\tan i_B \):

    \( \displaystyle \mu\cot i_B = \tan i_B\times\frac{1}{\tan i_B} = 1 \)

Why the other options are incorrect

• Option (1) \( \mu\sin i_B = 1 \) — Incorrect. This gives \( \sin i_B = 1/\mu \), which is the relation for the critical angle, not the polarising angle.

• Option (2) \( \mu\cos i_B = 1 \) — Incorrect. It does not correspond to any standard optical law.

• Option (4) \( \mu\tan i_B = 1 \) — Incorrect. It gives \( \tan i_B = 1/\mu \), the inverse of the true relation \( \tan i_B = \mu \).

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 – “… 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.”