The Brewster's angle for air to water transition is (Refractive index…

The correct answer is Option (3) → \( \tan^{-1}\left(\displaystyle\frac{4}{3}\right) \)

At the Brewster angle \( i_B \) the reflected light is completely plane polarised, and Brewster's law gives

    \( \displaystyle \tan i_B = \mu = \frac{\mu_{\mathrm{water}}}{\mu_{\mathrm{air}}} \)

For light going from air (\( \mu \approx 1 \)) into water:

    \( \displaystyle \tan i_B = \frac{4/3}{1} = \frac{4}{3} \;\Rightarrow\; i_B = \tan^{-1}\left(\frac{4}{3}\right) \)

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.”