When the angle of incidence on a material is 60°, the reflected light…

The correct answer is Option (3) → \( \sqrt{3} \times 10^{8} \)

The reflected light is completely polarised when light falls at the Brewster angle \( i_B \), and Brewster's law gives the refractive index of the material:

    \( \displaystyle n = \tan i_{B} = \tan 60^\circ = \sqrt{3} \)

The speed of light in the material is the speed in vacuum divided by the refractive index:

    \( \displaystyle v = \frac{c}{n} = \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{\sqrt{3}} = \sqrt{3}\times10^{8}\,\mathrm{m\,s^{-1}} \)

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 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.”