Given below are two statements: Statement I: If the Brewster's angle for the light propagating from air to glass is θB, then the Brewster's angle for the light propagating from glass to air is π/2 - θB Statement II: The Brewster's angle for the light propagating from glass to air is tan ⁻¹(μg) where μg is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below: (a) Both Statement I and Statement II are false (b) Both Statement I and Statement II are true (c) Statement I is false but Statement II is true (d) Statement I is true but Statement II is false

Answer: (d). Air to glass: tan θB = μg. Glass to air: tan θB' = (1)/(μg) = cot θB. So θB' = π/2 - θB: Statement I is true. Glass to air Brewster angle is tan ⁻¹(1/μg), not tan ⁻¹(μg): Statement II is false. So the answer is (d).

Given below are two statements: Statement I: If the Brewster's angle… | JEE Main 2023 (24 Jan, Shift 1)

Q38JEE Main 2023 · 24 JAN · SHIFT 1Medium

Polarisation

Given below are two statements:

Statement I: If the Brewster's angle for the light propagating from air to glass is $\theta_{\mathrm{B}}$ , then the Brewster's angle for the light propagating from glass to air is $\frac{\pi}{2} - \theta_B$

Statement II: The Brewster's angle for the light propagating from glass to air is $\tan^{-1}(\mu_{\mathrm{g}})$ where $\mu_{\mathrm{g}}$ is the refractive index of glass.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (d)

Air to glass: $\tan\theta_B = \mu_g$ . Glass to air: $\tan\theta_B' = \frac{1}{\mu_g} = \cot\theta_B$ .
So $\theta_B' = \frac{\pi}{2} - \theta_B$ : Statement I is true.
Glass to air Brewster angle is $\tan^{-1}(1/\mu_g)$ , not $\tan^{-1}(\mu_g)$ : Statement II is false.
So the answer is (d).

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Polarisation question from JEE Main 2023 (24 Jan, Shift 1)

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