A ray of light is incident at polarising angle θ on air-glass interface. If λa and λg are the wavelengths of light in air and glass respectively then (a) λa = λg cot θ (b) λg = λa cot θ (c) λa = λg tan ² θ (d) λg = λa tan ² θ

Answer: (b). Brewster's law: tan θ = μ = (λa)/(λg) λg = (λa)/(tan θ) = λacot θ So the answer is (b).

A ray of light is incident at polarising angle θ on air-glass… | MHT CET 2026 (19 Apr, Shift 2)

Q9MHT CET 2026 · 19 APR · SHIFT 2Medium

Polarisation

A ray of light is incident at polarising angle $\theta$ on air-glass interface. If $\lambda_a$ and $\lambda_g$ are the wavelengths of light in air and glass respectively then

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Answer: (b)

Brewster's law: $\tan\theta = \mu = \frac{\lambda_a}{\lambda_g}$
$\lambda_g = \frac{\lambda_a}{\tan\theta} = \lambda_a\cot\theta$
So the answer is (b).

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