Question

A ray of light is incident at polarising angle \(θ\) on air-glass interface. If \(λ_a\) and \(λ_g\) are the wavelengths of light in air and glass respectively then

• A. \(λ_a = λ_gcotθ\)
• B. \(λ_g = λ_acotθ\)
• C. \(λ_a = λ_gtan^2θ\)
• D. \(λ_g = λ_atan^2θ\)

Hint:

At the polarising angle tan(theta) equals the refractive index, which is also the ratio of wavelengths in air and glass.

Answer 1

The Correct Option is B

Step 1: Brewster's law
$\mu = \tan\theta_p$.

Step 2: Wavelength in a medium
$\lambda_g = \dfrac{\lambda_a}{\mu}$ because wavelength shrinks by the factor $\mu$ while frequency stays fixed.

Step 3: Substitute
$\lambda_g = \dfrac{\lambda_a}{\tan\theta} = \lambda_a\cot\theta$.

Step 4: Answer
Option (B).

Step 5: Numerical illustration
For ordinary glass with $\mu = 1.5$, the polarising angle is $\tan^{-1}1.5 = 56.3^\circ$. If the wavelength in air is 600 nm, then $\lambda_g = 600\cot56.3^\circ = 600/1.5 = 400$ nm, which is shorter than in air, as it should be in a denser medium. The option $\lambda_g = \lambda_a\tan^2\theta$ would give 1350 nm, which is longer than the wavelength in air, and that cannot happen in glass.

Final Answer:
lambda_g = lambda_a cot theta. This is option (B). \[ \boxed{\text{(B) }\lambda_g=\lambda_a\cot\theta} \]