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: Understand the concept
Brewster's law says that at the polarising angle \(\theta\), \(\tan\theta = \mu\), where \(\mu\) is the refractive index of glass relative to air.

Step 2: Link \(\mu\) with wavelength
The frequency does not change on entering glass, so \(\mu = \dfrac{c}{v} = \dfrac{\lambda_a}{\lambda_g}\).

Step 3: Combine
\[ \tan\theta = \frac{\lambda_a}{\lambda_g} \Rightarrow \lambda_g = \frac{\lambda_a}{\tan\theta} = \lambda_a\cot\theta \]

Step 4: Result
Option (B). Option (A) has \(\lambda_a\) and \(\lambda_g\) interchanged, and the \(\tan^2\theta\) options have a wrong power of the tangent.

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