A ray of light is incident at the polarising | vedclass

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Question

(B) According to Brewster's Law, the refractive index of the glass $(\mu)$ is given by $\mu = 	an 	heta$, where $	heta$ is the polarising angle

Vedclass · Accepted Answer

$\lambda_g = \lambda_a \cot 	heta$

Vedclass · Answer

(B) According to Brewster's Law, the refractive index of the glass $(\mu)$ is given by $\mu = 	an 	heta$, where $	heta$ is the polarising angle.
We know that the refractive index is also defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium: $\mu = \frac{v_a}{v_g}$.
Since the frequency $(f)$ of light remains constant when it travels from one medium to another, the speed of light is related to wavelength $(\lambda)$ by $v = f \lambda$.
Therefore, $\mu = \frac{f \lambda_a}{f \lambda_g} = \frac{\lambda_a}{\lambda_g}$.
Equating the two expressions for $\mu$, we get $	an 	heta = \frac{\lambda_a}{\lambda_g}$.
Rearranging this gives $\lambda_g = \frac{\lambda_a}{	an 	heta} = \lambda_a \cot 	heta$.

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

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