A ray of light is incident at the polarising | vedclass

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(A) According to Brewster's Law,the refractive index $\mu$ of the glass with respect to air is given by $\mu = 	an 	heta$,where $	heta$ is the pola

Vedclass · Accepted Answer

$\lambda_g = \lambda_a \cot 	heta$

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(A) According to Brewster's Law,the refractive index $\mu$ of the glass with respect to air is given by $\mu = 	an 	heta$,where $	heta$ is the polarising angle.We know that the refractive index $\mu$ is also defined as the ratio of the speed of light in air $(c_a)$ to the speed of light in glass $(c_g)$,which is $\mu = \frac{c_a}{c_g}$.Since the frequency $f$ of light remains constant when it travels from one medium to another,the speed of light is related to wavelength by $c = f \lambda$.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,we find $\lambda_a = \lambda_g 	an 	heta$. However,looking at the options provided,there seems to be a discrepancy in the standard form. Re-evaluating the relationship: $\mu = \frac{\lambda_a}{\lambda_g} = 	an 	heta$. Thus,$\lambda_g = \frac{\lambda_a}{	an 	heta} = \lambda_a \cot 	heta$.

$A$ 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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