A beam of light is incident on a glass plate | vedclass

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(A) According to Brewster's law,the refractive index $n$ is given by $	an 	heta_{B} = n$,where $	heta_{B}$ is the Brewster angle.Given $	heta_{B}

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{1}{\sqrt{6}}\right)$

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(A) According to Brewster's law,the refractive index $n$ is given by $	an 	heta_{B} = n$,where $	heta_{B}$ is the Brewster angle.Given $	heta_{B} = 60^{\circ}$,we have $n = 	an 60^{\circ} = \sqrt{3}$.Now,using Snell's law,$\frac{\sin i}{\sin r} = n$,where $i = 45^{\circ}$ is the angle of incidence and $r$ is the angle of refraction.Substituting the values,$\sin r = \frac{\sin 45^{\circ}}{n} = \frac{1/\sqrt{2}}{\sqrt{3}} = \frac{1}{\sqrt{6}}$.Therefore,the angle of refraction is $r = \sin^{-1}\left(\frac{1}{\sqrt{6}}ight)$.

$A$ beam of light is incident on a glass plate at an angle of $60^{\circ}$. The reflected ray is polarized. If the angle of incidence is $45^{\circ}$,then the angle of refraction is:

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