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Question

(C) According to Brewster's law,when light is incident at the Brewster's angle $\phi$ on a transparent medium,the reflected ray and the refracted ray

Vedclass · Accepted Answer

$90^\circ$

Vedclass · Answer

(C) According to Brewster's law,when light is incident at the Brewster's angle $\phi$ on a transparent medium,the reflected ray and the refracted ray are mutually perpendicular to each other.Let $i$ be the angle of incidence,$r$ be the angle of refraction,and $\phi$ be the Brewster's angle.Given $i = \phi$.From Snell's law,$\mu = \frac{\sin \phi}{\sin r}$.Also,by Brewster's law,$\mu = 	an \phi = \frac{\sin \phi}{\cos \phi}$.Equating the two expressions for $\mu$,we get $\frac{\sin \phi}{\sin r} = \frac{\sin \phi}{\cos \phi}$,which implies $\sin r = \cos \phi = \sin(90^\circ - \phi)$.Therefore,$r = 90^\circ - \phi$,or $\phi + r = 90^\circ$.The angle between the reflected ray and the surface is $90^\circ - \phi$. The angle between the refracted ray and the normal is $r$. The angle between the reflected ray and the refracted ray is $180^\circ - (90^\circ - \phi) - r = 90^\circ + \phi - r = 90^\circ + \phi - (90^\circ - \phi) = 90^\circ$ is incorrect logic; rather,the angle between the reflected ray and the refracted ray is $180^\circ - (90^\circ - \phi) - r = 180^\circ - (90^\circ - \phi) - (90^\circ - \phi) = 180^\circ - 90^\circ = 90^\circ$.

$A$ ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster's angle $\phi$. If $\mu$ represents the refractive index of glass with respect to air,then the angle between the reflected and refracted rays is:

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