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(C) According to Brewster's law,when the reflected and refracted rays are perpendicular to each other,the angle of incidence $i$ is equal to the Brews

Vedclass · Accepted Answer

$57$

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(C) According to Brewster's law,when the reflected and refracted rays are perpendicular to each other,the angle of incidence $i$ is equal to the Brewster's angle $i_B$.From Snell's law,$\mu = \frac{\sin i_B}{\sin r}$.Since the reflected and refracted rays are perpendicular,$i_B + r = 90^{\circ}$,which implies $r = 90^{\circ} - i_B$.Substituting this into Snell's law: $\mu = \frac{\sin i_B}{\sin(90^{\circ} - i_B)} = \frac{\sin i_B}{\cos i_B} = 	an i_B$.Given the refractive index of glass $\mu = 1.5$,we have $	an i_B = 1.5$.Therefore,$i_B = 	an^{-1}(1.5) \approx 57^{\circ}$.

Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other (in $°$)?

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