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Q.
A light ray falls on a glass surface of refractive index $\sqrt{3}$, at an angle $60^{\circ}$. The angle between the refracted and reflected rays would be:
Method (i)
By Snell's law
$1 \sin 60^{\circ}=\sqrt{3} \sin r $
$\frac{\sqrt{3}}{2}=\sqrt{3} \sin r $
$\sin r=\frac{1}{2} $
$r=30^{\circ}$
Angle between refracted and reflected ray is $90^{\circ}$
Method (ii)
Because angle of incidence is Brewster's angle so that angle between reflected and refracted ray is $90^{\circ}$
$\tan i_{p}=\mu=\sqrt{3}$
$i_{p}=60^{\circ}= i$
