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 :
A
$120^\circ$
B
$30^\circ$
C
$60^\circ$
D
$90^\circ$
The answer is D.
Solution
By Snell's law: $1 \times \sin 60^\circ = \sqrt{3}\sin r \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3}\sin r \Rightarrow \sin r = \frac{1}{2}$, so $r = 30^\circ$
Angle of reflection = $60^\circ$, so the angle between the reflected and refracted rays $= 180^\circ - 60^\circ - 30^\circ = 90^\circ$
(Alternatively, $\tan i_p = \mu = \sqrt{3} \Rightarrow i_p = 60^\circ$; the ray is incident at Brewster's angle, so the reflected and refracted rays are perpendicular.)
Conceptreflected and refracted rays at polarising angle
Ray Optics and Optical Instruments in past papers
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