A ray of light incident on a glass sphere (re | vedclass

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(D) Let the angle of incidence be $i$ and the angle of refraction be $r$. According to Snell's law at the first surface:$\frac{\sin i}{\sin r} = \mu =

Vedclass · Accepted Answer

$60$

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(D) Let the angle of incidence be $i$ and the angle of refraction be $r$. According to Snell's law at the first surface:$\frac{\sin i}{\sin r} = \mu = \sqrt{3} \implies \sin i = \sqrt{3} \sin r \quad \dots(1)$In the triangle formed by the center of the sphere and the two points of refraction/reflection,the triangle is isosceles because two sides are radii of the sphere. The angle of deviation at the first surface is $(i - r)$. The angle of reflection at the second surface is also $r$. Since the incident ray is parallel to the emergent ray,the total deviation $\delta$ must be $180^{\circ}$.The deviation at the first refraction is $(i - r)$.The deviation at the internal reflection is $(180^{\circ} - 2r)$.The deviation at the second refraction is $(i - r)$.Total deviation $\delta = (i - r) + (180^{\circ} - 2r) + (i - r) = 180^{\circ}$.$2i - 4r + 180^{\circ} = 180^{\circ} \implies 2i = 4r \implies i = 2r$.Substitute $i = 2r$ into equation $(1)$:$\sin(2r) = \sqrt{3} \sin r$$2 \sin r \cos r = \sqrt{3} \sin r$Since $\sin r 
eq 0$,we have $\cos r = \frac{\sqrt{3}}{2}$.Thus,$r = 30^{\circ}$.Since $i = 2r$,we get $i = 2 	imes 30^{\circ} = 60^{\circ}$.Therefore,the angle of incidence is $60^{\circ}$.

$A$ ray of light incident on a glass sphere (refractive index $\sqrt{3}$) suffers total internal reflection before emerging out exactly parallel to the incident ray. The angle of incidence was (in $^{\circ}$)

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