Four rays 1, 2, 3 and 4 are incident normally | vedclass

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(C) The prism is isosceles with $\angle Q = 120^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle P = \angle R = (180^{\circ} -

Vedclass · Accepted Answer

$3$ and $4$ only

Vedclass · Answer

(C) The prism is isosceles with $\angle Q = 120^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle P = \angle R = (180^{\circ} - 120^{\circ}) / 2 = 30^{\circ}$.The rays are incident normally on face $PQ$,so they enter the prism without deviation and strike face $PR$ at an angle of incidence $i = 30^{\circ}$.For a ray to emerge from face $PR$,it must satisfy the condition for refraction,which is $n \sin i If $n \sin i \geq 1$,total internal reflection occurs at face $PR$.Here,$i = 30^{\circ}$,so $\sin i = 0.5$.The condition for total internal reflection is $n 	imes 0.5 \geq 1$,which means $n \geq 2$.For rays $1$ $(n=1.85)$ and $2$ $(n=1.95)$,$n For rays $3$ $(n=2.05)$ and $4$ $(n=2.15)$,$n > 2$,so they undergo total internal reflection at face $PR$ and are directed towards face $QR$.Since the angle of incidence at face $QR$ for these reflected rays is less than the critical angle,rays $3$ and $4$ emerge from face $QR$.

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