A ray of light is incident normally on one fa | vedclass

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(C) $1$. The ray enters the prism normally,so it passes undeviated to point $P$ on the hypotenuse face.$2$. The angle of incidence at point $P$ is $i

Vedclass · Accepted Answer

Total internal reflection at point $P$ ceases if the refractive index of water is increased to $5 / 2\sqrt{3}$ by dissolving some substance.

Vedclass · Answer

(C) $1$. The ray enters the prism normally,so it passes undeviated to point $P$ on the hypotenuse face.$2$. The angle of incidence at point $P$ is $i = 60^{\circ}$.$3$. Applying Snell's law at the hypotenuse face: $\mu_p \sin(60^{\circ}) = \mu_w \sin(r)$,where $r$ is the angle of refraction inside the prism.$4$. $(5/3) 	imes (\sqrt{3}/2) = (4/3) \sin(r) \implies \sin(r) = (5\sqrt{3}/6) 	imes (3/4) = 5\sqrt{3}/8$.$5$. The angle of the ray with the base of the prism is $30^{\circ}$. The angle of incidence at the base is $i' = 90^{\circ} - (r + 30^{\circ}) = 60^{\circ} - r$.$6$. Applying Snell's law at the base: $\mu_p \sin(i') = \mu_w \sin(	heta_2)$.$7$. $(5/3) \sin(60^{\circ} - r) = (4/3) \sin(	heta_2)$.$8$. Using $\sin(60^{\circ}-r) = \sin 60^{\circ} \cos r - \cos 60^{\circ} \sin r = (\sqrt{3}/2) \sqrt{1 - (75/64)} - (1/2)(5\sqrt{3}/8) = (\sqrt{3}/2)(\sqrt{-11}/8) - 5\sqrt{3}/16$. Since $\sin r > 1$,total internal reflection occurs at $P$. Thus,the ray does not exit at the base.$9$. For statement $(C)$,$TIR$ at $P$ ceases when the angle of incidence $i=60^{\circ}$ equals the critical angle $C$. $\sin(60^{\circ}) = \mu_w / \mu_p \implies \sqrt{3}/2 = \mu_w / (5/3) \implies \mu_w = 5\sqrt{3}/6 = 5 / (2\sqrt{3})$. This is correct.

$A$ ray of light is incident normally on one face of a $30^{\circ}-60^{\circ}-90^{\circ}$ prism of refractive index $5/3$ immersed in water of refractive index $4/3$ as shown in the figure.

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