A transparent block A having refractive index | vedclass

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(C) For total internal reflection to occur at the top surface,the angle of incidence at that surface must be at least the critical angle $	heta_C$. F

Vedclass · Accepted Answer

$\sin ^{-1}(3 / 4)$

Vedclass · Answer

(C) For total internal reflection to occur at the top surface,the angle of incidence at that surface must be at least the critical angle $	heta_C$. From the geometry of the triangle formed inside the block,the angle of refraction $r$ at the first surface and the angle of incidence $	heta_C$ at the top surface satisfy $r + 	heta_C = 90^{\circ}$,so $r = 90^{\circ} - 	heta_C$. Applying Snell's Law at the first surface: $\mu_1 \sin 	heta = \mu_2 \sin r$. Substituting $r = 90^{\circ} - 	heta_C$: $\sin 	heta = \frac{\mu_2}{\mu_1} \sin(90^{\circ} - 	heta_C) = \frac{\mu_2}{\mu_1} \cos 	heta_C$. For total internal reflection,$\sin 	heta_C = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{1}{5/4} = \frac{4}{5}$. Thus,$\cos 	heta_C = \sqrt{1 - \sin^2 	heta_C} = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$. Substituting this into the equation for $\sin 	heta$: $\sin 	heta = \frac{1.25}{1.0} 	imes \frac{3}{5} = \frac{5}{4} 	imes \frac{3}{5} = \frac{3}{4}$. Therefore,$	heta = \sin^{-1}(3/4)$.

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