Consider the ray diagram for the refraction g | vedclass

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(B) For total internal reflection $(TIR)$ to occur at the vertical surface,the angle of incidence $i'$ at the vertical surface must be greater than or

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$\sin ^{-1}\left(\frac{3}{4}\right)$

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(B) For total internal reflection $(TIR)$ to occur at the vertical surface,the angle of incidence $i'$ at the vertical surface must be greater than or equal to the critical angle $C$.Given $\mu_1 = 1.0$ and $\mu_2 = 1.25 = \frac{5}{4}$.The critical angle $C$ is given by $\sin C = \frac{\mu_1}{\mu_2} = \frac{1}{1.25} = \frac{4}{5}$.So,$i' \ge C$,which means $\sin i' \ge \frac{4}{5}$.From the geometry of the ray diagram,the angle of refraction $r$ at the top horizontal surface and the angle of incidence $i'$ at the vertical surface are related by $r + i' = 90^{\circ}$,so $i' = 90^{\circ} - r$.For $TIR$,$i' \ge C \implies 90^{\circ} - r \ge C \implies r \le 90^{\circ} - C$.To find the maximum $	heta$,we need the maximum $r$,which occurs when $r = 90^{\circ} - C$.Using Snell's law at the horizontal surface: $1.0 	imes \sin 	heta = 1.25 	imes \sin r$.$\sin 	heta_{\max} = 1.25 	imes \sin(90^{\circ} - C) = 1.25 	imes \cos C$.Since $\sin C = \frac{4}{5}$,we have $\cos C = \sqrt{1 - (4/5)^2} = \frac{3}{5}$.Therefore,$\sin 	heta_{\max} = \frac{5}{4} 	imes \frac{3}{5} = \frac{3}{4}$.Thus,$	heta_{\max} = \sin^{-1}\left(\frac{3}{4}ight)$.

Consider the ray diagram for the refraction given below. The maximum value of angle $\theta$ for which the light suffers total internal reflection at the vertical surface is:

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