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(C) For total internal reflection to occur, light must travel from a denser medium to a rarer medium. Here, the speed of light in medium $A$ $(v_A = 2

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$	an^{-1}(\frac{8}{\sqrt{17}})$

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(C) For total internal reflection to occur, light must travel from a denser medium to a rarer medium. Here, the speed of light in medium $A$ $(v_A = 2.4 	imes 10^8 \ m/s)$ is less than in medium $B$ $(v_B = 2.7 	imes 10^8 \ m/s)$, so medium $A$ is denser.At the critical angle $c$, the angle of refraction is $90^\circ$. According to Snell's Law:$\mu_A \sin c = \mu_B \sin 90^\circ$Since $\mu = \frac{c_{light}}{v}$, we have $\frac{c_{light}}{v_A} \sin c = \frac{c_{light}}{v_B} 	imes 1$$\sin c = \frac{v_A}{v_B} = \frac{2.4 	imes 10^8}{2.7 	imes 10^8} = \frac{24}{27} = \frac{8}{9}$Now, using the trigonometric identity $	an c = \frac{\sin c}{\sqrt{1 - \sin^2 c}}$:$	an c = \frac{8/9}{\sqrt{1 - (8/9)^2}} = \frac{8/9}{\sqrt{1 - 64/81}} = \frac{8/9}{\sqrt{17/81}} = \frac{8}{\sqrt{17}}$Therefore, $c = 	an^{-1}(\frac{8}{\sqrt{17}})$.

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