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(C) According to Snell's Law at the interface between medium $2$ and $3$,the ray is incident at the critical angle $C$. Therefore,$\sin C = \frac{\mu_

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$\sin ^{-1}\left(\frac{13}{16}\right)$

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(C) According to Snell's Law at the interface between medium $2$ and $3$,the ray is incident at the critical angle $C$. Therefore,$\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.Since the angle of refraction at the first interface is $r$,and the ray is incident at the critical angle at the second interface,we have $r = C$.Applying Snell's Law at the first interface between medium $1$ and $2$:$\mu_1 \sin 	heta = \mu_2 \sin r$Substituting $\sin r = \sin C = \frac{1.3}{1.8}$:$1.6 	imes \sin 	heta = 1.8 	imes \left(\frac{1.3}{1.8}ight)$$1.6 	imes \sin 	heta = 1.3$$\sin 	heta = \frac{1.3}{1.6} = \frac{13}{16}$$	heta = \sin ^{-1}\left(\frac{13}{16}ight)$

$A$ ray of light refracts from medium $1$ into a thin layer of medium $2$,crosses the layer,and is incident at the critical angle on the interface between medium $2$ and $3$ as shown in the figure. If the angle of incidence of the ray is $\theta$,the value of $\theta$ is

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