A ray of light refracts from medium 1 into a | vedclass

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(C) Given: Refractive indices $\mu_1 = 1.6$, $\mu_2 = 1.8$, $\mu_3 = 1.3$.At the interface between medium $2$ and $3$, the ray is incident at the crit

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

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(C) Given: Refractive indices $\mu_1 = 1.6$, $\mu_2 = 1.8$, $\mu_3 = 1.3$.At the interface between medium $2$ and $3$, the ray is incident at the critical angle $C$. Therefore, $\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.Let $r$ be the angle of refraction in medium $2$. Since the ray is incident at the critical angle at the interface of $2$ and $3$, the angle of refraction $r$ at the first interface is equal to the critical angle $C$ (i.e.,$r = C$).Applying Snell's Law at the interface between medium $1$ and $2$:$\mu_1 \sin 	heta = \mu_2 \sin r$Since $r = C$, we have $\sin r = \sin C = \frac{1.3}{1.8}$.Substituting the values:$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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