A wide slab consisting of two media of refrac | vedclass

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(A) Given: $\sin 	heta > \frac{1}{n_1}$.Applying Snell's law at the $n_1-n_2$ interface: $n_1 \sin 	heta = n_2 \sin 	heta_2$,so $\sin 	heta_2 = \f

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$B, C, D$

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(A) Given: $\sin 	heta > \frac{1}{n_1}$.Applying Snell's law at the $n_1-n_2$ interface: $n_1 \sin 	heta = n_2 \sin 	heta_2$,so $\sin 	heta_2 = \frac{n_1}{n_2} \sin 	heta$.Applying Snell's law at the $n_2$-air interface: $n_2 \sin 	heta_2 = 1 \cdot \sin 	heta_3$,so $\sin 	heta_3 = n_2 \sin 	heta_2 = n_1 \sin 	heta$.Since $\sin 	heta > \frac{1}{n_1}$,we have $n_1 \sin 	heta > 1$,which implies $\sin 	heta_3 > 1$. This is impossible for a real angle $	heta_3$,meaning the ray undergoes total internal reflection at the $n_2$-air interface.$(A)$ If $n_2 = n_1$,$\sin 	heta_3 = n_1 \sin 	heta > 1$. The ray does not enter air. Statement $(A)$ is incorrect.$(B)$ If $n_2 $(C)$ If $n_2 > n_1$,the ray undergoes total internal reflection at the $n_2$-air interface. The ray will then be reflected back into $n_1$ after total internal reflection at the $n_2$-air interface. Statement $(C)$ is correct.$(D)$ If $n_2 = 1$,the ray undergoes total internal reflection at the $n_2$-air interface and is reflected back into $n_1$. Statement $(D)$ is correct.

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