The figure shows a ray incident at an angle i | vedclass

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(D) From Snell's Law,$\mu_2 \sin i = \mu_1 \sin r$,which gives $\frac{\mu_1}{\mu_2} = k = \frac{\sin i}{\sin r}$.$1$. When $k = 1$,$\mu_1 = \mu_2$,so

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All of the above.

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(D) From Snell's Law,$\mu_2 \sin i = \mu_1 \sin r$,which gives $\frac{\mu_1}{\mu_2} = k = \frac{\sin i}{\sin r}$.$1$. When $k = 1$,$\mu_1 = \mu_2$,so $\sin i = \sin r$,which implies $i = r$. Thus,$|r - i| = 0$. From the graph,this occurs at $k_2$,so $k_2 = 1$.$2$. As $k ightarrow \infty$,$\frac{\mu_1}{\mu_2} ightarrow \infty$,which implies $\sin r = \frac{\sin i}{k} ightarrow 0$,so $r ightarrow 0$. Then $|r - i| = |0 - i| = i = \pi / 3$. From the graph,this value is $	heta_2$,so $	heta_2 = \pi / 3$.$3$. Total Internal Reflection $(TIR)$ occurs when $k Since all statements are correct,the answer is $(D)$.

The figure shows a ray incident at an angle $i = \pi / 3$. The plot shows the variation of $|r - i|$ versus $k = \frac{\mu_1}{\mu_2}$,where $r$ is the angle of refraction. Based on the graph,which of the following statements is correct?

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