When a light ray is incident on a medium at a | vedclass

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(C) According to Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$,which implies $\frac{\sin r}{\sin i} = \frac{\mu_1}{\mu_2}$.From the given graph,the slope

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(C) According to Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$,which implies $\frac{\sin r}{\sin i} = \frac{\mu_1}{\mu_2}$.From the given graph,the slope is $	an 30^{\circ} = \frac{\sin r}{\sin i} = \frac{1}{\sqrt{3}}$.Therefore,$\frac{\mu_1}{\mu_2} = \frac{1}{\sqrt{3}}$,which means $\mu_2 = \sqrt{3} \mu_1$.Since the refractive index $\mu = \frac{c}{v}$,we have $\frac{\mu_1}{\mu_2} = \frac{v_2}{v_1} = \frac{1}{\sqrt{3}}$.This implies $v_1 = \sqrt{3} v_2 \approx 1.73 v_2$. Thus,statement $(B)$ is correct.The critical angle $i_c$ is defined by $\sin i_c = \frac{\mu_R}{\mu_D} = \frac{\mu_1}{\mu_2}$.Substituting the values,$\sin i_c = \frac{1}{\sqrt{3}}$. Thus,statement $(C)$ is correct.Therefore,the correct options are $(B)$ and $(C)$.

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