When light is incident on a medium at an angl | vedclass

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(B) From Snell's law,$\frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = {}_1\mu_2$.From the given graph,the slope is $\frac{\sin r}{\sin

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$b, c$

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(B) From Snell's law,$\frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = {}_1\mu_2$.From the given graph,the slope is $\frac{\sin r}{\sin i} = 	an 30^\circ = \frac{1}{\sqrt{3}}$.Therefore,$\frac{\sin i}{\sin r} = \sqrt{3}$.This implies $\frac{v_1}{v_2} = \sqrt{3}$,so $v_1 = \sqrt{3} v_2 \approx 1.73 v_2$. Thus,statement $(b)$ is correct.The relative refractive index of the first medium with respect to the second is ${}_2\mu_1 = \frac{\sin r}{\sin i} = \frac{1}{\sqrt{3}}$.The critical angle $i_c$ is given by $\sin i_c = \frac{\mu_2}{\mu_1}$ is incorrect; it should be $\sin i_c = \frac{	ext{rarer medium refractive index}}{	ext{denser medium refractive index}}$.Since $\frac{\sin i}{\sin r} = \sqrt{3} > 1$,the first medium is denser than the second. Thus,$\sin i_c = \frac{\mu_2}{\mu_1} = \frac{1}{\sqrt{3}}$. Thus,statement $(c)$ is correct.Therefore,the correct options are $(b)$ and $(c)$.

When light is incident on a medium at an angle $i$ and refracted into a second medium at an angle $r$,the graph of $\sin i$ vs $\sin r$ is as shown. From this,one can conclude that:

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