When light is incident on a medium at angle i | vedclass

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(D) From 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

Vedclass · Accepted Answer

Both $(b)$ and $(c)$.

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(D) From 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 $\mu = \frac{c}{v}$,we have $\frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = \sqrt{3} \approx 1.73$.Thus,$v_1 = 1.73 v_2$,which confirms statement $(b)$.The critical angle $i_c$ is given by $\sin i_c = \frac{\mu_{rarer}}{\mu_{denser}} = \frac{\mu_1}{\mu_2} = \frac{1}{\sqrt{3}}$,which confirms statement $(c)$.Therefore,both $(b)$ and $(c)$ are correct.

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

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