The figure shows a ray of light as it travels | vedclass

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(D) According to Snell's law,the refractive index of medium $B$ relative to medium $A$ $(n_{BA})$ is given by the ratio of the sine of the angle of in

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$\sqrt{3} / \sqrt{2}$

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(D) According to Snell's law,the refractive index of medium $B$ relative to medium $A$ $(n_{BA})$ is given by the ratio of the sine of the angle of incidence $(i)$ to the sine of the angle of refraction $(r)$: $n_{BA} = \frac{\sin i}{\sin r}$ From the figure,the light ray travels from medium $A$ to medium $B$. The angle of incidence $i$ is the angle between the incident ray and the normal in medium $A$. Given the angle with the normal is $60^\circ$,so $i = 60^\circ$. The angle of refraction $r$ is the angle between the refracted ray and the normal in medium $B$. Given the angle with the normal is $45^\circ$,so $r = 45^\circ$. Substituting these values into Snell's law: $n_{BA} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} 	imes \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$ Therefore,the refractive index of medium $B$ relative to medium $A$ is $\sqrt{3} / \sqrt{2}$.

The figure shows a ray of light as it travels from medium $A$ to medium $B$. The refractive index of medium $B$ relative to medium $A$ is:

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