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A ray of monochromatic light is incident on the plane surface of separation between two media \(X\) and \(Y\) with angle of incidence 'i' in medium \(X\) and angle of refraction ' \(r\) ' in medium \(Y\). The given graph shows the relation between sin \(i\) and \(\sin r\). If \(V_x\) and \(V_{\mathrm{Y}}\) are the velocities of the ray in media \(X\) and \(Y\) respectively, then which of the following is true?

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Verified Answer
The correct answer is:
\(\mathrm{V}_{\mathrm{X}}=\sqrt{3} \mathrm{~V}_Y\)
Hint : 
\(\begin{aligned} & \mu_1 \sin i=\mu_2 \sin r \\ & \frac{C}{V_X} \sin i=\frac{C}{V_Y} \sin r \\ & \frac{\sin i}{\sin r}=\frac{V_X}{V_Y}=\sqrt{3} \\ & V_Y=\frac{V_X}{\sqrt{3}} \\ & V_X=\sqrt{3} V_Y\end{aligned}\)
\(y=m x\)
\(\sin r=\left(\tan 30^{\circ}\right) \sin \mathrm{i}\)
\(\frac{\sin i}{\sin r}=\frac{1}{\tan 30^{\circ}}=\frac{\sqrt{3}}{1}\)

\(\begin{aligned} & \mu_1 \sin i=\mu_2 \sin r \\ & \frac{C}{V_X} \sin i=\frac{C}{V_Y} \sin r \\ & \frac{\sin i}{\sin r}=\frac{V_X}{V_Y}=\sqrt{3} \\ & V_Y=\frac{V_X}{\sqrt{3}} \\ & V_X=\sqrt{3} V_Y\end{aligned}\)

\(y=m x\)
\(\sin r=\left(\tan 30^{\circ}\right) \sin \mathrm{i}\)
\(\frac{\sin i}{\sin r}=\frac{1}{\tan 30^{\circ}}=\frac{\sqrt{3}}{1}\)
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