An electromagnetic wave, whose wave normal makes an angle of \(45^{\circ}\) with the vertical, travelling in air strikes a horizontal liquid surface. While travelling through the liquid it gets deviated through \(15^{\circ}\). What is the speed of the electromagnetic wave in the liquid, if the speed of electromagnetic wave in air is \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\) ? \(\left(\sin 30^{\circ}=0 \cdot 5, \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)\)

The angle of incidence is \(\mathrm{i}=45^{\circ}\)
The angle of deviation is \(\delta=15^{\circ}\)
The angle of refraction is
\(\begin{aligned}
& \delta=i-r \\
& r=i-\delta \\
& r=45^{\circ}-15^{\circ} \\
& r=30^{\circ}
\end{aligned}\)
The relation between the speed of the wave in the medium, the angle of incidence and the angle of refraction is given by the formula:
\(\begin{aligned}
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\sin \mathrm{r}}{\sin \mathrm{i}} \\
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\sin 30^{\circ}}{\sin 45^{\circ}} \\
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{1}{\sqrt{2}}
\end{aligned}\)
\(\begin{aligned} & \mathrm{v}_2=\frac{\mathrm{v}_1}{\sqrt{2}} \\ & \mathrm{v}_2=\frac{3 \times 10^8}{\sqrt{2}} \\ & \mathrm{v}_2=2.1 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}\)