A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of m R from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in figure.

Applying $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$, firstly at the plane surface and then at the curved surface. For the plane surface, we get

$$\begin{gathered} \frac{1.5}{A{I}_1}-\frac{1}{(-mR)}=\frac{1.5-1}{\infty }=0 \{\because R\to \infty \} \\ \Rightarrow A{I}_1=-(1.5mR) \end{gathered}$$

For the curved surface, since the final image is formed at infinity, so we get

$$\begin{gathered} \frac{1}{\infty }-\frac{1.5}{-(1.5mR+R)}=\frac{1-1.5}{-R} \\ \Rightarrow \frac{1.5}{(1.5m+1)R}=\frac{0.5}{R} \\ \Rightarrow 3=1.5m+1 \\ \Rightarrow \frac{3}{2}m=2 \\ \Rightarrow m=\frac{4}{3} \end{gathered}$$