A quarter cylinder of radius R and refractive index 1.5 is placed...

Refraction at plane surface (1), $n_1=1, n_2=1.5=\frac{3}{2}, u=-m R, v=?$ Using formula, $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$$\begin{aligned} & \because \quad R=\infty \text{\text{ for } plane surface. } \\ & \because \frac{3 / 2}{v}-\frac{1}{(-m R)}=0 \\ & \therefore \quad v=-\frac{3}{2} m R\end{aligned}$$$Refraction at spherical surface (2)$$\begin{aligned} & u=-\left(\frac{3}{2} m R+R\right) \text{\text{ and } } R=-R, v=\infty, \\ & n_1=\frac{3}{2}, n_2=1\end{aligned}$$\frac{1}{\infty}-\frac{3 / 2}{\left[-\left(\frac{3}{2} m R+R\right)\right]}=\frac{1-3 / 2}{-R}$$\begin{aligned} & \Rightarrow \quad \frac{3 / 2}{\frac{R(3 m+2)}{2}}=\frac{1 / 2}{R} \\ & \Rightarrow \quad \frac{6}{3 m+2}=1 \Rightarrow 3 m+2=6 \\ & m=\frac{4}{3} \\ & \end{aligned}$$$