Two concave refracting surfaces of equal radii of curvature R and refractive indices $({\mu }_2=1.5)$ face each other in air $({\mu }_1=1.0)$ and a point object O is placed as shown in figure. What is the separation between the images of O formed by each refracting surface?

For image formed by the surface on the right side

$$\begin{gathered} {\text{μ}}_1=1,{\text{μ}}_2=1.5,\text{u}=\frac{-\text{R}}{2},{\text{R}}_1=-\text{R, }\frac{{\text{μ}}_2}{\text{v}}-\frac{{\text{μ}}_1}{\text{u}}=\frac{{\text{μ}}_2-{\text{μ}}_1}{\text{R}}, \frac{1.5}{\text{v}}+\frac{2}{\text{R}}=\frac{1.5-1}{-\text{R}} \\ \Rightarrow \text{v}=\frac{-3\text{R}}{5} \end{gathered}$$

The image is at a distance $(\text{R}-\frac{3\text{R}}{5})=0.4\text{R}$ from the centre P towards the righthand side. For the surface on the left hand side $\text{u}=\frac{-3\text{R}}{2},{\text{R}}_2=-\text{R},{\text{μ}}_1=1,{\text{μ}}_2=1.5$ So, $\frac{1.5}{\text{v}}+\frac{2}{3\text{R}}=\frac{1.5-1}{-\text{R}}$ or $\text{v}=\frac{-9\text{R}}{7}$

The image is at a distance of $(\frac{9\text{R}}{7}-\text{R})=\frac{2\text{R}}{7}$ from the centre P towards the right hand side.

∴ The distance between the two images is $=0.4\text{R}-\frac{2\text{R}}{7}=0.114\text{R}$

Therefore option (c) is correct.