Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of $O$, formed by each refracting surface is :

Question

Aicharya · Accepted Answer

For B
$\begin{aligned}
& \frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}+\frac{1}{\frac{R}{2}}=\frac{0.5}{-\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}=-\frac{1}{2 \mathrm{R}}-\frac{2}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}=\frac{-5}{2 \mathrm{R}} \Rightarrow V_B=-0.6 \mathrm{R}
\end{aligned}$
For A
$\begin{aligned}
& \frac{1.5}{V}+\frac{2}{3 R}=\frac{0.5}{-R} \\ & \frac{1.5}{V}=-\frac{1}{2 R}-\frac{2}{3 R} \\ & \frac{1.5}{V}=-\frac{7}{6 R} \\ & V_A=-\frac{9}{7} R
\end{aligned}$
Distance between images
$=2 \mathrm{R}-\left(0.6 \mathrm{R}+\frac{9}{7} \mathrm{R}\right)=0.114 \mathrm{R}$

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air…

Solution