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 :

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0.411R

0.124R

0.214R

0.114R

answer is B.

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Detailed Solution

For B
$\begin{aligned} \begin{array}{ll} \frac{μ_{2}}{V} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} \\ \frac{1.5}{V} + \frac{1}{\frac{R}{2}} = \frac{0.5}{- R} \\ \frac{1.5}{V} = - \frac{1}{2 R} - \frac{2}{R} \\ \frac{1.5}{V} = \frac{- 5}{2 R} \Rightarrow V_{B} = - 0.6 R \end{array} \\ For A \\ \begin{aligned} \frac{1.5}{V} + \frac{2}{3 R} = - \frac{0.5}{R} \\ \frac{1.5}{V} = - \frac{1}{2 R} - \frac{2}{3 R} \\ \frac{1.5}{V} = - \frac{7}{6 R} \\ V_{A} = - \frac{9}{7} R \end{aligned} \end{aligned}$
Distance between images
$= 2 R - (0.6 R + \frac{9}{7} R) = 0.114 R$
option (2)

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