Question 142 of 295
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 , formed by each refracting surface is :
Show answer and solution
Answer: (D) 0.114 R
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
$\begin{aligned}
& \frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
& \frac{1.5}{\mathrm{~V}}+\frac{1}{\frac{R}{2}}=\frac{0.5}{-\mathrm{R}} \\
& \frac{1.5}{\mathrm{~V}}=-\frac{1}{2 \mathrm{R}}-\frac{2}{\mathrm{R}} \\
& \frac{1.5}{\mathrm{~V}}=\frac{-5}{2 \mathrm{R}} \Rightarrow V_B=-0.6 \mathrm{R}
\end{aligned}$
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
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