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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2025 · 29 January · Evening Shift · Physics · Ray Optics

Two concave refracting surfaces of equal radii of curvature and...

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in
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 OO, formed by each refracting surface is :

Options

  1. A0.214 R
  2. B0.411 R
  3. C0.124 R
  4. D0.114 R

Correct answer

D. 0.114 R

Step-by-step solution

For B μ2 V−μ1u=μ2−μ1R1.5 V+1R2=0.5−R1.5 V=−12R−2R1.5 V=−52R⇒VB=−0.6R\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 1.5V+23R=0.5−R1.5V=−12R−23R1.5V=−76RVA=−97R\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 =2R−(0.6R+97R)=0.114R=2 \mathrm{R}-\left(0.6 \mathrm{R}+\frac{9}{7} \mathrm{R}\right)=0.114 \mathrm{R}
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