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Physics · Geometrical Optics

JEE Main 2025 — 29 January, Evening Shift — Question 25

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 figure
  1. Option A:

    0.214 R

  2. Option B:

    0.114 R

    Correct
  3. Option C:

    0.411 R

  4. Option D:

    0.124 R

Answer: B

Step-by-step solution

For B μ2v−μ1u=μ2−μ1R\frac{\mu_{2}}{\mathrm{v}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}} 1.5 V+1R2=0.5−R\frac{1.5}{\mathrm{~V}}+\frac{1}{\frac{\mathrm{R}}{2}}=\frac{0.5}{-\mathrm{R}}

1.5 V=−12R−2R\frac{1.5}{\mathrm{~V}}=-\frac{1}{2 \mathrm{R}}-\frac{2}{\mathrm{R}}

1.5 V=−52R⇒VB=−0.6R\frac{1.5}{\mathrm{~V}}=\frac{-5}{2 \mathrm{R}} \Rightarrow \mathrm{V}_{\mathrm{B}}=-0.6 \mathrm{R} For A 1.5 V+23R=0.5−R\frac{1.5}{\mathrm{~V}}+\frac{2}{3 \mathrm{R}}=\frac{0.5}{-\mathrm{R}}

1.5 V=−12R−23R\frac{1.5}{\mathrm{~V}}=-\frac{1}{2 \mathrm{R}}-\frac{2}{3 \mathrm{R}}

1.5V=−76R\frac{1.5}{V}=-\frac{7}{6 R} VA=−97R\mathrm{V}_{\mathrm{A}}=-\frac{9}{7} \mathrm{R}

Distance between images =2R−(0.6R+97R)=0.114R=2 R-\left(0.6 R+\frac{9}{7} R\right)=0.114 R option (2)

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Refraction in a Medium with Variable Refractive Index

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