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Physics · Geometrical Optics

JEE Main 2025 — 23 January, Morning Shift — Question 53

A spherical surface of radius of curvature R, separates air from glass (refractive index =1.5 ). The centre of curvature is in the glass medium. A point object ' O ' placed in air on the optic axis of the surface, so that its real image is formed at ' I ' inside glass. The line OI intersects the spherical surface at P and PO=PI. The distance PO equals to-

  1. Option A:

    5R

    Correct
  2. Option B:

    3R

  3. Option C:

    2R

  4. Option D:

    1.5R

Answer: A

Step-by-step solution

PO=u=−x\mathrm{PO}=\mathrm{u}=-\mathrm{x} PI=v=x\mathrm{PI}=\mathrm{v}=\mathrm{x} PO=PI\mathrm{PO}=\mathrm{PI} μ2v−μ1u=μ2−μ1R\frac{\mu_{2}}{\mathrm{v}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}} 1.5x+1x=12R\frac{1.5}{x}+\frac{1}{x}=\frac{1}{2 R} 52x=12R\frac{5}{2 \mathrm{x}}=\frac{1}{2 R} X=5RX=5 R

Solution figure

Answer key and solution verified before publishing.

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

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