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Physics · Geometrical Optics

JEE Main 2025 — 4 April, Morning Shift — Question 45

When an object is placed 40 cm away from a spherical mirror an image of magnification 12\frac{1}{2} is produced. To obtain an image with magnification of 13\frac{1}{3}, the object is to be moved

  1. Option A:

    20 cm towards the mirror

  2. Option B:

    20 cm away from the mirror

  3. Option C:

    80 cm away from the mirror

  4. Option D:

    40 cm away from the mirror

    Correct

Answer: D

Step-by-step solution

m=ff−um=\frac{f}{f-u}

12=ff−(−40)\frac{1}{2}=\frac{f}{f-(-40)} f=40f=40 4040−(+μ)=13\frac{40}{40-(+\mu)}=\frac{1}{3} u=−80u=-80

Move 40 cm away from mirror.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Reflection of Light at Curved Surfaces and Spherical Mirrors

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