Physics · Geometrical Optics

JEE Main 2025 — 7 April, Evening Shift — Question 62

A mirror is used to produce an image with magnification of 14\frac{1}{4} If the distance between object and its image is 40 cm , then the focal length of the mirror is \qquad .

  1. Option A:

    10.7 cm

    Correct
  2. Option B:

    15 cm

  3. Option C:

    12.7 cm

  4. Option D:

    10 cm

Answer: A

Step-by-step solution

here ∣vu∣=14\left|\frac{v}{u}\right|=\frac{1}{4}

u=4vu=4 v

and u+v=40u+v=40

so u=32,v=8u=32, v=8

1f=1v+1u\frac{1}{f}=\frac{1}{v}+\frac{1}{u} F=10.66 cm≈10.7 cmF=10.66 \mathrm{~cm} \approx 10.7 \mathrm{~cm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Geometrical Optics
Topic
Reflection of Light at Curved Surfaces and Spherical Mirrors