Start free

Physics · Electrostatics

JEE Main 2024 — 29 January, Shift 2 — Question 37

If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm , the focal length of the mirror must be :

  1. Option A:

    15 cm

  2. Option B:

    -12 cm

  3. Option C:

    -10 cm

    Correct
  4. Option D:

    10/3 cm10 / 3 \mathrm{~cm}

Answer: C

Step-by-step solution

m=2=−vu\mathrm{m}=2=\frac{-\mathrm{v}}{\mathrm{u}}

2=−(15−u)−u2=\frac{-(15-\mathrm{u})}{-\mathrm{u}}

2u=15−u2 \mathrm{u}=15-\mathrm{u}

3u=15⇒u=5 cm3 \mathrm{u}=15 \Rightarrow \mathrm{u}=5 \mathrm{~cm}

v=15−u=15−5=10 cm\mathrm{v}=15-\mathrm{u}=15-5=10 \mathrm{~cm}

1f=1v+1u\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}

=110+1(−5)=1−210=−110=\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10}

f=−10 cmf=-10 \mathrm{~cm}

Solution figure

Answer key and solution verified before publishing.

Practise Electrostatics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Electric Charge and Coulomb's Law

More Electrostatics questions from this paper

If the distance between object and its two times magnified virtual… | JEE Main 2024 PYQ with Solution · DhiX AI