JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2024
- Hard
- MCQ
If the distance between object and its two times magnified virtual image produced by a curved mirror is \(15 \mathrm{~cm}\), the focal length of the mirror must be _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(-10 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{m}=2=\frac{-\mathrm{v}}{\mathrm{u}} \)
\( 2=\frac{-(15-\mathrm{u})}{-\mathrm{u}} \)
\( 2 \mathrm{u}=15-\mathrm{u} \)
\( 3 \mathrm{u}=15 \Rightarrow \mathrm{u}=5 \mathrm{~cm} \)
\( \mathrm{v}=15-\mathrm{u}=15-5=10 \mathrm{~cm} \)
\( \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \)
\( =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \)
\( \mathrm{f}=-10 \mathrm{~cm}\)

\( 2=\frac{-(15-\mathrm{u})}{-\mathrm{u}} \)
\( 2 \mathrm{u}=15-\mathrm{u} \)
\( 3 \mathrm{u}=15 \Rightarrow \mathrm{u}=5 \mathrm{~cm} \)
\( \mathrm{v}=15-\mathrm{u}=15-5=10 \mathrm{~cm} \)
\( \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \)
\( =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \)
\( \mathrm{f}=-10 \mathrm{~cm}\)

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