Q141 / 180 · KCET 2014 · Physics
An object is placed at \( 20 \mathrm{~cm} \) in front of a concave mirror produces three times magnified real
image. What is focal length of the concave mirror?
image. What is focal length of the concave mirror?
- Correct answerA\( 15 \mathrm{~cm} \)
- B\( 6.6 \mathrm{~cm} \)
- C\( 10 \mathrm{~cm} \)
- D\( 7.5 \mathrm{~cm} \)
Solution
Given, position of object \( =u=20 \mathrm{~cm} \); image formed \( =3 \) times magnified and real
Using \( m=\frac{-v}{m} \), we get
\( 3=\frac{-v}{20} \Rightarrow v=-60 \mathrm{~cm} \)
Using mirror formula, \( \frac{1}{f}=\frac{1}{v}+\frac{1}{w} \), we get
\( \frac{1}{f}=\frac{1}{-60}+\frac{1}{-20}=\frac{-1-3}{60}=\frac{-4}{60}=\frac{-1}{15} \)
Therefore, focal length of concave mirror \( =15 \mathrm{~cm} \)
Using \( m=\frac{-v}{m} \), we get
\( 3=\frac{-v}{20} \Rightarrow v=-60 \mathrm{~cm} \)
Using mirror formula, \( \frac{1}{f}=\frac{1}{v}+\frac{1}{w} \), we get
\( \frac{1}{f}=\frac{1}{-60}+\frac{1}{-20}=\frac{-1-3}{60}=\frac{-4}{60}=\frac{-1}{15} \)
Therefore, focal length of concave mirror \( =15 \mathrm{~cm} \)
Hence, the correct answer is option: 'A'.