KCET · Physics · Ray Optics
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?
- A \( 15 \mathrm{~cm} \)
- B \( 6.6 \mathrm{~cm} \)
- C \( 10 \mathrm{~cm} \)
- D \( 7.5 \mathrm{~cm} \)
Answer & Solution
Correct Answer
(A) \( 15 \mathrm{~cm} \)
Step-by-step Solution
Detailed explanation
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 \( =15 \mathrm{~cm} \)
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