KCET · Physics · Ray Optics
An object is placed at a distance of \( 20 \mathrm{~cm} \) from the pole of a concave mirror of focal length \( 10 \)
\( \mathrm{cm} \). The distance of the image formed is
- A \(+20 \mathrm{~cm} \)
- B \( +10 \mathrm{~cm} \)
- C \( -20 \mathrm{~cm} \)
- D \( -10 \mathrm{~cm} \)
Answer & Solution
Correct Answer
(C) \( -20 \mathrm{~cm} \)
Step-by-step Solution
Detailed explanation
Using mirror formula, we have
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Here, \( u=-20 \mathrm{~cm} ; f=-10 \mathrm{~cm} \).
Therefore, \( \frac{1}{-10}=\frac{1}{v}+\frac{1}{-20} \)
\[
\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{10}=\frac{-1}{20}
\]
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Here, \( u=-20 \mathrm{~cm} ; f=-10 \mathrm{~cm} \).
Therefore, \( \frac{1}{-10}=\frac{1}{v}+\frac{1}{-20} \)
\[
\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{10}=\frac{-1}{20}
\]
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