Q139 / 180 · KCET 2018 · Physics
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
\( \mathrm{cm} \). The distance of the image formed is
- A\(+20 \mathrm{~cm} \)
- B\( +10 \mathrm{~cm} \)
- Correct answerC\( -20 \mathrm{~cm} \)
- D\( -10 \mathrm{~cm} \)
Solution
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, \( \frac{1}{-10}=\frac{1}{v}+\frac{1}{-20} \)
\[ \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{10}=\frac{-1}{20} \]
Hence, the correct answer is option: 'C'.