CUET · PHYSICS · PYQ PAPER 2025
The distance between an object and its virtual image of magnification \( \frac{1}{2} \) as produced by a lens is 20 cm.
The focal length of the lens is
- A \(+40 cm\)
- B \(-20 cm\)
- C \(+20 cm\)
- D \(-40 cm\)
Answer & Solution
Correct Answer
(D) \(-40 cm\)
Step-by-step Solution
Detailed explanation
\(m = \frac{v}{u} = \frac{1}{2}\) \(u = 2v\) For a virtual image from a lens, \(u\) and \(v\) are both negative, and the image is closer to the lens than the object: \(|u| - |v| = 20\text{ cm}\) \(|2v| - |v| = 20 \Rightarrow |v| = 20\text{ cm}\) \(v = -20\text{ cm}\)…
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