AP EAMCET · PHYSICS · Ray Optics
A thin converging lens of focal length \(25 \mathrm{~cm}\) forms a sharp image of an object on a screen placed at a distance of \(75 \mathrm{~cm}\) from the lens. Later the screen is moved closer to the lens by a distance \(25 \mathrm{~cm}\). The distance through which the object is to be shifted so that its image on the screen is sharp again is
- A \(50 \mathrm{~cm}\) towards the lens
- B \(50 \mathrm{~cm}\) away from the lens
- C \(12.5 \mathrm{~cm}\) towards the lens
- D \(12.5 \mathrm{~cm}\) away from the lens
Answer & Solution
Correct Answer
(D) \(12.5 \mathrm{~cm}\) away from the lens
Step-by-step Solution
Detailed explanation
According to the question, Given, focal length, \(f=25 \mathrm{~cm}\) and distance between image of an object and screen, \(v=75 \mathrm{~cm}\) Now, By lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) [ \(\because\) Because screen is moved closer to the lens.]…
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