Optics Question 600

Question: An object is placed at a distance of $ f/2 $ from a convex lens. The image will be

[CPMT 1974, 89]

Options:

A) At one of the foci, virtual and double its size

B) At 3f / 2, real and inverted

C) At 2f, virtual and erect

D) None of these

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Answer:

Correct Answer: A

Solution:

$ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} $ (Given $ u=\frac{-f}{2} $ )

$ \Rightarrow $ $ \frac{1}{f}=\frac{1}{v}+( \frac{1}{f/2} )\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f} $

$ \Rightarrow $ $ \frac{1}{v}=\frac{-1}{f} $ and $ m=\frac{v}{u}=\frac{f}{f/2}=2 $ So virtual at the focus and of double size.



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