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.
BETA
