Optics Question 82

Question: A thin rod of length $ f/3 $ lies along the axis of a concave mirror of focal length $ f. $ One end of its magnified image touches an end of the rod. The length of the image is

[MP PET 1995]

Options:

A) $ f $

B) $ \frac{1}{2}f $

C) $ 2f $

D) $ \frac{1}{4}f $

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Answer:

Correct Answer: B

Solution:

If end A of rod acts an object for mirror then it’s image will be A’ and if $ u=2f-\frac{f}{3}=\frac{5f}{3} $ so by using $ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} $

$ \Rightarrow \frac{1}{-f}=\frac{1}{v}+\frac{1}{\frac{-5f}{3}} $

$ \Rightarrow v=-\frac{5}{2}f $

$ \therefore $ Length of image $ =\frac{5}{2}f-2f=\frac{f}{2} $



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