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) $\frac{1}{f} = \frac{1}{2}$
D) $ \frac{1}{4}f $
Show Answer
Answer:
Correct Answer: B
Solution:
If end A of rod acts as an object for mirror then its 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} $
BETA
