question_answer

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]

A)            \[f\]                                         

B)            \[\frac{1}{2}f\]

C)            \[2f\]                                       

D)            \[\frac{1}{4}f\]

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 \[=\,\frac{5}{2}f-2f=\frac{f}{2}\]