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 \[=\,\frac{5}{2}f-2f=\frac{f}{2}\]You need to login to perform this action.
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