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  • . An infinitely long rod lies along with the axis of a concave mirror of focal length . The near end o

. An infinitely long rod lies along with the axis of a concave mirror of focal length \mathrm{f}. The near end of the rod is at a distance \mathrm{u> f} from the mirror. Its image will have a length:

Option: 1

\mathrm{\frac{f^2}{u-f}}


Option: 2

\mathrm{\frac{u f}{u-f}}


Option: 3

\mathrm{\frac{\mathrm{f}^2}{\mathrm{u}+\mathrm{f}}}


Option: 4

\mathrm{\frac{u f}{u+f}}


Answers (1)

best_answer

From the relation,
\begin{array}{lll} \mathrm{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}} & \text { or } & \mathrm{\frac{1}{v}-\frac{1}{u}=\frac{1}{-f}} \\\\ \mathrm{\frac{1}{v}=\frac{1}{u}-\frac{1}{f}} & \text { or } &\mathrm{ v=\left(\frac{u f}{f-u}\right)} \end{array}
Since, \mathrm{u> f,\ v} is negative
\mathrm{\text { or } \quad|v|=\left(\frac{u f}{u-f}\right)>f}
The end which is at infinity will have its image at focus.
\mathrm{\therefore } Length of image, \mathrm{ L=|v|-f=\frac{f^2}{u-f} }

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chirag

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