An infinitely long rod lies along the axis of a concave mirror of focal length $f$. The nearer end of the…

An infinitely long rod lies along the axis of a concave mirror of focal length $f$. The nearer end of the rod is at a distance $u,(u>f)$ from the mirror. Its image will have a length
  1. $\frac{u f}{u+f}$
  2. $\frac{u f}{u-f}$
  3. $\frac{f^2}{u+f}$
  4. $\frac{f^2}{u-f}$

Solution

Using mirror formula, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ or, $\frac{-1}{f}=\frac{-1}{u}+\frac{1}{v}[\therefore \mathrm{u} \& \mathrm{f}$ negative for concave mirror $\begin{aligned} & \Rightarrow \frac{1}{v}=\frac{1}{u}-\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{f-u}{u f} \\ & v=\frac{u f}{u-f} \end{aligned}$ Length of image, $\begin{aligned} & \mathrm{L}=|v|-|f|=v=\frac{u f}{u-f}-f \\ & \mathrm{~L}=\frac{u f-f(u-f)}{u-f} \\ & =\frac{u f-u f+f^2}{u-f}=\frac{f^2}{u-f} \end{aligned}$

Asked in: AP EAMCET 2016

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