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Question

(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is replaced by $-u$ and $f$ by $-f$

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

Vedclass · Answer

(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Using sign convention,$u$ is replaced by $-u$ and $f$ by $-f$.So,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$.Rearranging for $v$,we get $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f-u}{uf}$.Thus,$v = \frac{uf}{f-u}$.The magnitude of the image position is $|v| = \left| \frac{uf}{f-u} \right| = \frac{uf}{u-f}$.The rod extends from $u$ to $\infty$. The image of the nearer end is at $v_1 = \frac{uf}{u-f}$.The image of the farther end (at $\infty$) is at the focus $v_2 = f$.The length of the image is $L = |v_1 - v_2| = \left| \frac{uf}{u-f} - f \right|$.$L = \left| \frac{uf - f(u-f)}{u-f} \right| = \left| \frac{uf - uf + f^2}{u-f} \right| = \frac{f^2}{u-f}$.

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:

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