An infinitely long rod lies along the axis of | vedclass

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(A) 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 $f$ is repl

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

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(A) 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 $f$ is replaced by $-f$.Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$,which gives $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.So,$v = \frac{uf}{f - u} = -\frac{uf}{u - f}$.The far end of the rod is at infinity $(u = \infty)$,so its image is formed at the focus $(v = f)$.The near end of the rod is at distance $u$,so its image is at distance $v = \frac{uf}{u - f}$ from the pole.The length of the image is the difference between the positions of the two images: $L = |v - f|$.$L = |\frac{uf}{u - f} - f| = |\frac{uf - f(u - f)}{u - f}| = |\frac{uf - uf + f^2}{u - f}| = \frac{f^2}{u - f}$.

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

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