A thin rod of length f/3 lies along the axis | vedclass

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(B) Let the rod be placed between $u = 2f - f/3 = 5f/3$ and $u = 2f$ from the concave mirror. The end of the rod at $u = 2f$ forms an image at $v = 2f

Vedclass · Accepted Answer

$\frac{1}{2}f$

Vedclass · Answer

(B) Let the rod be placed between $u = 2f - f/3 = 5f/3$ and $u = 2f$ from the concave mirror. The end of the rod at $u = 2f$ forms an image at $v = 2f$. For the other end of the rod at $u = 5f/3$,we use the mirror formula $\frac{1}{f_{mirror}} = \frac{1}{v} + \frac{1}{u}$. Using sign convention,$f_{mirror} = -f$ and $u = -5f/3$. $\frac{1}{-f} = \frac{1}{v} + \frac{1}{-5f/3} \Rightarrow \frac{1}{v} = \frac{3}{5f} - \frac{1}{f} = \frac{3-5}{5f} = -\frac{2}{5f}$. Thus,$v = -2.5f$. The length of the image is the distance between the two image positions: $|2.5f - 2f| = 0.5f = f/2$.

$A$ thin rod of length $f/3$ lies along the axis of a concave mirror of focal length $f$. One end of its magnified image touches an end of the rod. The length of the image is

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