Question

a thin rod of length f/3 is placed along the principal axis of a concave mirror of focal length f such that its image which is a real and elongated just touches one of its end of the rod what is its magnification

Answer 1

Since the image of B is formed at B′ itself, therefore,

B is situated at the center of curvature, that is, at a distance 2f from the pole.

$\sf {PA=2f-\frac{f}{3}=\frac{5f}{3} }$

For point A

$\sf {u=-\frac{5}{3} }\\\\\\\\sf {v=?}$

$\sf {Using\ \ \frac{1}{u} +\frac{1}{v} =\frac{1}{f} }$

$\sf {=\frac{1}{\frac{-5f}{3} }+ \frac{1}{v}= \frac{1}{-f} }\\\\\\\sf {=\frac{1}{v} =-\frac{1}{f}+\frac{3}{5f} }\\\\\\\sf {=\frac{1}{v}=\frac{-5+3}{5f}=\frac{-2}{5f} }\\\\\\\sf {=-2.5f}$

$\sf {Image\ length =2.5f-2f=0.5f=\frac{f}{2} }\\\\\\\sf {Magnification =\frac{\frac{f}{2} }{\frac{f}{3} }=1.5 }$