Question

a pole 4m long is laid along the principal axis of a convex mirror of focal length 1m.the end of the pole nearer the mirror is 2m from it.find the length of the image of the pole

Answer 1

Distance of the near end of the pole from mirror is given by

$u = -2m$

focal length of the mirror will be

$f = 1m$

now by mirror formula we can say

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

now plug in all values

$\frac{1}{v} + \frac{1}{-2} = \frac{1}{1}$

$v = \frac{2}{3} m$

now the distance of other end of the rod will be given as

$u =- (2 + 4) = -6 m$

now again by mirror formula

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

now plug in all values

$\frac{1}{v} + \frac{1}{-6} = \frac{1}{1}$

$v = \frac{6}{7} m$

now the length of image will be given by

$L_{image} = v - v'$

$L_{image} = \frac{6}{7} - \frac{2}{3}$

$L_{image} = 0.19 m = 19 cm$

so length of the rod is 19 cm long.