Question

A rod of length $10$ cm lies along the principal axis of a concave mirror of focal length $10$ cm in such a way that its end closer to the pole is $20$ cm away from the mirror. The length of the image is

• A. $10$ cm
• B. $15$cm
• C. $2.5$cm
• D. $5$cm

Answer 1

The correct option is D $5$cm

REF.Image

focal length = -10 cm

$\therefore$ Radius of curvature = 2f =-20 cm

Length of rod AB = 10 cm

Distance of the end A= 20 cm = R

Since the hear end 'A' of rod is at the center of curvature

it's image $A^1$ will also be at 'c'.

Distance of the end B of the rod = 20+10 =30 cm.

$\therefore$ u= -30 cm (from the Cartesian sign convention)

f = -10 cm

v = ?

from mirror formula

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

$\Rightarrow -\frac{1}{10}=-\frac{1}{30}+\frac{1}{v}\Rightarrow \frac{1}{v}=-\frac{20}{300}$

$\therefore v=-15cm$

Length of image = $A^1{B}^1=20-15=5cm$

Rod AB placed at a given distance from concave mirror.