Question

An object of height 6 cm is placed on the principal axis of a concave mirror of focal length f at a distance of 4f. The length of the image will be

Answer 1

Step 1: Given that:

Height of the object($h_o$) = $6cm$

Focal length of the concave mirror= $-f$

Object distance($u$)= $-4f$

(Note:

1. The distances taken left to the pole of a spherical mirror are taken as negative.
2. And the distances taken right to the pole of a mirror are taken as positive.
3. The focal length of a concave mirror is taken as negative as the focus of the concave mirror lies left to the pole of the mirror.
4. The object distance is always taken as negative as it is placed left to the pole of the mirror.)

Step 2: Formula used:

The mirror formula is given by;

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

The magnification of the mirror is given by:

$m=\frac{h_i}{h_o}=-\frac{v}{u}$

Where,

$h_o=Objectdistance$

$h_i=Imagedistance$

$u=objectdistance$

$v=imagedistance$

$f=focallength$

Step 3: Calculation of image distance:

$\frac{1}{v}+\frac{1}{-4f}=\frac{1}{-f}$

$\frac{1}{v}-\frac{1}{4f}=-\frac{1}{f}$

$\frac{1}{v}=-\frac{1}{f}+\frac{1}{4f}$

$\frac{1}{v}=\frac{-4+1}{4f}$

$\frac{1}{v}=\frac{-3}{4f}$

$v=-\frac{4f}{3}$

Negative sign indicates that the image is formed in front of the mirror.

Step 4: Calculation of size of image:

$\frac{h_i}{h_o}=-\frac{v}{u}$

$\frac{h_i}{6cm}=-\frac{\frac{-4f}{3}}{-4f}$

$h_i=-6\times \frac{4f}{3}\times \frac{1}{4f}$

$h_i=-2cm$

(Negative sign indicates that the image is real and inverted.)

Thus,

The length of the image will be $2cm$