Question

When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, -$\frac{1}{2}$ . Where should the object be placed to get a magnification of, - $\frac{1}{5}$ ?

Answer 1

Step 1: Given that:

For a concave mirror

Object distance(u) = $-50cm$

magnification(m)= $-\frac{1}{2}$

Magnification($m^{\prime }$)= $-\frac{1}{5}$

Image distance, $v$

Object height, $h_o$

Image height, $h_i$

(Note:

1. All the distances taken to the left of a spherical mirror are taken as negative.
2. Thus, the object distance will be negative.
3. Negative magnification suggests that the image is real and inverted.)

Step 2: Formula used:

for a spherical mirror

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

Mirror formula is given by;

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

Step 3: Calculation of focal length of the concave mirror:

$-\frac{1}{2}=-\frac{v}{-50cm}$

$2v=-50cm$

$v=-25cm$

Using mirror formula;

$\frac{1}{f}=\frac{1}{-25cm}+\frac{1}{-50cm}$

$\frac{1}{f}=-\frac{1}{25}-\frac{1}{50}$

$\frac{1}{f}=\frac{-2-1}{50}$

$\frac{1}{f}=\frac{-3}{50}$

$f=-\frac{50}{3}cm$

Step 4: Calculation of the position of the object when magnification is

$-\frac{1}{5}$:

$-\frac{1}{5}=-\frac{v}{u}$

$5v=u$

$v=\frac{u}{5}$

Using this value in mirror formula, we get

$\frac{1}{-\frac{50}{3}cm}=\frac{1}{\frac{u}{5}}+\frac{1}{u}$

$-\frac{3}{50}=\frac{5}{u}+\frac{1}{u}$

$-\frac{3}{50}=\frac{6}{u}$

$-3u=300$

$u=-100cm$

Thus,

The object must be placed at 100cm in front of the concave mirror to get the magnification of $-\frac{1}{5}$ .