Question

The size of an image of an object by a mirror having a focal length of 20 cm is observed to be reduced to $\frac{1}{3}rd$ of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer 1

Step 1: Given data:

magnification, $m=\frac{1}{3}$

(Note: Positive magnification describes that the image is virtual and erect.)
Focal length, f = 20 cm

Step 2: Finding the type of mirror:

1. A convex mirror forms a virtual, erect and diminished image.
2. As the magnification in the given case suggests that the image is diminshed therefore the mirror is a convex mirror.

Step 3: Finding the object distance

We know,

Magnification is given as
$m=-\left(\frac{v}{u}\right)$

Putting all the values
$\frac{1}{3}=-\left(\frac{v}{u}\right)$

$-3v=u$

$v=-\frac{u}{3}$

sign convention

Mirror formula is given by,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Or,

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

Putting all the values, we get

$\frac{1}{\frac{-u}{3}}+\frac{1}{u}=\frac{1}{20cm}$

$\frac{-3}{u}+\frac{1}{u}=\frac{1}{20}$

$\frac{-2}{u}=\frac{1}{20}$

$u=-40cm$

Thus,

The object must be placed at $40cm$ in front of the convex mirror and the image is virtual and erect.