An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

Solution

We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm

We know, using the mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\Rightarrow$ $v = \frac{uf}{u - f} = \frac{(- 30) (+ 20)}{- 30 - 20} = + 12 cm$

The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror.

Now, size if the object, O = 0.5 cm
sixe of the image = I
Magnification, $m = \frac{I}{O} = - \frac{v}{u} = - \frac{12}{- 30}$
$\Rightarrow$ $I = \frac{2}{5} \times O = \frac{2}{5} \times 0.5 = + 0.2 cm$

Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size

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Wave Optics

An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

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