An object 0.2 cm high is placed 15 cm from a concave mirror length 5 cm. Find the position and size of the image.

Solution

We have,

$object distance, u = - 15 cm Focal length, f = - 5 cm Image distance, v = ?$

Using formula,

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

$\Rightarrow v = \frac{uf}{u - f} = \frac{(- 15) (- 5)}{- 15 + 5} = - 7.5 cm$

The image is formed at a distance of 7.5 cm in front of mirror.
Now,
Magnification, $m = \frac{I}{O} = - \frac{v}{u}$

where,
I is the image size and,

O is the object size.

$\frac{I}{O} = - \frac{(- 7.5)}{(- 15)} I = - 0.1 cm$

The negative sign indicates that the image is inverted and is diminished in size.

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Wave Optics

An object 0.2 cm high is placed 15 cm from a concave mirror length 5 cm. Find the position and size of the image.

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