A 2 cm high object is placed at a distance of 20 cm from a concave mirror. A real image is formed 40 cm from the mirror. Calculate the focal length of the mirror and the size of the image.

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Solution

Given:
$u = - 20 c m$
$v = - 40 c m$

We know Mirror formula
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{f} = \frac{1}{- 20} + \frac{1}{- 40}$
$f = - 13.3 c m$
Focal length of the mirror is 13.3 cm.

By using magnification formula:
$m = \frac{h_{i}}{h_{o}} = \frac{- v}{u}$
$\frac{h_{i}}{2} = \frac{- (- 40)}{- 20}$
$h_{i} = - 4 c m$
So, an inverted image of 4 cm is formed.

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A 2 cm high object is placed at a distance of 20 cm from a concave mirror. A real image is formed 40 cm from the mirror. Calculate the focal length of the mirror and the size of the image.

Given: u=−20 cm v=−40 cm We know Mirror formula 1f=1v+1u 1f=1−20+1−40 f=−13.3 cm Focal length of the mirror is 13.3 cm. By using magnification formula: m=hiho=−vu hi2=−(−40)−20 hi=−4 cm So, an inverted image of 4 cm is formed.