Find the size, nature and position of image formed when an object of size $1 c m$ is placed at a distance of $15 c m$ from a concave mirror of focal length $10 c m$ ?

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Solution

$u = - 15$ cm
$f = - 10$ cm
$h = 1$ cm
From mirror formula
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 15}$
$v = - 30$ cm
The image is formed at a distance $30$ cm on the left side of the mirror from the pole.
$m = \frac{h^{'}}{h} = \frac{- v}{u}$
$h^{'} = - 1 \times \frac{- 30}{- 15}$
$h = - 2$ cm
The image size is $2$ cm and it is real and inverted.

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Find the size, nature and position of image formed when an object of size 1cm is placed at a distance of 15 cm from a concave mirror of focal length 10cm?

u=−15cmf=−10cmh=1cmFrom mirror formula1f=1u+1v1v=1f−1u1v=1−10−1−15v=−30cmThe image is formed at a distance 30cm on the left side of the mirror from the pole.m=h′h=−vuh′=−1×−30−15h=−2cmThe image size is 2cm and it is real and inverted.

Find the size, nature and position of image formed when an object of size $1 c m$ is placed at a distance of $15 c m$ from a concave mirror of focal length $10 c m$ ?