Question 15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and the nature of the image.

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Solution

Object distance, $u = - 27 c m$
Focal length, $f = - 18 c m$
Height of the object, $h_{0} = 7.0 c m$
According to mirror formula,
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$\frac{1}{v} = - \frac{1}{18} + \frac{1}{27} = - \frac{1}{54}$
Image distance, $v = - 54 c m$
Screen must be placed at a distance of 54 cm from the mirror in front of it.
$\frac{h_{i}}{h_{0}} = - \frac{v}{u}$
$\frac{h_{i}}{7} = \frac{54}{- 27} = - 2$
Height of the image, $h_{i} = - 2 \times 7 = - 14 c m$
Thus, the image is of 14 cm length and is inverted image.

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Question 15 An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and the nature of the image.

Question 15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and the nature of the image.