Question

If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of image.



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Answer 1

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The position of the image is inverted and the height of the image is 5

Given:

Object height = 10 cm  

Distance of concave mirror = 36 cm

Focal length = 12 cm

To Find:

The position “nature of image” or the “height of the image” =?  

Solution:

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

$\frac{1}{v}=\frac{1}{-12}-\frac{1}{-36}$

$\frac{1}{v}=\frac{-(3)+(1)}{36}$

$\frac{1}{v}=-\frac{2}{36}$

$\frac{1}{v}=-\frac{1}{18}$

$\text{Magnification}=\frac{\text {Image height}}{\text {Object height}}$

$\Rightarrow \text{Image height}=\frac{\text { Magnification }}{\text { object height }}$

$\Rightarrow \text{Image height} =\frac{10 \times-(-18)}{-36}$

$=-\frac{10}{2}$

$\therefore \text { Image height }=-5$

The object will be inverted.

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