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# An erect image and erect image 3 times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. Calculate the position of the object.

Last updated date: 20th Jun 2024
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Hint To solve this we must know the formula of the magnification of mirror and the mirror formula of the concave mirror. Where we will get the relation between the object distance and the image distance from the mirror which we can further use in the mirror formula to find position of object.

Complete step by step solution
Given
Image is three times the size of object.So,
$m = 3$
Where m is the magnification of a mirror.
Magnification: It is the ratio of image distance to the object distance. As the object is always above the principal axis, the height of the object is always positive.
$R = 36cm$
Where R is radius of curvature
We can find the focal length as focal length is the half of radius of curvature
So
$f = \dfrac{R}{2} = - 18cm$
Focal length is negative because it is concave mirror
$f$ is the focal length of the mirror
Also
$m = - \dfrac{v}{u}$
Where m is the magnification of the mirror,
v is the distance of the image from the mirror
u is the distance of the object from the mirror
As $m = 3$ (given)
Therefore,
$v = - 3u$
Now as we know the mirror formula for concave mirror is
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Using the relation $v = - 3u$ and putting value of focal length we get,
$\Rightarrow$ $\dfrac{1}{{ - 3u}} + \dfrac{1}{u} = \dfrac{1}{{ - 18}}$
$\Rightarrow$ $\dfrac{{ - 1 + 3}}{{3u}} + \dfrac{1}{u} = \dfrac{1}{{ - 18}}$
$\Rightarrow$ $\dfrac{2}{{3u}} = \dfrac{{ - 1}}{{18}}$
$\Rightarrow$ $3u = - 36$
$\therefore$ $u = - 12cm$

Hence the position of the object is $12cm$ in front of the mirror

Note Remember while solving the problem always consider the correct sign conventions. Sign conventions are different for the different mirrors. In the case of the lens, the mirror formula can not be used, in this case lens formula is only used.