Question

An object 2cm high produces a real image 3cm high, when placed at a distance of 15cm from a concave mirror. Calculate the position of the image.

Answer 1

Magnification is defined as the ratio of the image size to the object size which is equal to the negative of distance of image from the mirror to the distance of object from the mirror. $m = \dfrac{{{h_1}}}{{{h_2}}} = - \dfrac{v}{u}$, where h1 and h is the height of image and object respectively. v and u are the distance of image and object from the mirror.

Complete step by step answer:
Given, the height of the object ${h_2} = 2cm$.
Height of image ${h_1} = 3cm$.
The height of the image is negative as the image formed is real and inverted.
Distance of the object from the concave mirror $u = 15cm$.
Let the distance of image from the mirror be v.
We know that magnification for a concave mirror is the ratio of the height of the image to the height of the object which is equal to the negative of the distance of image from the mirror to the distance of object from the mirror, $m = \dfrac{{{h_1}}}{{{h_2}}} = - \dfrac{v}{u}$.
Substituting the value of height of image height of object and the distance of object from the mirror, we get
$m = \dfrac{3}{2} = - \dfrac{v}{{15}} \\\ \therefore v = - \left( {\dfrac{3}{2} \times 15} \right) = - 22.5cm \\\$
Therefore the value of distance of image from the mirror is 22.5cm and is formed on the same side as the object.

Note: Conventions for the distance measures should be kept in consideration.
Distance is always measured from the pole of the mirror. Distance in the direction of incident rays are taken as positive and the distance opposite to the incident ray are taken as negative. Height above the principal axis is taken as positive and below the principal axis is taken as negative.