Question

An object is placed
(i) $10 cm$
(ii) $5 cm$
In front of a concave mirror of radius of curvature 15 cm. Find the position, nature and magnification of the image in each case.

Answer 1

In this question, we need to determine the position, nature and magnification of the image formed when the object is placed in front of a concave mirror of radius of curvature 15 cm. For this, we will use the relation between the focal length and radius of curvature. Also, we will use the relation between the focal length, object distance and the image distance from the plane of the mirror.

Complete step by step answer:
Given in concave lens using sign notation, the radius of curvature $R = - 15cm$
Hence, the focal length of the lens is given as
$f = \dfrac{R}{2} = \dfrac{{ - 15}}{2} = - 7.5cm$
Now when the object is placed at 10cm from the mirror by using the mirror relation given as
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} - - (i)$
(i)The distance of the object from the mirror will be

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\\ \Rightarrow\dfrac{1}{v} = - \dfrac{1}{{7.5}} - \left( { - \dfrac{1}{{10}}} \right) \\\ \Rightarrow\dfrac{1}{v} = - \dfrac{1}{{7.5}} + \dfrac{1}{{10}} \\\ \Rightarrow\dfrac{1}{v} = - \dfrac{1}{{30}} \\\$$

Therefore the distance of the image from the mirror will be $v = - 30cm$
We can say that the image is real and is at a distance of 30cm from the mirror.
Now we know that the magnification of the mirror is the ratio of the distance of the image from the mirror to the distance of the object from the mirror which is given by the formula
$m = \dfrac{v}{u} - - (ii)$
Hence the magnification of the mirror will be

$$m = \dfrac{{ - v}}{u} \\\ \Rightarrow m= \dfrac{{ - 30}}{{10}} \\\ \therefore m=- 3 \\\$$

This means that the image is inverted and is 3 times magnified.
Now when the object is placed at 5cm from the mirror
(ii) Again by using the equation (i), we will find the distance of the image formed

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\\ \Rightarrow\dfrac{1}{v} = - \dfrac{1}{{7.5}} - \left( { - \dfrac{1}{5}} \right) \\\ \Rightarrow\dfrac{1}{v} = \dfrac{1}{{15}} \\\$$

Therefore the distance of the image from the mirror will be
$v = 15cm$
We can say that the image is behind the mirror
Now find the magnification of the mirror by using equation (ii)
$m = \dfrac{v}{u}$
Hence the magnification of the mirror will be

$$m = \dfrac{{ - v}}{u} \\\ \Rightarrow m= \dfrac{{ - 15}}{{ - 5}} \\\ \therefore m= 3 \\\$$

Hence the image formed is virtual and erect.

Note: Nature of image in a mirror can be determined when the position and the distance of the image from the mirror are known to us, and the distance of the object is determined by the mirror formula where the distance of the object and the focal length of the mirror is known.