Question

An object 3cm high is placed at a distance of 9cm in front of a concave mirror of focal length 18cm. find the position, nature and the size of image formed.

Answer 1

the given spherical mirror is a concave mirror and the object is placed 3 cm height from the principal Axis so due to the sign convention it remains positive therefore ho=3cm
and the next part of the question makes understand that the object is placed 9 CM from the concave mirror. due to sign convention the measurement becomes negative therefore object distance, u=- 9 CM
and also the focal length of a concave mirror is to be taken negative due to the sign convention f=-18cm
so here we know the values of height of the object ,object distance and focal length
we know the formula , 1\f=1\v+1\u
1\-18=1\v+1\-9
6-3\-54=1\v
v=-18 cm
the magnification of the image can be found by using the formula - v\u
m,=-(-18)\-9
=18\-9
=-2
as the magnification of the mirror is -2 the image is real and inverted as the magnification number is more than 1 the image is magnified.
therefore the size of the image is more than that of the size of the object
and given mirror forms a real inverted and magnified image


thank you hope it helps:)

Answer 2

Explanation:

It is given that,

Size of the object, h = 3 cm

Object distance, u = -9 cm

Focal length of the concave mirror, f = -18 cm

Let v is the image distance. Using the mirror's formula to find its position. It can be calculated as :

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

$\dfrac{1}{v}=\dfrac{1}{-18}-\dfrac{1}{-9}$

v = +18 cm

Magnification of the mirror is calculated as :

$m=\dfrac{-v}{u}=\dfrac{h'}{h}$, h' is the size of image

$m=\dfrac{-18}{-9}=\dfrac{h'}{3}$

h' = 6 cm

So, the image is formed at a distance of 18 cm behind the mirror. The formed image is virtual. Hence, this is the required solution.