Question

A 4cm tall object is placed at a distance of 5 cm from a concave mirror perpendicular to the principal
axis. If its radius of curvature is 20 cm, determine the position, size and nature of the image formed.

Answer 1

Given :

• Radius of curvature = 20 cm

• Object Distance = 5 cm

• Height of the object = 4 cm

To find :

• Position of the Image or Image distance (v).

• Size of the image (h.i)

• Nature of the image.

Solution :

Before moving further , let's calculate the focal length of the mirror.

We know that ,

$\bf{F = \dfrac{R}{2}}$

Using the above formula and substituting the values in it, we get :

$:\implies \bf{F = \dfrac{20}{2}}$

$:\implies \bf{F = 10}$

$\therefore \bf{Focal\:length\:(f) = 10\:cm}$

Hence, the focal length of the mirror is 10 cm.

To find the Image distance :

We know the mirror formula , i.e,

$\boxed{\bf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}$

Where ,

• f = Focal length
• v = Image distance
• u = Object Distance

[Note :- In a concave mirror the focal length and the object distance is always taken as negative)

Now using the mirror formula and substituting the values in it, we get :

$:\implies \bf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

$:\implies \bf{\dfrac{1}{(-10)} = \dfrac{1}{v} + \dfrac{1}{(-5)}}$

$:\implies \bf{\dfrac{1}{(-10)} - \dfrac{1}{(-5)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{(-10)} + \dfrac{1}{5} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1 - 2}{(-10)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{(-1)}{(-10)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{(\not{-}1)}{(\not{-}10)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{10} = \dfrac{1}{v}}$

$:\implies \bf{10 = v}$

$\boxed{\therefore \bf{Image\:Distance\:(v) = 10\:cm}}$

Hence , the image distance of the mirror is 10 cm.

To find the height of the image :

First let us find the Magnification of the image.

We know that ,

$\bf{m = - \bigg(\dfrac{v}{u}\bigg)}$

Where,

• m = Magnification
• u = Object Distance
• v = Image Distance

Now , using the above formula and substituting the values in it, we get :

$:\implies \bf{m = - \bigg(\dfrac{v}{u}\bigg)}$

$:\implies \bf{m = - \bigg(\dfrac{(-10)}{(-10)}\bigg)}$

$:\implies \bf{m = (-1)}$

$\boxed{\therefore \bf{Magnification\:(m) = (-1)}}$

Hence, the Magnification is (- 1).

Now , using the other formula for magnification i.e,

$\bf{m = \dfrac{h_{I}}{h_{o}}}$

Where,

• m = Magnification
• hi = Height of the object
• ho = Height of the image

Using the formula and substituting the values in it, we get :

$:\implies \bf{m = \dfrac{h_{I}}{h_{o}}}$

$:\implies \bf{(-1) = \dfrac{h_{I}}{10}}$

$:\implies \bf{(-1) \times 10 = h_{I}}$

$:\implies \bf{(-10) = h_{I}}$

$\boxed{\therefore \bf{Image\:height\:(v) = (-10)\:cm}}$

Hence, the height or size of the image is (-10) cm.

Nature :- The image is real , inverted and magnified.