Question

An object of height 2 cm is placed at a distance
20 cm in front of a concave mirror of focal length
12 cm. Find the position, size and nature of the image​

Answer 1

Given :

• Height of object (o) = 2 cm
• Object distance (u) = 20 cm
• Focal length (f) = 12 cm

To Find :

• Position of image
• Size of image
• Nature of image

Formula used :

• Mirror formula :

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

Here, v is image distance, u is object distance and f is the focal length.

Solution :

For a concave mirror,

As the object is placed infront so object distance will be taken as negative. And its focal length will also be negative. Thus,

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

$\dfrac{1}{v} +( - \dfrac{ 1}{20} ) = - \dfrac{1}{12}$

$\dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{20}$

$\dfrac{1}{v} = \dfrac{ - 20 + 12}{20 \times 12}$

$\dfrac{1}{v} = - \dfrac{1}{30}$

$v = - 30 \: cm$

Now,

We know that, for magnification (m)

$\dfrac{i}{o} = \dfrac{-v}{u}$

Here, i is height of image and o is height of object.

Hence,

$= > \dfrac{i}{2} = - (\dfrac{ - 30}{ 20}) \\ \\ = > \dfrac{i}{2} = ( \dfrac{3}{2}) \\ \\ = > i = 3\: cm$

Therefore,

• Position of image is beyond the centre of curvature.

• Size of image is 3 cm.

• Nature of image is enlarged, real and inverted.