Question

An object is kept in front of a mirror of focal length 30cm. The image is three times the size of the object. Calculate the two possible distances of the object from the mirror

Answer 1

The possible positions of the image are 40 cm  and 20 cm respectively.

Explanation:

Focal length = f = 30 cm

• M=3   i.e  
• q/p=3

Case 1: When q is positive

q = 3 p

Position of object=p=?

Using spherical mirror formula

1/f=1/p+1/q

1/30=1/p+1/3p

1/30=(3+1)/3p

1/30=4/3p

3 p=120

p=40

Position of object is 40 cm from mirror.

Case 2:   When q is negative then

q = -3 p

Position of object=p=?

Using spherical mirror formula

1/f=1/p+1/q

1/30=1/p+1/-3p

1/30=(3-1)/3p

1/30=2/3p

3p=60

p=20

Position of object is 20 cm from mirror.

Also learn more

an object is placed at a distance of 40 cm from a concave lens of focal length 20 cm . find the nature and position of the image ?

https://brainly.in/question/7181642

Answer 2

The two possible distances of the object from the mirror are 40 cm and 20 cm.

Given:

Focal length = f = -30 cm

$h_{i}=3 h_{o}$

To find:

Two possible distances of the object = ?

Solution:

If the  image is three times the size of the object, then there are two cases,

Case 1: For real image

Magnification is given by the formula:

$m=-\frac{v}{u}=-3$

$v=3 u$

On using mirror's formula, we get,

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

$\frac{1}{-30}=\frac{1}{3 u}+\frac{1}{u}$

$u=\frac{-120}{3} \ \mathrm{cm}$

Therefore, u = - 40 cm

Case 2: For virtual image

Magnification is given by the formula:

$m=\frac{-v}{u}=3$

$v=-3 u$

On using mirror's formula, we get,

$\frac{1}{-30}=\frac{1}{-3 u}+\frac{1}{u}$

$u=\frac{-60}{3} \ \mathrm{cm}$

Therefore, u = - 20 cm