Question

An object is 20cm in front of convex mirror if the image height is 1/5 times the object height determine 1. The image distance 2. The focal length 3. The properties of image

Answer 1

Answer:

1. Image distance = - 4 cm

2. Focal length = - 5 cm

3. Image is erect, imaginary and smaller than the object.

Explanation:

Given that:

Object is 20 cm in front of convex mirror.

Height of image is $\frac{1}{5}$ times the object height.

$d_o = 20\ cm$

Magnification, M = $\frac{1}{5}$

Formula for magnification is given as:

$M = -\dfrac{d_i}{d_o}$

Where $d_i$ is the Distance of image from the mirror

Putting the values of M and $d_o$ to find $d_i$:

$\dfrac{1}5 = -\dfrac{d_i}{20}\\\Rightarrow d_i= -\dfrac{1}{5}\times 20\\\Rightarrow d_i= -4\ cm$

Answer 1: Image distance is -4 cm.

Let us use mirror formula to find the Focal length.

$\dfrac{1}{f} =\dfrac{1}{d_o} +\dfrac{1}{d_i}$

where 'f' is the focal length of mirror.

$\Rightarrow \dfrac{1}{f} =\dfrac{1}{20} +\dfrac{1}{-4}\\\Rightarrow \dfrac{1}{f} =\dfrac{1}{20} -\dfrac{1}{4}\\\Rightarrow \dfrac{1}{f} =\dfrac{1-5}{20}\\\Rightarrow \dfrac{1}{f} =\dfrac{-4}{20}\\\Rightarrow \dfrac{1}{f} =\dfrac{-1}{5}\\\Rightarrow f =-5\ cm$

Answer 2: So, focal length is -5 cm.

Answer 3: The Distance of image is negative, so image is imaginary, small (as suggested by magnification) and erect.

Answer 2

object distance =20

and magnification =1/5

magnification = -v/u

1/5= -v/20

v = -1/5x20

v =-4

image distance= -4

1/f =1/v+1/u

1/f =1/-4+1/+20

f=-5