Question

magnification of an image is -1/3 of the object. focal length is 10cm of concave mirror. find position,size and nature of the image

Answer 1

Answer:

The image will form 13.33cm in front of the mirror and it will be real and will be diminished .

Explanation:

Let us consider the following

•the object distance be u

• the image distance be v

• the focal length be f

• the magnification be m

Formulae to be used

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

and

$m = \frac{ h_{2} }{ h_{1} } = - \frac{v}{u}$

where h₁ and h₂ are respectively the object and image sizes

Given data ,

m = -1/3

and f = -10cm

$m \: = \frac{ - v}{u} \\ \implies - \frac{1}{3} = - \frac{v}{u} \\ \implies \frac{1}{3} = \frac{v}{u} \\ \implies u = 3v$

Now using this value in mirror formula we have

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \implies \frac{1}{v} + \frac{1}{3v} = \frac{1}{ - 10cm} \\ \implies \frac{1 + 3}{3v} = \frac{1}{ - 10cm} \\ \implies \frac{4}{3v} = \frac{1}{ - 10cm} \\ \implies3v = - 40cm \\ \implies v = - 13.33cm$

Therefore , the image distance is -13.33cm i.e. image is formed 13.33cm away from mirror and the negative sign indicates that the image is formed in front of the mirror . Again , the image formed is real.