Question

the focal length of a concave mirror is 30cm, find the position of the object in front of the mirror so that the size of the image is three times the size of the object.. please help me out with this one

Answer 1

f = -30 cm
let position of object = u cm
position of image = v cm
magnification = -3 
magnification is taken as negative as real image is formed.

m = -v/u
⇒ -3 = -v/u
⇒ 3 = v/u
⇒ 3u = v
⇒ v = 3u 

we know that

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \Rightarrow \frac{1}{3u} + \frac{1}{u} = \frac{1}{-30} \\ \\ \Rightarrow \frac{1+3}{3u}= \frac{1}{-30} \\ \\ \Rightarrow \frac{4}{3u}= \frac{1}{-30} \\ \\ \Rightarrow 4 \times (-30)=3u \\ \\ \Rightarrow u= \frac{4 \times (-30)}{3}=-40cm$

The object is at a distance 40cm in front of the mirror.

Answer 2

f = -20 cm
magnification m = +3 (erect image)  or  -3 (inverted image)

case 1)  erect virtual image
 m = 3 = -v/u  => v = -3 u
  1/f = 1/v + 1/u =>    -1/30 = -1/3u  + 1/u = 2/3u
      u = -20 cm

case 2)  inverted real image.
   m = -3 = -v/u  => v = 3u
   -1/30 = 1/3u + 1/u = 4/3u
         =>  u = -40 cm