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2015-03-23T22:43:09+05:30

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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.
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2015-03-24T01:49:18+05:30

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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     

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