an object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. the focal length of the mirror is 15 cm. at what distance from the mirror should a screen be kept so as to get a clear image ? what will be the size and nature of the image ?

Hi Khushi Sinha, let us follow mirror formula 1/u + 1/v = 1/f 
Given u = 25 cm, f = 15 cm
Now v = uf /(u - f)
==> v = 25 * 15 / 10 = 37.5 cm.
Screen has to be placed at a distance 37.5 cm to get the sharp image
Magnification = v/u = 37.5/25 = 1.5
Hence size of image will be 1.5 * 7 = 10.5 cm 
Nature of image :
Real
Inverted
Enlarged
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Your answer is here
1) 1/V + 1/U = 1/F
     1/V = 1/F - 1/U
     1/V = 1/-15 - 1/-25
     1/V = 1/-15 + 1/25
     1/V = 5 - 3/75
     1/V = 2/-75
     V = 75/2
     V = -3.75cm
     H2/H1 = V/W
     H2/7 = -(-37.5)
     H2/7 = 37.5/25
     H2 = 37.5/250 x 7
     H2 = - 10.5
     Hence    Prove.
  • 1
Using the formula :-
1/V + 1/U = 1/F
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