An object of height 2cm is placed at a distance of 15 cm in front of a concave lens of focal lenght 10cm find the nature position and size of the image formed
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1/f = 1/v - 1/u
-1/10 = 1/v - (-1/15)
-1/10 = 1/v + 1/15
-1/10 - 1/15 = 1/v
1/v = (-3 - 2 )/30
1/v = -5/30
v = -6 cm.
magnification = v/u = h2/h1
= -6/-15 = h2/2
= h2 = 4/5 cm.
v = -6
magnification = 2/5
size of image = 4/5
nature = virtual, erect and diminished
-1/10 = 1/v - (-1/15)
-1/10 = 1/v + 1/15
-1/10 - 1/15 = 1/v
1/v = (-3 - 2 )/30
1/v = -5/30
v = -6 cm.
magnification = v/u = h2/h1
= -6/-15 = h2/2
= h2 = 4/5 cm.
v = -6
magnification = 2/5
size of image = 4/5
nature = virtual, erect and diminished
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