An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.
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u=-50cm
f=-20cm
1/f=1/v-1/u
1/-20=1/v-1/-50
1/-20=1/v+1/50
1/v=1/-50-1/20
1/v=-2-5/100
v=100/-7
v=-14.2cm
therefore position of the image is 14.2cm from the optic Centre on the same side of the lens
Nature:
m=v/u
m=-100/7÷-20
m=5/7=0.714
hence the image is virtual, erect and diminished image
f=-20cm
1/f=1/v-1/u
1/-20=1/v-1/-50
1/-20=1/v+1/50
1/v=1/-50-1/20
1/v=-2-5/100
v=100/-7
v=-14.2cm
therefore position of the image is 14.2cm from the optic Centre on the same side of the lens
Nature:
m=v/u
m=-100/7÷-20
m=5/7=0.714
hence the image is virtual, erect and diminished image
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Explanation:
The nature and approximate position is trivial. Concave lenses always produce virtual, upright, reduced images closer to the lens than the object.
Use the Lens Formula (here with Real Is Positive convention)
1/v + 1/u = 1/f
1/v = 1/f - 1/u = -1/30 - 1/50 = -5/150 - 3/150 = -8/150
v = -150/8 or 18.75 from the lens, virtual.
M = v/u = -150/8 / +50 = -3/8
(Negative here indicates erect image)
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