A person fails to see clearly an object placed at a distance less than 60cm from his eye. lens of what power should be used by him so as to eliminate this defect
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as he could bot able to see the objects less than 60 cm from his eyes it means his crystalline lens is not able focus the light rays it means its near point is more than 25 cm . there fore the person has to make his near point 25 cm . as the lens is not able to focus so he needs a converging lens of appropriate power so as to give the required converging power to make the image form on the retina.
u = 60cm =0.6m
V= 25 cm = 0.25m
power = 1/F = 1/U +( - 1/V)
=> POWER = 1/6/10 +(-1/25/100)
= 10/6+ (-100/25)
= 1.66666+(-4)
= 3.67 d
u = 60cm =0.6m
V= 25 cm = 0.25m
power = 1/F = 1/U +( - 1/V)
=> POWER = 1/6/10 +(-1/25/100)
= 10/6+ (-100/25)
= 1.66666+(-4)
= 3.67 d
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