If near point of eye is x metres for a defective eye and the far point is y metres for the same eye. Find the power of lens required.
CLASS 10 -HUMAN EYE
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Step-by-step explanation:
Near point - 25cm
Far point - infinity
In cas of a Myopic eye,
U(object distance) = infinity
V(image distance) = 25cm
So,
1/v - 1/u = 1/f
1/(-25)= 1/f
f= (-25)cm
P= 1/f -> 1/(-25)= (-0.4)cm= -4m....
Similarly, for hypermetropic eye.....
Take U= (-25)cm and V = dis. of the image....
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