Physics

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The near and the far points of a short-sighted are 8 cm and 20 cm from the eyeball.
a) What power of corrective lens is needed for this eye to see distant objects clearly?
b)What will the resulting near point then be?(Assume that the lens is placed very close to the eye)

Answers so far:
a) 1/f = 1/u + 1/v
1/f = 1/0.08 + 1/0.20
1/f = 17.5
P = 1/f
P = 17.5 D

b) 1/∞ + 1/-0.2 = 1/f
f = 0

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