the near point of hypermetropic
eye is 100 cm calculate the focal length and power of the lens to correct his defect
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Answer:
here , u= -25 cm , v= -100cm
1/v - 1/u = 1/f
1/f = - 1/100 -(-1/25) = -0.01 + .04 = .03
f= 1/ 0.03
f= 100/3 = +33.33 cm
Explanation:
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