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What is the far point of a person whose eyes have a relaxed power of 52.1 D? Assume the lens-to-retina distance is 2.00 cm fa

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* the power of eye is given, p= dot di » di = acm do=? P = 52.1 D. * from Ean O. do = P- di = 52.1 - do : 1 = 2.1 0.02taking reciprocal on both side, de do = 0.436 m & Thus, the farthest object - the person can view clearly knowing the power oSo, Far point : 0.476 m

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