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Effect of a diverging lense on a nearsighted patient
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[QUOTE="Chris_Physics, post: 4987207, member: 527924"] [B]The Problem: I calculated that a "patient" needs a diverging lense with P=-2 diopters. The patient has a power of accomodation of 2 diopters. This means, that the near point of the patient is 1/2 diopters = 50 cm. Is this correct?[/B] Relevent equations: 1/f = 1/s + 1/s' 1/s(min) = power of accomodation My attempt [B]Now, what is the nearpoint of the patient with the glasses on? I thought the powers would add, but 2+-2 =0, so that doesn't make sense. Unless the new near point is 1/4 diopters = 25 cm, but this doesn't seem right. I'm not sure how to be thinking about this problem[/B] [/QUOTE]
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Effect of a diverging lense on a nearsighted patient
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