# Effect of a diverging lense on a nearsighted patient

1. Jan 25, 2015

### Chris_Physics

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?

Relevent equations:
1/f = 1/s + 1/s'
1/s(min) = power of accomodation

My attempt
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

2. Jan 27, 2015

### BvU

Hello Chris, welcome to PF. :) Please do use the template :( It may be one of the reasons you got such a late reply (although I admit I didn't notice your post before .. ).

I take it the patient needs the glasses to see far away. (I do). So his far away point (I'm no expert, just a wearer) is 50 cm. So his eye(s) starts off with 2 diops and can accomodate 2 more. Glasses + eye = 0 diops $\rightarrow$ 20/20 vision all over the horizon. Make sense ?

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