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Suppose a person wears the specs of power -6D, the far point for that person is 16cm. But the near point of normal eye is 25cm.

So, is the near point of a person with myopia (having far point less than the near point of normal person) also changed ?

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up vote 1 down vote accepted

Let us say this myopic patient has amplitude of accommodation of 8D. The patient accommodates 8D beyond far point, thus his near point is 1/6+8=0.07m (7.1cm).

If we compare the case with emmetropic eye with accommodation amplitude of 8D, the near point is 1/8=0.12m (12cm).

So, yes, near point of myopic person is different from an emmetropic one. But, you are wrong thinking that near point of "normal" eye is 25cm. The near point is age-and-lens dependent, thus near point is individual one and should be calculated considering amplitude of accommodation.

Look at this picture of age-related accommodation changes -

enter image description here

Look at the scheme of far/near points (near point here is just an example)

enter image description here

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Hi!Thank u for answering.I could not understand the first calculation. Why did u take 1/14 instead of 1/8 ? Also - would the near point of myopic eye change after correction of far point ? – biogirl Apr 7 '14 at 8:27
That is very simple: in nonaccommodative state the eye in example has 6D and to this power the adds its accommodation power of 8D for nearest point, thus the near point is at 1/6D+8D... – Ilan Apr 7 '14 at 8:37
I forget to answer about correction. By correction we convert a myopic eye to "emmetropic", thus with correction the near point is different (more far from the eye). – Ilan Apr 7 '14 at 8:54
Thanks a lot . Now I understand. – biogirl Apr 7 '14 at 8:57

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