# Medical physics - Correcting presbyopia

#### SassyBean

Hi all,
I have another problem, this time involving the lens equation.
(1/f= 1/v+ 1/u)
v and u are in Meters and 1/f is the power in diopters (D)
I have solved some problems where the near point of an eye is correct and the far point needs correcting and where the far point is correct (infinite) and the near point needs adjusting, however the last question has me stumped.

It seems as though I need to modify both the near point and far point simultaneously but this also seems strange as it specifies I need not change the near point but instead maintain it and bring objects into focus at that distance.
My working so far has been to calculate the difference from a healthy eye in the near and far points separately and combine the answer to give an overall lens strength adjustment, but I don't connect this with answering the near point issue correctly and would therefore like some guidance:

An elderly person with presbyopia has a near point of 0.40m and a far point of 4.0m

Calculate the power of spectacle lens required to enable objects at the near point to be seen clearly

My working :

Near point:
Actual ( 1 / 0.4 ) + ( 1 / 0.02 ) = 52.5 D
Required ( 1 / 0.25 ) + ( 1 / 0.02 ) = 52 D
Difference = 0.5 D

+

Far point:
Actual ( 1 / 4 ) + ( 1 / 0.02 ) = 50.25 D
Required ( 1 / ∞ ) + ( 1 / 0.02 ) = 50 D
Difference = -0.25 D

0.5 – 0.25 = 0.25 D

Lens power required = 0.25 D