How Do You Calculate the Near Point Distance for an Unaided Eye?

[ejonesdj]

Hello everyone,
Really struggling to answer a practice exam question on the eye, and was wandering if anyone could help, here's the question:

"The unaided eye of a student has a power of 59.0D when viewing an object at infinity. When using her spectacle lens, the near point is 25cm from her eye.

Calculate the distance of the near point for her unaided eye.
The power of the student's glasses lens is 1.5D"

I do have the markscheme however I'm very confused as to how they have got their answer.

Thanks in advance,
Ed

 

[ejonesdj]

I understand the first part of the markscheme - where the 1/f=1/u +1/v formula (where f= focal length, u= object-lens distance and v= image-lens distance) which gives the object-lens distance by substituting 59D as the power (1/f).
So:
59D = 1/infinity +1/u, thus giving u= 1.69cm

The next part of the markscheme confuses me:
It says that you need to work out the power of the lens and eye together by doing the following:
Pe + 1.5 = 1/0.25 + 1/0.017
Pe = 61.5D

Im struggling to see how that works...
The markscheme continues to give the final answer stating:
61.5 = 1/u + 59
giving u = 40cm
Im also not sure what's going on here..
If someone could explain what there doing that would be great,
thanks in advance
Ed