Determing the new Near Point after correcting myopia with glasses

[Cheezay]

Homework Statement

1.Without glasses, your uncle Albert can see things clearly only if they are between 28.4 cm and 137 cm from his eyes. What power eyeglass lens will correct your uncle's myopia? Assume the lenses will sit 2.47 cm from his eyes. Give answer in diopters. Do not enter unit.
=-7.42×10^-1
2.What is your uncle's near point when wearing these glasses?

My professor left this additional note: Part B asks ''What is your uncle's near point when
wearing these glasses?''. When you calculate the
new near point assume it is the distance between
the object and uncle Albert's glasses. The correct
answer is then always accepted by Lon-Capa.

Homework Equations

The Attempt at a Solution

I've figured out the first part of the question, with the correct answer listed. However I'm not sure how to set up an equation for the 2nd part. Do I use the thin lens equation,
(1/f)=(1/di)+(1/do), using my answer for partA to find f, and 28.4 for di, and solve for do?

 

[Steve4Physics]

This is a reply to an old (13+ years at time of writing) question, in case it helps someone encountering it.

First, can I note that the question gives the eye-lens distance to the nearest 0.01cm (2.47cm). This is silly, especially as the far-point is only given to the nearest cm (137cm). However to answer the question asked…
___________________________________

Yes, your proposed method is correct. But you don't need to "find f" as you should already have it as part of your working from part a).

So for part b):

- use the focal length of the lens found in part a);

- set the (virtual) image-distance to dᵢ = -(28.4cm -2.47cm) = -25.93cm; this places the virtual image at the uncorrected near point;

- find the corresponding object-distance, dₒ, with the usual thin lens equation;

This gives the required distance between “the object and uncle Albert's glasses”. (The distance from object to uncle Albert is dₒ+2.47cm, but this is not required by the question.)