Near Point Without Spectacles: 1m

[SwaggerJohn]

If the far point of a nearsighted person is 1m in front of their eyes:

a) What power should their spectacles have, so that they may see distinctly as object at infinity.

b) If when wearing these glasses, their near point is at 0.25m, where would it be without them?

So for a)
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1/f= 1/s + 1/s'
1/f = 1/infinity + 1/(-1)
=> f = -1m

So dioptric power is D = 1/f = 1/(-1) = -1 diopter

for b)
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any clues?

 

[SwaggerJohn]

Figured it out:

use the same formula with f=-100 as found in a) and s=0.25 and then solve for s' which yields a value of -0.05.