Optics - Diopter farsighted + shortsigted

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A man with both farsightedness and shortsightedness has a natural viewing range of 70cm to 100cm. When using reading glasses, calculations show that his maximum view range could be between 25cm and 38.889cm. There is debate on whether the limit for his view range should be set at 100cm or infinity, with some arguing that 100cm is more appropriate since it reflects his maximum natural vision. The correct approach involves determining the power of the reading lens to ensure that objects at 25cm are focused at 70cm, and objects at his far point create images at 100cm. Understanding these calculations is crucial for determining the effective viewing range with corrective lenses.
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Homework Statement


A man is both farsighted and shortsighted so his natural correct view is between 70cm to 100cm.
What will be his view range if he puts on his reading glasses?


Homework Equations



\frac{1}{u}+\frac{1}{v} = \frac{1}{f} = D

The Attempt at a Solution


I calculated diopter for his farsightness and came with D=2.571 (1/0.25 - 1/0.70 = 2.571
Now, I wanted to know what will be the his maximum range and have come to the equation:
\frac{1}{u}-\frac{1}{\infty} = 2.571
And the solution for u is 38.889cm, so the view range will be between 25cm to 38.889cm.
Now, there is another opinion in our class that when I use the equation to reveal the limit to his view range I should put 1/100cm instead of \frac{1}{\infty} because the man can see to 100 cm at max and not to the extent of the horizon.

What is the correct way and why?

Thank you
 
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the reading lens should have power such that for object at near point of vision (25cm) its image should form at 70cm ... so that eye lens can focus it.
find power of it that way ...
now for far point ... when object is at u ... its image should be at 100cm ... so solve for u (answer)
 
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