Geometric optics (near point problem)

[Augustine Duran]

Homework Statement

A person with a near point of 100 cm , but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. If the lenses of the old pair have a power of +2.55 diopters , what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.00 cm in front of his eye?

Homework Equations

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1/f = 1/s + 1/s'

The Attempt at a Solution

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im having trouble figuring out what's the image distance and object distance. I am assuming his original near point (100cm) is the image distance. If i plug in -98cm (subtracting the 2.0 cm from his eye) and 39.2 as the focal length that means ill be solving for S, which is confusing me since s is for object distance yet there's no object? wouldn't his new near point also be an image distance?

 

[TSny]

The near point is the closest distance (measured from the eye) that an object can be placed and still be seen clearly.