Two lab partners, Mary and Paul, are both farsighted. Mary has a near point of 6

Question

Two lab partners, Mary and Paul, are both farsighted. Mary has a near point of 63.4 cm from her eyes and Paul has a near point of 110 cm from his eyes. Both students wear glasses that correct their vision to a normal near point of 25.0 cm from their eyes, and both wear glasses 1.80 cm from their eyes. In the process of wrapping up their lab work and leaving for their next class, they exchange glasses (Mary leaves with Paul's glasses and Paul leaves with Mary's glasses). When they get to their next class, find the following.

(a) Determine the distance to the closest object that Mary can see clearly (relative to her eyes) while wearing Paul's glasses.

Try splitting the problem into two steps. First, find the refractive power of each pair of glasses and then the closest object that each student can clearly view while wearing the other student's glasses. Think carefully about whether a particular object or image distance should be relative to the glasses or relative to the eyes. Give careful thought to the appropriate algebraic sign (±) for all object and image distances. cm

(b) Determine the distance to the closest object that Paul can see clearly (relative to his eyes) while wearing Mary's glasses.

How can you modify the result used to determine the closest object Mary can see clearly to determine the closest object Paul can see clearly?

The answers I already tried are 27.7 for part A and 20.0 for part B which I got from a chegg posting but they were incorrect.

Explanation / Answer

(a) Lets first calculate the eyeglasses focal point for Mary
G = 1.8 cm
NP = 63.4 cm
Di = NP - G = (63.4 - 1.8) cm = 61.6 cm
O = 25cm = required to see clearly with glasses
Do = O - G = (25 - 1.8) cm = 23.2 cm
(1/F) = (1/Do) - (1/Di) = (1/23.2) - (1/61.6)
F = 37.22 cm

Now, lets first calculate the eyeglasses focal point for Paul
G = 1.8 cm
NP = 110 cm
Di = NP - G = (110 - 1.8) cm = 108.2 cm
O = 25cm = required to see clearly with glasses
Do = O - G = (25 - 1.8) cm = 23.2 cm
(1/F) = (1/Do) - (1/Di) = (1/23.2) - (1/108.2)
F = 29.53 cm
(a)

Mary without glasses : near point (NP) unchanged, Di unchanged, use F from Paul's glasses
(1/F) = (1/Do) - (1/Di)
(1/Do) = (1/F) + (1/Di)
(1/Do) = (1/29.53) + (1/61.6)
Do = 19.96 cm

Paul without glasses : near point (NP) unchanged, Di unchanged, use F from Mary's glasses
(1/F) = (1/Do) - (1/Di)
(1/Do) = (1/F) + (1/Di)
(1/Do) = (1/37.22) + (1/108.2)
Do = 27.69 cm

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