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 6.7 cm from her eyes and Paul has a near point of 130 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 get their glasses exchanged (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 closest object that Mary can see clearly (relative to her eyes) while wearing Paul's glasses.
m

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

Explanation / Answer

Given that,

Near point of Mary = Np(m) = 6.7 cm that of Paul = Np(p) = 130 cm

Normal near point = Np(n) = 25

Do(paul) = 25 - 1.8 = 23.2 cm

Di(paul) = 130 - 1.8 = -128.2 cm

Power of Paul = P(paul) = 3.53 d

Di(Mary) = 6.7 - 1.8 = - 4.9 cm = -0.049 m

We know that,

P = 1/Di + 1/Do

we can caluclate Do(Mary) as follows:

3.53 = -1/0.049 + 1/Do(Mary)

Do(Mary) = 4.18 cm

(a) Do'(Mary) = 4.18 + 1.8 = 5.98 cm

Similarly ,

(b)Do(paul) = -7

Do'(Paul) = -7. + 1.8 = -5.2 cm

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