Two lab partners, Mary and Paul are both farsighted. Mary has a near point of 6.
ID: 1779003 • Letter: T
Question
Two lab partners, Mary and Paul are both farsighted. Mary has a near point of 6.6 cm from her eyes and Paul has a near point of 126 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.
(b) Determine the closest object that Paul can see clearly (relative to his eyes) while wearing Mary's glasses.
*******NOT .04 m or .057
Explanation / Answer
given, mary and paul are farsighted
Near point of mary, M = 6.6 cm
near point of paul , P = 126 cm
normal near point N = 25 cm
distacne of glasses from eyes, d = 1.8 cm
a. let focal length of paul's glasses be Fp
then object distance u = -(N - d) = -23.2 cm
image distance = v = -(P - d) = -125.2 cm
hence
from thin lens formula
1/v - 1/u = 1/f
-1/125.2 + 1/23.2 = 1/Fp
Fp = 28.476 cm
let the closest object mary can see be at distance x from her eyes
then object distance, u = -(x - d) = -(x - 1.8)
image distance v = -(M - d) = -4.8 cm
hence
1/Fp = 1/v - 1/u
1/28.476 = -1/4.8 + 1/(x - 1.8)
x = 5.9076 cm
b. let focal laneght of mary's glasses be Fm
then object distance u = -(N - d) = -23.2 cm
image distance v = -(M - d) = -4.8 cm
hence
from thin lens formula
1/v - 1/u = 1/f
-1/4.8 + 1/23.2 = 1/Fm
Fm = -6.052 cm
let closest object paul can see be at distance x from his eyes
then u = -(x - d) = -(x - 1.8)
v = -(P - d) = -125.2 cm
hence
1/Fm = 1/v - 1/u
-1/6.052 = -1/4.8 + 1/(x - 1.8)
x = 25.00255 cm
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