Physics 205B Quiz 2, spring semester 2018
Cuesta College, San Luis Obispo, CA
A Physics 205B student with hyperopia has an uncorrected near point of 0.65 m. The refractive power of the contact lens used to correct this student's vision is:
(A) +0.40 D.
(B) +0.65 D.
(C) +1.5 D.
(D) +2.5 D.
Correct answer (highlight to unhide): (D)
The Physics 205B student is farsighted, and do = 0.65 m is the nearest object distance that this student's unaided eye can see. The thin lens equation for this student's unaided eye is then:
(1/do) + (1/di) = (1/f1),
where f1 is the focal length of the student's (accommodated) cornea/lens, and di is the distance from the student's cornea/lens to the retina at the back of the eye, where the (inverted) real image is projected.
This student would like to see things at an object distance of do' = 0.25 m when wearing contact lenses, where the (') indicates this is the corrected farthest object distance. The thin lens equation for this student's eye with contacts is then:
(1/do') + (1/di) = (1/f1) + (1/f2),
where f1 is the focal length of the student's (accommodated) cornea/lens, f2 is the focal length of the contact lens, and di is the distance from the student's cornea/lens to the retina at the back of the eye, where the (inverted) real image is projected.
Since the student's cornea/lens to retina distance is constant, di is the same for both equations; similarly the students student's (relaxed) cornea/lens focal length f1 is also constant. We can eliminate these two quantities from both equations by subtracting the first equation from the second:
    [(1/do') + (1/di) = (1/f1) + (1/f2)]
– [(1/do) + (1/di) = (1/f1)]
                                                                   
    (1/do') – (1/do) = (1/f2),
such that the refractive power of the contact lens is then:
P2 = (1/f2),
P2 = (1/do') – (1/do),
P2 = (1/(0.25 m)) – (1/(0.65 m)) = + 2.4615384615... m–1 = +2.5 D,
where the units of diopters (D) is equal to inverse meters (m–1).
(Response (A) is (0.65 m – 0.25 m); response (C) is the absolute value of the inverse of the student's near point (0.65 m), which if it was –1.5 D would be the prescription for myopia with a far point of 0.65 m. Note that an incorrect calculation (0.65 m – 0.25 m)–1 would also result in the correct response, to two significant figures.)
Section 30882, 30883
Exam code: quiz02P0wR
(A) : 4 students
(B) : 4 students
(C) : 21 students
(D) : 5 students
Success level: 15%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.22
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