Physics 205B Quiz 2, spring semester 2019
Cuesta College, San Luis Obispo, CA
An object 1.0 cm in height located 8.0 cm in front of a diverging lens produces an upright image that is 0.75 cm in height. The focal length of this diverging lens is:
(A) –24 cm.
(B) –11 cm.
(C) –6.0 cm.
(D) –3.4 cm.
Correct answer (highlight to unhide): (A)
From the linear magnification equation:
m = hi/ho = –di/do,
the ratio between the image distance di and object distance do (8.0 cm) can be determined from the given image height hi (+0.75 cm) and object height ho (1.0 cm):
hi/ho = –di/do,
(+0.75 cm)/(1.0 cm) = –di/(8.0 cm),
–(8.0 cm)·(+0.75 cm)/(1.0 cm) = di = –6.0 cm.
Substituting this result for di into the thin lens equation will then obtain a value for the focal length f, as the object distance do is given:
(1/do) + (1/di) = (1/f),
(1/(8.0 cm)) + (1/(–6.0 cm)) = (1/f),
–0.04166666667... cm–1 = (1/f),
–24 cm = f,
which is negative, as expected for a diverging lens.
(Response (B) is do·ho/hi; response (C) is di; response (D) is di·do/(do – di).)
Sections 30882, 30883
Exam code: quiz02BTLn
(A) : 17 students
(B) : 4 students
(C) : 7 students
(D) : 13 students
Success level: 41%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.90
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