Physics 205B Quiz 2, spring semester 2013
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 23.63
Placing an object 15 cm to the left of a f = –15 cm diverging lens will produce a(n) __________ image.
(A) diminished, real.
(B) enlarged, real.
(C) diminished, virtual.
(D) enlarged, virtual.
(E) (No image would be produced.)
Correct answer (highlight to unhide): (C)
Solving for the location of the image di:
(1/do) + (1/di) = (1/f),
(1/di) = (1/f) – (1/do),
(1/di) = (1/(–15 cm)) – (1/(+15 cm)) = –(2/(15 cm)),
di = –(15 cm)/2 = –7.5 cm,
where the negative sign by convention makes this a virtual image located to the left side of the lens (the same side of the lens as the original object). The linear magnification m is given by:
m = hi/ho = –di/do = –(– 7.5 cm)/(15 cm) = +0.50,
which means that this virtual image is diminished, being one-half of the size of the original object (and will be upright, due to the positive sign).
Section 30882
Exam code: quiz02hYp0
(A) : 2 students
(B) : 1 student
(C) : 13 students
(D) : 8 students
(E) : 9 students
Success level: 39%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.56
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