Physics 205B Quiz 2, spring semester 2020
Cuesta College, San Luis Obispo, CA
An object 1.0 cm in height is placed 16 cm in front of a converging lens with a focal length of +25 cm. The resulting image is:
(A) upright, diminished.
(B) upright, enlarged.
(C) inverted, diminished.
(D) inverted, enlarged.
(E) (No image would be produced.)
Correct answer (highlight to unhide): (B)
Solving for the location of the image di:
(1/do) + (1/di) = (1/f),
(1/di) = (1/f) – (1/do),
(1/di) = (1/(+25 cm)) – (1/(+16 cm)) = –0.0225 cm–1,
di = 1/(–0.0225 cm–1) = –44.444... cm,
where the negative sign by convention makes this a virtual image located to the right side of the lens (the side of the lens opposite the original object). The linear magnification m is given by:
m = hi/ho = –di/do = –(–44.444... cm)/(16 cm) = +2.777...,
which means that this virtual image is upright, due to the positive sign, and is enlarged, being about 2.8 times the size of the original object.
Sections 30882, 30883
Exam code: quiz02B3rD
(A) : 1 student
(B) : 21 students
(C) : 3 students
(D) : 10 students
(E) : 0 students
Success level: 60%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.83
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