20170304

Physics quiz question: comparison of image distance with diverging lens focal length

Physics 205B Quiz 2, spring semester 2017
Cuesta College, San Luis Obispo, CA

An object is placed in front of a f = –20 cm diverging lens, which produces an upright image that is one-tenth the size of the object. The absolute value of the distance from the lens to the image is __________ 20 cm.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information given.)

Correct answer (highlight to unhide): (A)

From the linear magnification equation:

m = hi/ho = –di/do,

the ratio between the object distance do and image distance di can be determined (where the linear magnification is +(1/10), as the image is diminished ("one-tenth the size of the object") and upright):

+(1/10) = –di/do,

do = –10·di.

Substituting this result for do into the thin lens equation will then obtain a value for the image distance di, as the focal length f is given:

(1/do) + (1/di) = (1/f),

(1/(–10·di)) + (1/di) = (1/f),

(9/(10·di)) = (1/f),

di = (9/10)·f = (9/10)·(–20 cm) = –18 cm,

which means that the image will be located to the left of the diverging lens (same side as the object), 2 cm closer than the diverging lens' left focal point.

Sections 30882, 30883
Exam code: quiz02J4sZ
(A) : 20 students
(B) : 3 students
(C) : 7 students
(D) : 0 students

Success level: 67%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45

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