## 20180222

### Physics quiz question: image height produced by diverging lens focal length

Physics 205B Quiz 2, spring semester 2018
Cuesta College, San Luis Obispo, CA

An object 1.0 cm in height is placed 10 cm in front of a f = –15 cm diverging lens. The absolute value of the image height is:
(A) 0.17 cm.
(B) 0.60 cm.
(C) 0.67 cm.
(D) 1.5 cm.

Correct answer (highlight to unhide): (B)

The image location can be determined from the thin lens equation:

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

(1/di) = (1/f) – (1/do),

di = ((1/f) – (1/do))–1,

di = ((1/(–15 cm)) – (1/(+10 cm))–1 = –6.0 cm,

Then from the linear magnification equation:

m = hi/ho = –di/do,

the image height can be determined, given the object distance do and image distance di:

hi/ho = –di/do,

hi = –ho·di/do,

hi = –(+1.0 cm)·(–6.0 cm)/(+10 cm) = +0.60 cm.

(Response (A) is the absolute value of (1/f) – (1/do); response (C) is the absolute value of do/f; and response (D) is the absolute value of f/do.)

Sections 30882, 30883
Exam code: quiz02P0wR
(A) : 1 student
(B) : 20 students
(C) : 5 students
(D) : 8 students

Success level: 59%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63