Physics quiz question: refractor telescope

Physics 205B Quiz 2, Fall Semester 2010
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 24.51

The objective lens of an astronomical telescope forms an image of a distant object at the focal point of the eyepiece, which has a focal length of +1.5 cm. If the two lenses are 80.0 cm apart, what is the angular magnification?
(A) –54.
(B) –52.
(C) –17.
(D) –4.8.

Correct answer: (B)

The objective lens of a telescope, for an object at infinity, produces a real image at its primary focal point. An eyepiece, used as a simple magnifier by a relaxed eye, will need to have this intermediate real image at its second focal point in order to achieve the maximum angular magnification. This means that the primary focal point of the objective lens is located at the secondary focal point of the eyepiece, and thus the separation distance L between the lenses is:

L = 80.0 cm = f_o + f_e.

Since the focal length of the eyepiece is given (f_e = +1.5 cm), then the focal length of the objective can be substituted in using the above information into the angular magnification equation for a telescope:

M = -f_o/f_e = -(L - f_e)/f_e = -(80.0 cm - (+1.5 cm))/(+1.5 cm) = -52.333... = -52,

to two significant figures.

Response (A) is -(L + f_e)/f_e; response (C) is -N/f_e (where N is the nominal near point of 0.25 m); response (D) is -L*f_e/N.

Student responses
Section 70856
(A) : 2 students
(B) : 7 students
(C) : 1 student
(D) : 1 student

Success level: 64%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.80