Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 24.5
[20 points.] An upright object is located 14.0 cm in front of a diverging lens with focal length –10.0 cm. Behind the diverging lens is a converging lens of focal length +16.0 cm. (Drawing below is not to scale.) Find the minimum separation distance between the diverging lens and converging lens such that the final image will be inverted, with respect to the original object. Show your work and explain your reasoning.
Solution and grading rubric:
- p = 20/20:
Correct. Determines that the first diverging lens will produce an upright (virtual) image located 5.8 cm in front of it. For the second converging lens to produce an inverted (real) image, its object must be located at least 16.0 in front of it. Since the image from the first lens (5.8 cm in front of it) will be the object for the second lens, the second lens must be placed at a minimum distance of 10.2 cm behind the first lens in order for the second lens to produce a final inverted (real) image. - r = 16/20:
Nearly correct, but includes minor math errors. May have not sufficiently explained the conditions for the second lens to create an inverted (real) image in terms of its object and focal length positions. - t = 12/20:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least determined the location of the image produced by the first lens, and some attempt at constraining the location of the second lens. - v = 8/20:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. - x = 4/20:
Implementation of ideas, but credit given for effort rather than merit. - y = 2/20:
Irrelevant discussion/effectively blank. - z = 0/20:
Blank.
Grading distribution:
Section 30882
Exam code: finalv0L7
p: 2 students
r: 1 student
t: 2 students
v: 2 students
x: 0 students
y: 1 student
z: 0 students
A sample "p" response (from student 2414):
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