Solution: Physics 123A, Exam 2, Problem 1
1. [25 points] An erect object (see diagram below) is placed at a distance in front of a converging lens equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror with a focal length f2. The lens and mirror are separated by a distance of d = 2(f1 + f2). Justify each of your answers.
(a) [8 points] Use the thin-lens equation to find the location, nature (real or virtual), and magnification of the image produced by the lens.
From the lens equation, 1/o + 1/i = 1/f1 so i = o f1/(o- f1) = 2 f12/f1 = 2f1. Therefore, i = 2 f1 and the magnification is m = -i/o = -2f1/2f1 = -1. The image is real (since i>0) and inverted (since m<0) and is a distance 2f1 beyond the lens.
(b) [9 points] Taking the image produced by the lens as the new object, use the mirror equation to find the location, nature, and magnification of the image produced by the mirror.
We convert the image from part (a) to a new object, which has an object distance o' from the mirror. Then o' = 2(f1 + f2) - i = = 2(f1 + f2) - 2f1 = 2f2 . From the mirror equation, 1/o' + 1/i' = 1/f2 so i' = o'f2/(o' - f2) = 2f22/f2 = 2f 2 . Therefore, i' = 2f2 and the magnification is m' = -i'/o' = -2f2/2f2 = -1. The image is real (since i'>0) and is re-inverted (since m'<0) so that it is upright like the original object. It is a distance 2f2 in from of the mirror, and therefore at the same position as the first image, but upright instead of inverted.
(c) [8 points] Draw a set of rays on the diagram which indicate the positions and orientations of both images discussed above.
(See the diagram above.)