a. Two lenses, separated by a distance of 22.1 cm, are used in combination. The

Question

a. Two lenses, separated by a distance of 22.1 cm, are used in combination. The first lens has a focal length of +26.0 cm; the second has a focal length of -15.0 cm. An object, 1.91 mm long, is placed 2.42 cm before the first lens. What is the intermediate image distance relative to the first lens? Take negative to be to the left, and positive to be to the right.
b.What is the final image distances relative to the second lens? Take negative to be to the left, and positive to be to the right.
c. What is total magnification
d. What is the height of the final image?

Explanation / Answer

Part A)

For the first lens

1/f = 1/p + 1/q

1/26 = 1/2.42 + 1/q

q = -2.67 cm   which means its 2.67 cm to the left of the lens

Part B)

Since the q value for part a is 2.67 cm to the left of the lens, and the lenses are separated by a distance of 22.1 cm, the object distance for the second lens is 24.77 cm

1/-15 = 1/24.77 + 1/q

q = -9.34 cm which means it is 9.34 cm to the left of the second lens

Part C)

M for the first lens = -q/p = -(-2.67)/(2.42) = 1.10

M for the second lens = -(-9.34)/(24.77) = .377

Total magnification is (1.10)(.377) = .415

Part D)

M = h'/h

.415 = h'/1.91

h' = .792 mm

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