A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin le

Question

A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal length is 5.01 cm. Behind lens A there is another thin lens, lens B, with focal length 25.9 cm. The two lenses share a common axis and are 62.5 cm apart. Is the image of the firefly that lens B forms real or virtual? How far from lens B is this image located (expressed as a positive number)? What is the height of this image (as a positive number)? Is this image upright or inverted with respect to the firefly?

Explanation / Answer

Start with the first lens and apply 1/f = 1/p + 1/q

1/5.01 = 1/13.7 + 1/q

q = 7.90 cm

Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus...

1/25.9 = 1/54.6 + 1/q

q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real (your first answer)

Also, using that information, you have the second answer, which is 49.3 cm

The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

h' = 3.05 mm (your third answer)

For the fourth answer, since the overall magnification is positive, the final image is upright

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