When two lenses are used in combination, the first one forms an image that then

Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.10cm -tall object is 54.0cm to the left of a converging lens of focal length 40.0cm . A second converging lens, this one having a focal length of 60.0cm , is located 300cm to the right of the first lens along the same optic axis.

Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0cm .

answer =

154,3.14

B.

I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

154,3.14

Explanation / Answer

from lens formula

A) 1/v - 1/u = 1/f
u=-54, f=40
1/v=1/40-1/54
154.285cm
v=+154cm (154cm to the right of the 1st lens).

magnification=v/u=154/54=2.851
so image height is 2.851x1.10 = +3.14 cm (erect).

21/v - 1/u = 1/f
u=- (300-23 )cm= -277 cm, f=60.
1/v=1/60-1/277
v=4380/43=49.32cm (to the left of the 2nd lens, so the image is virtual).
magnification=v/u=-49.32/277=-0.178
so image height hv = -0.178 x 3.14 = .5582 cm (inverted).

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