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 1.00 cm -tall object is 55.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis.

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

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.

Exercise 34.41 Part A 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 1.00 cm -tall object is 55.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens this one having a focal length of 60.0 cm, is located 300 cm to the right of the first lens along the same optic axis Find the location and height of the image (call it 11) formed by the lens with a focal length of 40.0 cm Enter your answer as two numbers separated with a comma cm Submit My Answers Give Up Part B 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 enses Enter your answer as two numbers separated with a comma cm Submit My Answers Give Up

Explanation / Answer

A) 1/u+1/v=1/f.
u=55, f=40
1/v=1/40-1/55
v=+146.66cm (146.66cm to the right of the 1st lens).

(Note: this distance seems quite far, but the object is quite near the focal point; if it had been at the focal point the image would have been formed at +infinity.)

magnification=v/u=146.66/55=2.66
so image height is 2.66x 1 = +2.66 cm (erect).

B) 1/u+1/v=1/f.
u=300-146.66=153.34 , f=-60.
1/v=1/-60-1/100
300/v=-5-3=-8
v=-300/8=-43.13cm (to the left of the 2nd lens, so the image is virtual).
magnification=v/u=-43.5/153.34=-0.2836
so image height hv = -0.2836 x 2.66 = -0.7543 cm (inverted).

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