help- -PartA Constants When two lenses are used in combination- the fi.docx

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The document describes a combination of two converging lenses finding the location and height of the final image. A 1.30 cm tall object is located 58.0 cm from the first lens with a focal length of 40.0 cm, forming an image located 128.9 cm from the lens that is inverted and 2.89 cm tall. This image then serves as the object for the second lens located 300 cm away with a focal length of 60 cm, forming a final image located 44.42 cm from the second lens that is inverted and 0.75 cm tall.

Formation

help?
?PartA Constants When two lenses are used in combination, the first one forms an mage that
then serves as the object for the second lens. The magniication of the combination is the ratio of
the height of the final mage to the height of the object. A 1.30 cm -tall object is 58.0 cm to the
lem 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 focalength of 40.0
cm Enter your answer as two numbers separated with a comma cm Submit x Incorrect: Try
Again; 6 attempts remaining Part B l1 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 Enter your answer as two numbers separated with a comma cm Submit
Solution
a) from the realtion
1/f = 1/p+1/q
1/40 = 1/58+1/q
1/q = 1/40-1/58
q = 128.9 cm
magnification m = h'/h = -q/p
h' = -1.3*128.9/58 = -2.89 cm
image is inverted
b) 1/(-60) = 1/(300-128.9)+1/q
1/q = -1/60-1/171.1
q = -44.42 cm

h'/h = -q/p
h' = 2.89*-44.42/171.1 = -0.75

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