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Consider an object with s=12cm that produces an image with s1=15cm. Note that whenever you are working with a physical o...

Consider an object with s=12cm that produces an image with s1=15cm. Note that whenever you are working with a physical object, the object distance will be positive (in multiple optics setups, you will encounter "objects" that are actually images, but that is not a possibility in this problem). A positive image distance means that the image is formed on the side of the lens from which the light emerges. (1)Find the focal length of the lens that produces the image described in the problem introduction using the thin lens equation. Express your answer in centimeters, as a fraction or to three significant figures.(2)What is the magnification of the lens?
science   Advanced-Physics   
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Answer #1
1/12 + 1/15 = 1/f
f = 6.67 cm
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Answer #2

Part A

f=6.67 cm

Part B

Converging

Part C

m=-1.25

Part D

real and inverted

Part E

virtual

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Answer #3

1 / s + 1 / s' = 1 / f

is the equation for calculating the focal length of the lens

suppose s=+12 and s'=+15
then

1 / 12 + 1 / 15 = 1 / f

f = 6.667 cm

Hope that helps
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