A converging lens of focal length 19.2 cm is separated by 55.3 cm from a converg

Question

A converging lens of focal length 19.2 cm is separated by 55.3 cm from a converging lens of focal length 4.72 cm. Calculate the position (relative to the second lens) of the final image of an object placed 39.9 cm in front of the first lens.

If the height of the object is 3.10 cm, what is the height of the final image (A negative height indicates an inverted image)?

If the two lenses are now placed in contact with each other and the object is 5.32 cm in front of this combination, where will the image be located?

Explanation / Answer

Part A)
For the first lens

1/f = 1/p + 1/q

1/19.2 = 1/39.9 + 1/q

q = 37 cm

Since the lenses are 55.3 cm apart, that image is 55.3 - 37 = 18.3 cm in front of the second lens and is the object for that lens

So 1/f = 1/p + 1/q

1/4.72 = 1/18.3 + 1/q

q = 6.36

Thus the final image is 6.36 cm behind the second lens

Part B)
To find the toal magnification..

M = -q/p times -q/p

M = -37/39.9 times -6.36/18.3

M = .322

The image height is .322(3.1) = .999 cm (Just about 1 cm tall)

Part C)
The net focal length is from 1/F = 1/f + 1/f

1/F = 1/19.2 + 1/4.72

F = 3.79

Then, 1/f = 1/p + 1/q

1/3.79 = 1/5.32 + 1/q

q = 13.2 cm

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