A converging lens of focal length 18.8 cm is separated by 56.4 cm from a converg

Question

A converging lens of focal length 18.8 cm is separated by 56.4 cm from a converging lens of focal length 3.11 cm. Calculate the position (relative to the second lens) of the final image of an object placed 41.9 cm in front of the first lens.

If the height of the object is 2.87 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 5.62 cm in front of this combination, where will the image be located?

Explanation / Answer

by lens equation

1 / f = 1 / v + 1 / u

1 / 18.8 = 1 / 41.9 + 1 / u

u = 34.1 cm

this image will act as an abject for second lens

so distance of object for second lens = 56.4 - 34.1

distance of object for second lens = 22.3 cm

1 / 3.11 = 1 / 22.3 + 1 / u

u = 3.614 cm

position of the final image = 3.614 cm

magnificaton of first lens = -distance of image / distance of object

magnificaton of first lens = -34.1 / 41.9

magnificaton of second lens = -distance of image / distance of object

magnificaton of second lens = -3.614 / 22.3

total magnification = magnfication of first lens * magnification of second lens

total magnification = (-34.1 / 41.9) * (-3.614 / 22.3)

(-34.1 / 41.9) * (-3.614 / 22.3) = height of image / height of object

(-34.1 / 41.9) * (-3.614 / 22.3) = height of image / 2.87

height of image = 0.378534 cm

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