Question

A lens of focal length +15.0 cm is 10.0 cm to the left of a second lens of focal length -15.0 cm.

Where is the final image of an object that is 27.0 cm to the left of the positive lens? Enter the image distance with respect to the right lens. Follow the sign convention.(Express your answer to three significant figures.)

How does the image's size compare to this of the original object? Find the magnification m=hI/hOm=hI/hO. (Express your answer to three significant figures.)

Answer 1

Using the lens equation for first converging lens

1/f = 1/u + 1/v

u = object distance = +27 cm

f = focal length = +15 cm

v = image distance = ?

1/v = 1/15 - 1/27

v = 15*27/(27 - 15) = +33.75 cm

Now this image will be +33.75 cm right from the converging lens, Image's distacne from diverging lens will be

u1 = 10 - 33.75 = -23.75 cm = object distnace for diverging lens

f1 = focal length of diverging lens = -15 cm

v1 = image distance = ?

1/v1 = -1/15 - 1/(-23.75)

v1 = 23.75*15/(15 - 23.75) = -40.7 cm

v1 = -40.7 cm (-ve sign means image to the left of diverging lens (lens on the right))

Part B.

Magnification is given by:

M = M1*M2 = (-v/u)*(-v1/u1)

M = (-33.75/27)*(-(-40.7)/(-23.75))

M = 2.14

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