(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two converging lenses, the first with focal length 40 cm and the second with focal length 33 1/3 cm, are separated by 10.0 cm. An object of height 6.0 cm is placed 20.0 cm in front of the first lens. What are the size and orientation of the final image?

A. 6.0 cm, upright

B. 6.0 cm, inverted

C. 24.0 cm, upright

D. 24.0 cm, inverted

E. 1.5 cm, inverted

2. Relevant equations

Lens #1:

F1 = 40 cm

Do1 = 20 cm

Lens #2

F2 = 33 1/3 cm = (100 / 3) cm

3. The attempt at a solution

First, I found the image after the first lens as follows:

1 / di1 = 1 / f1 - 1 / do1 = 1 / 40 cm - 1 / 20 cm = -1 / 40 cm

di1 = -40 cm

Negative Image Distance should imply it is a virtual image on the same side as the object.

The 2nd lens is 10 cm in front of the 1st lens, so the distance between the virtual image and the 2nd lens should be 10 cm + 40 cm = 50 cm

do2 = 50 cm

Now, I attempt to find the image after the 2nd lens as follows:

1 / di2 = 1 / f2 - 1 / do2 = 1 / (100 / 3) cm - 1 / 50 cm = 1 / 100 cm

di2 = 100 cm

Now, I attempted to find the magnification.

M1 = - di1 / do1 = - (-40) / 20 = 2

M2 = - di2 / do2 = - 100 / 50 = -2

Mn = M1 x M2 = 2 x -2 = -4

Does this mean the image should be four times as small?

The answer key states the answer is 24.0 cm. However, if it was four times as small it would be 1.5 cm.

Could somebody point me in the right direction?

Thanks in advance as it is greatly appreciated.

Sincerely,

Travis Walters

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# Homework Help: Lens Optics Question

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