Geometric Optics: A Diverging and Converging Lense in Series

1. Nov 3, 2009

cwatki14

A romantic, lighted candle is placed 39 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 21 cm beyond the converging lens. Find the focal length of the diverging lens.

-1/S + 1/S' = 1/f
Here's my approach which led me to the wrong answer:
I started by just looking a the converging lense. The image's location is 21 to the right of the lense, or at +21cm. The primary focal point of the converging lense should be to the left, or at -10cm. Here we solve for S using the equation I listed about.
-1/S=1/f - 1/S'
-1/S= 1/10 - 1/21
So S= -19.09cm

Then I take into account that the diverging and converging lense are 5 cm apart. So I am going to subtract 5 from the S solved about.
So S' for the diverging lense= 14.090cm. Since I have to create a new coordinate system, this image is to the left of the diverging lense, so it is at -14.090cm.
Then I use -1/S + 1/S' = 1/f again to solve for f.
so -1/-39cm + 1/-14.090cm= 1/f
I solved for f, and got f= -22.0598, which makes sense because the image should be closer than the focal point, but it's wrong...

Any ideas?

Last edited: Nov 3, 2009