A diverging lens with a focal length of -18 cm and a converging lens with a foca

Question

A diverging lens with a focal length of -18 cm and a converging lens with a focal length of 15 cm have a common central axis. Their separation is 15 cm. An object with height 2.0 cm is 10 cm in front of the diverging lens, on the common central axis.

The final image is produced by the second lens. Where is this image, as measured from the second lens?

50.00 cm
You are correct.


Take the orientation of the object to be positive.

What is the height of the final image?

-->cant figure this part out

Explanation / Answer

The total magnification is the multiplied result of the individual magnifications of the individual lenses. You already figured out all of the image and object distances to get the final 50 cm distance, now you just have to use the magnification formula for each part

For the diverging M = -q/p = -(-6.43)/(10) = .643

For the converging M = -q/p = -(50)/(21.43) = -2.33

Total magnification is (-2.33)(.643) = -1.50

To find the final height, multiply the magnification time the original height

h' = (-1.5)(2) = -3 cm

The negative is just sign convention letting you know the final image is inverted. So the final height is

3 cm, and the image is inverted

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