A diverging lens has a focal length of 20.0 cm. Locate the images for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.
I did this using the converging lens. What is the difference when using a diverging lens?
Quick Physics Question? Please Help!?
the image formed by a converging lens is real therefore the formula becomes 1/f=1/p+1/q, however if it is a diverging lens the image ain't real and thus the image distance is virtual also the formula becomes.... -1/f=1/p-1/q
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