Okay for the most part I understand the concepts involved but I'm having some trouble with this question.(adsbygoogle = window.adsbygoogle || []).push({});

The distance between an object and its upright image is 20.0 cm. If the magnification is 0.500, what is the focal length of the lens that is being used to form the image?

Now first I understand since this is an upright image formed (virtual) by the lens, that it is either a Double-Convex (converging) lens with the object between the first focal point (F_1) and the front side of the lens, or it is a Double-Concave (diverging) lens with the object placed anywhere on the front side of the lens.

My guess is that it is a Double Concave lens since the question states the distance between the object and the upright image is +20.0 cm. Since it's positive does this mean that the image formed is to the right (closer to the front of the lens than the object is)?

If so, if it was a -20.0 cm would this mean that it is a Double-Convex (converging) lens with the object between F_1 and the front of the lens, and producing a virtual image to the left of the object?

I'm assuming where I'm confused what type of lens it is in order to use the correct "sign conventions". And the object's distance from the focal point is not given..

What I did was:

M = -q/p

0.500 = -(20.0 cm)/p

p = -(20.0 cm)/0.500

p = -40.0 cm

Also how do I know if 20.0 cm is "p" or "q"? I know that p is the object distance and q is the image distance, but if I have the distance between p and q, how do I know which is which? :yuck:

Then I used 1/p + 1/q = 1/f to find the focal point.

Any help is greatly appreciated.

Thanks in advance.

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# Homework Help: Sign conventions of Lenses

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