3) if you have a convex (converging) lens (f#12 cm) that produced an image with a magnification of -2.5, how far must the object be placed from the lens?
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b) Draw a ray-tracing diagram of this situation below (label all points in cm)

Explanation:

step 1. since the magnification is negative this means the object is real and is farther away from the lens than the focal point

step 2. 1/f = 1/o + 1/i where f is the focal length, o is the object distance from the lens and i is the image distance from the lens

step 3. 1/12 = 1/o + 1/i, -i/o = -2.5 (these are our 2 equations or system

step 4. 1/12 = 1/o + 1/2.5o

step 5. o = 12 + 12/2.5 = 16.8cm

step 6. i = 2.5o = 2 5(16) = 42cm

what is the question because that is not a question.