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Use the equation, 1/f = 1/di + 1/do

So, di/f = 1+1/m (m = magnitude)

If we are trying to find twice the distance, m=2

Therefore, di/f = 1+1/2

di/f = 1.5

Take the focal length out and plug it into the problem and you have, di = 8.5*1.5

So finally, di = 12.75

The piece of paper would have to be placed at 12.75cm in order for the real image to appear twice as far as the object.

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Q: If you have a concave mirror with focal length of 8.5 cm where would you place a piece of paper so the image projected onto it is twice as far from the mirror as the object is?
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