Mirrors and Magnification

Real and Virtual Images

Q) Two students are performing an experiment where they are asked to find the image position relative to a concave mirror (radius of curvature 0.5m) that creates an image three times the size of the object. Afterwards, they realize that they have each got different answers – how can they both be correct?

A) The solution might lie in the fact that the sign of the magnification was not specified – the three times larger image could have been either upright or inverted – as long as we can find object positions that yield both these results:

M=+/- 3 = -i/o

i = +/- 3 o

Using the “mirror maker’s equation� (the mirror is converging, thus the focal length is positive)

1/f (= 2/r = 4) = 1/i + 1/o

= 1/i +/- 1/(3o)

= (3+/-1)/3o

So, if o(+/-) are the plus and minus solutions:

4 = 4/(3o(+)) or 4=2/(3o(-))

o(+)=1/3m or o(-)=1/6m

These two object positions thus give our two three times larger images: o(+) corresponds to i(+)=3o(+) – a positive image position and thus a real image (with M=-3). o(-) corresponds to i(-)=3o(-) – a negative image position and thus a virtual image (with M=+3). So we have found that there is a way for the two students to be correct with different answers in this case.