Thursday, February 16, 2006

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.

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