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Homework Help: Lateral Magnification

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data

    If lateral magnification is negative what does that mean?

    2. Relevant equations

    m = -di/do

    3. The attempt at a solution

    From what I understand that focal distance for converging devices (concave mirrors and convex lens) is always positive. So this would make m="-" since di= "+". Also that a negative lateral magnification would mean less power. My question are:

    1) Is the above correct?

    2) If the object was within the focal point the image would be virtual and upright, what would happen to power then?

    3) If I increase the focal distance would that mean I would decrease the power of the lens or mirror based on this formula: P=1/f?

    4) Based on this formula: P=1/f = 1/di + 1/do, the farther the object is from the lens or mirror the greater power is?
  2. jcsd
  3. Aug 10, 2008 #2


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    Homework Helper

    For a single lens or mirror: When [itex]d_i[/itex] is positive, then [itex]m[/itex] would be negative; but [itex]d_i[/itex] can be negative, in which case [itex]m[/itex] would turn out to be a positive number.

    If you have a system of lenses or mirrors, then [itex]d_o[/itex] can also be negative.

    The power of a lens is a property of that lens; it's the inverse of the focal length (having units of diopters when focal length is in meters). So the power of a particular lens is a constant.

    Just to repeat, the power would be the same.

    However, go ahead and try out this case. Set [itex]f=10[/itex] and [itex]d_o[/itex] be anything less than 10, for example. Solve for [itex]d_i[/itex], and then [itex]m[/itex]. What is the sign of [itex]m[/itex]?

    Do the same for [itex]f=10[/itex] and do be anything greater than 10, and find the sign of [itex]m[/itex].

    So what does the sign of [itex]m[/itex] mean?

    Okay, but to change the focal length you have to change the radius of curvature of the sides of the lenses; so effectively you have a different lens altogether.

    Do you now see why this is not true?
  4. Aug 10, 2008 #3
    Thanks alphysicist. I see what you are saying.
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