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Homework Help: Concave mirror problem

  1. Jun 3, 2004 #1
    Well, here is the problem:

    A girl has her 2-cm high eye 12 cm directly in front of a concave mascara mirror or focal length 18 cm. Where is the object located, and what is the magnification?

    I'm really lost on this one.

    EDIT: Sorry, I just realized this would be better suited for the homework help k12 forum.
  2. jcsd
  3. Jun 3, 2004 #2

    Doc Al

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    Staff: Mentor

    mirror equation

    I presume that her eye is the object and that you are trying to find where the image is.

    You'll need the lens/mirror equation:
    [tex]\frac{1}{f} = \frac{1}{o} + \frac{1}{i}[/tex]

    And you'll also need the linear magnification:
    [tex]M = -\frac{i}{o}[/tex]

    To understand what these mean, and how to use the sign conventions, read your text. Give it a try.
  4. Jun 3, 2004 #3
    Just in case your text uses the same variable names as the one we used to study this, Doc Al' [tex]o[/tex] signifies the distance of the object from the mirror and [tex]i[/tex] signifies the distance of the image from the mirror. Sometimes they are replaced by [tex]u[/tex] and [tex]v[/tex] repsectively, so your formula might look like this:

    [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
  5. Jun 3, 2004 #4
    Okay, thanks chen and doc al.
    I didn't have those equations, and didn't have access to the text (end of the year, books had been turned in already).

    So I did use the right equation...good.

    Thanks Chen for the variables thing, honestly, thanks.
  6. Jun 4, 2004 #5
    No problem. Since you already returned your books, you might want to a take a look at HyperPhysics:
    It provides a clear and concise explanation of many basic subjects in physics (and math).
  7. Jun 4, 2004 #6
    Don't forget, when doing concave mirrors/converging lens, focal length is always positive.
    When doing convex mirrors/diverging lens, focal length is always negative.

    *Just in case you were doing similar problems involving mirrosrs
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