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Math Help - mirrors

  1. #1
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    Smile mirrors

    okay here are the equations;

    An object is 17.3 cm from a concave mirror of focal length 8.10 cm. (a) What is the image distance? State your answer with the correct sign. (b) What is the magnification? (c) Is the image upright or inverted? (i know what the answer is, i just dont get how to get it.)

    You wish to use a concave mirror to produce a virtual image that is three times the size of an object. (a) Which will be closer to the mirror, the image or the object? (b) If the object is 22.0 cm from the mirror at this magnification, what is the radius of curvature of the mirror?

    last one;

    A mirror of focal length 26.6 cm creates an image with magnification −0.350. What is the image distance?

    if anyone could explain how to do it to me then that would be great!
    many thnxs to those who help!!! i really appreciate it.
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  2. #2
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    Quote Originally Posted by soojin89 View Post
    okay here are the equations; .... which equations?


    An object is 17.3 cm from a concave mirror of focal length 8.10 cm. (a) What is the image distance? State your answer with the correct sign. (b) What is the magnification? (c) Is the image upright or inverted? (i know what the answer is, i just dont get how to get it.)

    ...
    Let o denote the object distance, O the size of the object, i the image distance, I the size of the image and f the focal distance. Then you have to use the equations:

    \dfrac1o+\dfrac1i=\dfrac1f

    and

    \dfrac IO = \dfrac io

    Plug in the values you know:

    \dfrac1{17.3}+\dfrac1i=\dfrac1{8.1}~\implies~i\app  rox 15.2

    m = \dfrac{15.2}{17.3} \approx 0.8786
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  3. #3
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    Quote Originally Posted by soojin89 View Post
    ...

    You wish to use a concave mirror to produce a virtual image that is three times the size of an object. (a) Which will be closer to the mirror, the image or the object? (b) If the object is 22.0 cm from the mirror at this magnification, what is the radius of curvature of the mirror?

    ...
    If the image is virtual then i must be negative. From the question you know:

    \dfrac{-i}{o} = 3~\implies~i=-3o . Therefore i = -66 cm.

    Use the first formula of my previous post:

    \dfrac1{22}+\dfrac1{-66} = \dfrac1f~\implies~ f = 33

    A spheric mirror has (under certain conditions!) the property:

    r = 2f

    Therefore r = 66 cm
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