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Optics Problems

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    When one mirror of a michelson interforemeter is translated by 0.0114cm, 523 fringes are observed. Calculate the wavelength of the light.


    2. Relevant equations

    wavelength = 2dm/m

    3. The attempt at a solution

    wavelength = 2x0.0114 / 523
    = 4.39 ^10

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    1. The problem statement, all variables and given/known data

    An object 2.5cm high is placed 12 cm in front of a thin lens focal length +3cm. How can i work out the image distance, and the magnification, and nature of the image.

    2. Relevant equations

    1/f = 1/v - 1/u

    M = v/u

    3. The attempt at a solution

    Image distance i have worked out to be 4cm.

    Magnification = v/u = 4/12 = 0.33?

    I am unsure what a "nature of the image means"


    Thanks for any help, is much appreciated
     
  2. jcsd
  3. Mar 4, 2009 #2

    lanedance

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    Homework Helper

    try drawing a ray diagram to scale to check your calcs, easy & quick plus helps you understand what the equation is doing

    the nature of the image usually consists of 2 things

    inverted - upside down relative to object
    real or virtual - reall means it is on the opposite side of the lens form the object

    once again they will be easily apprent from your ray diagram & once you have done a few of these wioll have a feel for what the nature of the image will be
     
  4. Mar 5, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Welcome to PF :smile:

    Looks good so far ...
    Hmmm, looks like you entered something wrong into your calculator. Does it make sense for a wavelength of light to be such a large number? Also ... what are the units here???

    Should be
    1/f = 1/v + 1/u
    but it was probably a simple typo, as you got the correct answer of 4 cm.

    Should be
    M = -v/u

    and a negative answer will indicate that the image is inverted.

    Yes, good.
    Almost correct, just need the negative sign in the formula :smile:

    They probably want to know if the image is real or virtual ... your text book or class notes should discuss what that means ...
     
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