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Optics Homework- 11th grade

  1. Jun 10, 2014 #1
    So I originally posted a thread for my optics homework listing every question from a worksheet AND my attempts. Apparently, it's seen as spam so I have tried and am now posting the second question.

    1. The problem statement, all variables and given/known data
    A concave mirror has a focal length of 40 cm. Determine the object position for which the resulting image is upright and four times the size of the object. Construct a ray diagram for this situation.

    GIVEN DATA
    Focal length= 40 c.m


    2. Relevant equations
    (1/do)+(1/di)=1/f

    3. The attempt at a solution

    I can't really attempt it because I barely have any information. How can I determine the object position with only the focal length.

    And what exactly are do and di?
     
  2. jcsd
  3. Jun 10, 2014 #2

    BiGyElLoWhAt

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    What happens to the light as it passes through the mirror?
     
  4. Jun 10, 2014 #3

    BiGyElLoWhAt

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    lens*
     
  5. Jun 10, 2014 #4
    elementozzy
    do and di usually mean the distance from the object to the vertex of the mirror and the distance of the image to the vertex of the mirror, respectively.

    Try something now.

    Also, do you know how the relation between do and di is related to the ratio between the height of the object and the height of its image?


    BiGyeILoWhAt
    Why did you "correct" your post to lens? You could just have edited your first post.
    Also, according to what the problem states, it is a mirror.
     
  6. Jun 10, 2014 #5

    BiGyElLoWhAt

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    Lol yea I did, I don't know why, I was at work, and posted in a hurry, I guess I thought I misread it. Oh well.
     
  7. Jun 12, 2014 #6
    Well I'm attempting to first fill in the equations that I have with as much data as possible.

    Also, I forgot to add in another equation. m= -di/do

    So I now have

    (1/do)+(1/di)= 1/40

    and since the magnification is 4x ("four times the size of the object")

    I have 4= -di/do

    what now?
     
  8. Jun 12, 2014 #7
    Oh!!!

    Maybe, do I set up a system of equations?

    Just tried that----- came up with nothing.....
     
  9. Jun 12, 2014 #8
    About the system of equations. It is the right way to go.

    ALSO, you have 2 different equations and 2 variables to solve for. Just go for it, everything you posted is correct.
     
  10. Jun 17, 2014 #9
  11. Jun 17, 2014 #10
    You should definitely learn how to solve a system simple as that.

    It is for your own good. It is not a matter of ours.

    [tex]\begin{cases}\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{40}\\4=-\frac{d_i}{d_o}\end{cases}[/tex]
    [tex]4=-\frac{d_i}{d_o}\Rightarrow{}d_i=-4d_o[/tex]
    [tex]\frac{1}{d_o}-\frac{1}{4d_o}=\frac{1}{40}[/tex]
    [tex]\frac{3}{4d_o}=\frac{1}{40}\Rightarrow{}4d_o=120\Rightarrow{}d_o=30[/tex]
    [tex]d_i=-4(30)=-120[/tex]

    Then, give the answer with the proper dimensions.

    [tex]\begin{cases}d_o=30\;\text{cm}\\d_i=-120\;\text{cm}\end{cases}[/tex]
     
  12. Jun 18, 2014 #11
    Thanks but I'm not sure how to apporach the second part of the question.....

    "Construct a ray diagram for this situation."
     
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