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Concave mirror problem!

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A light bulb is placed 10.7 cm in front of a concave mirror. When the concave mirror is replaced by a plane mirror in the same location, the bulb's image moves 4.00 cm closer to the mirror. Calculate the focal length of the concave mirror.

    I am given: do: 10.7
    and 4.00 is Di


    2. Relevant equations
    Mirror equation: 1/di+1/do=1/f


    3. The attempt at a solution

    I know I have to use the mirror equation so I did. However, I also added 10.7 and 4.00 to get a new di.

    1/10.7+1/14.7=1/f
    Eventually, I get 6.19 which is wrong! Anyone would like to help me see what is wrong?
     
  2. jcsd
  3. Feb 1, 2015 #2

    haruspex

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    Which side of the mirror is the new image? (It's not made clear.)
    What's the sign convention for object and image distances when dealing with mirrors?
    (What would your equation produce for the flat mirror?)
     
  4. Feb 1, 2015 #3

    The only information I am given is the one I have provided. I am told it is a concave mirror- that's it.
     
  5. Feb 1, 2015 #4

    haruspex

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    In order to write the equation you used, you have to make an assumption about whether the object and image are on the same side of the mirror or on opposite sides. Which did you assume? Is your equation correct for that assumption?
     
  6. Feb 1, 2015 #5
    I was thinking the image should be real and on the same side which is why my Di would not be negative.1/10.7+1/14.7=1/f which is why I get
    eventually, I get 6.19. Are you trying to imply that my di is negative and my equation should be 1/10.7-1/14.7=1/f
     
  7. Feb 1, 2015 #6

    haruspex

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    I'm saying it's a strong possibility. Note that the question says
    It doesn't say "is 4cm closer" or "moves to 4cm closer and on the opposite side". That suggests to me it moves by 4cm, not by 10.7+14.7=25.4 cm.
     
  8. Feb 1, 2015 #7
    So you would think it be a simple mirror equation then: 1/10.7+1/4.0=1/f so 2.94
     
  9. Feb 1, 2015 #8
    but 2.94 cm is off. I am so confused!
     
  10. Feb 1, 2015 #9

    haruspex

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    No, we're discussing what sign is appropriate to use in your original equation. The distances are correct, but are the image and object the same sign or opposite signs?
     
  11. Feb 1, 2015 #10
    I think they will be the same size
     
  12. Feb 1, 2015 #11
    I think they will be the same sign.
     
  13. Feb 1, 2015 #12

    haruspex

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    Why?
     
  14. Feb 1, 2015 #13
    Because it's a concave mirror. The image is also real. Earlier I was thinking they would be different signs as 4.00 cm is moving closer. But then the concave part made me think the sign would be the same.
     
  15. Feb 1, 2015 #14

    haruspex

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    Read http://en.wikipedia.org/wiki/Curved_mirror#Image_2
     
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