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Mirror question

  1. May 24, 2005 #1
    Hi!

    I am having trouble with this question:

    You are trying to photograph a bird sitting on a tree branch, but a tall hedge is blocking your view. However, as the drawing shows, a plane mirror reflects light from the bird into your camera. If x = 4.3 m and y = 4.5 m in the drawing, for what distance must you set the focus of the camera lens in order to snap a sharp picture of the bird's image?

    I know that the angle of incidence and reflection are the same. I have tried making right triangles, etc, but I can't get the right answer :(
     

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  3. May 24, 2005 #2

    quasar987

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    In words, what do you think the correct lenght is?
     
  4. May 24, 2005 #3
    i think the length is equal to the hypo. of the right triangle formed between the camera and the mirror
     
  5. May 24, 2005 #4

    quasar987

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    I think so too.

    To find this lenght, set the unknown side of the small triangle (not the hypothenuse, the other one) equal to z. Find two different expressions for [itex]tan\theta[/itex] (the reflection angle). Compare them to find z, and then find theta, and then find the answer.
     
    Last edited: May 24, 2005
  6. May 24, 2005 #5
    so, would one expression be tan (theta) = z / 4.3 ?

    how do you find the other expression?
     
  7. May 24, 2005 #6

    quasar987

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    What you wrote is good for z = the "opposite side" of the LARGE triangle. Set z equal to the "opposite side" of the small one.

    The one whose corners are bird-obstacle-mirror
     
  8. May 24, 2005 #7
    so for the small triangle it would be tan (theta) = z / 2.1 ?
     
  9. May 24, 2005 #8

    quasar987

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    yeah

    for the other one, I'm sure you can find it, since you can easily express the "opposite side" of the large triangle in terms of z too.
     
  10. May 24, 2005 #9
    here is what i tried:

    tan (theta) = (4.5 - z)/2.1

    tan (theta) = z / 4.3

    I found z to be 3.02, and theta to equal 35.1

    then I found the hypo. to be 5.25, I submitted this answer but it is not the correct one, where am I going wrong?
     
  11. May 24, 2005 #10

    OlderDan

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    So far you have been working with two different z. It would be best to give them different labels. Perhaps z_u for the upper triangle and z_l for the lower triangle. Since you have two expressions for tan (theta) you can set them equal. That will give you a relationship between z_u and z_l. You need another relationship to solve for the z values. What is their sum?

    You can do this problem a different way if you know how a plane mirror forms images. Where is the image of the bird? How far is it from the camera?

    Edit

    I had not seen your latest post when I posted this. I see you now have taken the different z into consideration. Looks like you have only found one hypotenuse. You need them both.
     
    Last edited: May 24, 2005
  12. May 24, 2005 #11

    quasar987

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    The problem is in your first equation. Hint: you got the numerator right. lol
     
  13. May 24, 2005 #12
    the image of the bird is located the same distance in front of the mirror as it is behind the mirror - 2.1 m

    is that right?
     
  14. May 24, 2005 #13
    i don't understand what is wrong the equation, this is for the small triangle (bird-hedge-mirror) right?
     
  15. May 24, 2005 #14

    quasar987

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    Oops, I hadn't noticed that you switched the definition of z to "size of opposite side of large triangle". Ok, then I would say everything is okay...

    Do you see what's wrong OlderDan?
     
  16. May 24, 2005 #15

    OlderDan

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    The distance from the camera to the bird is the sum of the two hypotenuses. You need to find both of them and add. It appears to me only one has been calculated so far.
     
  17. May 24, 2005 #16
    so would it be the larger hypo. (5.25)

    and the smaller hypo. (2.57)

    added together (7.82)-- is that right?
     
  18. May 24, 2005 #17

    quasar987

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    Ooh.. we though the answer would simply be the hypo. of the large triangle.
     
  19. May 24, 2005 #18
    we have 9 tries to submit answers, i submitted 7.82 as my last try and it was right

    THANKS so much for everyones help, physics can drive me crazy sometimes!
     
  20. May 24, 2005 #19

    OlderDan

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    Looks good, but be sure to include your units. You could also get this result from

    [tex]\sqrt{(x + 2.1m)^2 + y^2}[/tex]

    because the image of the bird would be on the other side of the mirror at a distance of 2.1m. One big triangle with the hypotenuse from the camera to the bird's image would give this result
     
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