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Resolution physics help

  1. Apr 23, 2006 #1
    the headlights of a car are seperated by a distance of 1.4m. at what distance would these be resolved as 2 seperate sources by a lens of diameter 5cm if a wavelength of 500nm is being used ?

    using theta=1.22 * wavelenth / D I am stuck as I dont know what angle theta is or how to work it out.
     
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  3. Apr 23, 2006 #2

    Hootenanny

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    You need to find theta using the formula you stated above.

    ~H
     
  4. Apr 23, 2006 #3
    i have tried that but it doesnt work.
     
  5. Apr 23, 2006 #4

    Hootenanny

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    Well, can you show me your working and then perhaps I can point out where you have gone wrong?

    ~H
     
  6. Apr 23, 2006 #5

    Gokul43201

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    Did you draw a picture ? Theta is the angle subtended by the object (ie: the 2 headlights) at the center of the lens.
     
  7. Apr 23, 2006 #6
    using theta= 1.22 wavelength / b gives 1.22*(500*10^-9) / 0.05 = 1.22*10^-5
    so 1.22*10^-6 = 1.22 wavelength / D

    so D = 1.22*(500*10^-9) / (1.22*10^-5) = 0.05

    the answer i have found in the book is 115km so I am stuck as how to get to that
     
  8. Apr 23, 2006 #7

    nrqed

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    Usually, the angle is small enough that one uses [itex] \theta \approx tan(\theta) =[/itex] separation between the sources /distance to the sources
     
  9. Apr 23, 2006 #8

    Hootenanny

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    D is the thinkness of the lense. Once you have obtained theta, you need to use trig to find the distance. (As Gokul hinted at)

    ~H
     
  10. Apr 23, 2006 #9

    nrqed

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    Yu are going in circle!! Of course you will find D=0.05, this is what you used to start with! (I don't know why your exponent changed from -5 to -6 though)

    You need another equation relating to the distance to the sources (which is NOT D) and I gave it in my other post.
     
  11. Apr 23, 2006 #10

    Gokul43201

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    When you are given an equation, the first thing you must understand id what the various terms in the equation represent. You are using the same equation with two different meanings attached to the same term (D), and hence getting the same value for two different quantities.

    If you merely attempt to juggle numbers in an equation so as to end up with the number that is the answer, you will gain no understanding of the physics involved.

    In the equation you've written down, what do the different symbols represent ?
     
    Last edited: Apr 23, 2006
  12. Apr 23, 2006 #11

    nrqed

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    Just to avoid possible confusion for the OP: D is the diamter of the circular slit used (where slit may stand for the pupil of a person or a telecope lens or the slit of a pinhole camera)
     
  13. Apr 23, 2006 #12

    Hootenanny

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    Sorry, thinkess was the wrong word. I apologise.

    ~H
     
  14. Apr 23, 2006 #13
    D is the distance from the headlights to the lens and b is the diameter of the lens
     
  15. Apr 23, 2006 #14

    Hootenanny

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    So we're all singing off the same hym sheet;

    [tex]\theta = \frac{1.22\lambda}{D}[/tex]

    Where D is the diameter of the lense.

    Your other equation D = 1.22*(500*10^-9) / (1.22*10^-5) is wrong.

    Try drawing a diagram like Gokul suggests. HINT: Use triangles.

    ~H
     
  16. Apr 23, 2006 #15

    Gokul43201

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    Do you not see the problem with that ? Okay, let's start over :

    1. Write down the equation from the OP.

    2. An equation is nothing but a statement relating various quantites, expressed in a concise mathematical fashion, using symbols to represent these quantities. In the equation above, what quantities do the different symbols represent ?

    3. Now using these quantities, rewrite the statement of the equation in the form of a sentence.

    You need to do this with every equation you come across, else you have not really understood what it is saying.
     
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