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Simple Diffraction question

  1. Mar 15, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    A laser is a light source that emits a diffraction-limited beam (like waves diffracting through a wide slit) of diameter 2 mm. Ignoring any scattering due to the earth's atmosphere, calculate how big a light spot would be produced on the surface of the moon, 240,000 miles away. Assume a wavelength of approximately 600 nm.

    2. Relevant equations
    d = 2mm
    L = 240,000 miles
    λ = 600nm

    3. The attempt at a solution

    I am using a small angle approximation where Θ = λ/d from dsinΘ = λ

    And so, converting the proper units, we have
    Θ = (6e-7)/0.002 = 0.0003

    Angle is a dimensionless unit, so this seems to be correct.

    Now, if I wanted to find how big the light spot is, do I simple do
    tanΘ = x/240,000 => 240,000*tan(0.0003) = x = 72 miles

    Looking good?
     
  2. jcsd
  3. Mar 15, 2016 #2

    BvU

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    Is that the angle for the width of the central spot ?
     
  4. Mar 15, 2016 #3

    RJLiberator

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    Yes, I believe that is true. It is the angle for the width of the central maximum.
     
  5. Mar 15, 2016 #4

    BvU

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    These guys think differently. Depends on how you define the width, I suppose. I had in mind this is the expression for the angle for the first minimum and was inclined to multiply by 2.
     
  6. Mar 15, 2016 #5

    RJLiberator

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    So, the 240,000 miles plays no part in this question then? eh?
     
  7. Mar 15, 2016 #6

    RJLiberator

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    Er, my answer seems to be incorrect. A laser would create a 72 mile bright spot on the moon? That doesn't seem reasonable. mmm.
     
  8. Mar 15, 2016 #7

    BvU

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    It does play a role and you did that correctly. My hunch is they want the distance between the two minima on either side. And a 144 miles spot isn't all that big when seen from earth (namely a viewing angle of ## \arctan 0.0006## :smile:).
     
  9. Mar 15, 2016 #8

    RJLiberator

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    BvU, you were absolutely correct. He accepted both answers, but he did mention the factor of 2.

    Thank you for the help.
     
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