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Laser Physics - Diameter of a Spot on the Moon from a Laser

  1. May 21, 2017 #1
    Was not sure whether to post here on in the advanced section, since it is part of a final year undergraduate module, yet seems like a pretty simple problem (though I still need help! haha).

    1. The problem statement, all variables and given/known data

    A frequency doubled Nd:YAG laser (##\lambda = 532##nm) with an initial beam radius of 10cm is shone at the moon, what will be the diameter of the spot? Distance to moon is ##3.84 \times 10^8##m

    2. Relevant equations
    ##w^2(z)=w_0^2 \left(1+\frac{z^2 \lambda^2}{\pi^2 w_0^4} \right)##
    ## \frac{\theta}{2} = \frac{w(z)}{z}## for large ##z##

    3. The attempt at a solution
    I first did the problem by just plugging in the numbers into the equation for ##w^2(z)## and taking the square root, because in my notes I have written that ##w(z)## is the beam's radius and got a value of ##8.06##m for ##w(z)##.

    But then I also saw that I have also written in my notes that ##w^2(z)## is called "spot size" and since it has same units as area I then assumed that since ##A=\pi \frac{D^2}{4}## then the diameter must be ##\sqrt{\frac{4w^2(z)}{\pi}}## and got a value of ##9.1##m.

    Also was not sure about the frequency doubled bit of question, I assumed that the (532nm) given for the wavelength had already taken this into account? Perhaps not and I need to half the wavelength?

    I think I probably over thought it a bit much. I have not written my explicit workings as it is just putting in number and the TeX is tedious but will type it up if needed.

    Any help appreciated :)
     
    Last edited: May 21, 2017
  2. jcsd
  3. May 21, 2017 #2

    phyzguy

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    Science Advisor

    (1) w is just the spot diameter, not the spot area.
    (2) You can just use the 532 nm as the wavelength - the frequency doubling is already included in this.

    I think your answer is way off - you probably have not done the units correctly. Why don't you walk us through how you arrived at 8.06 m and let's see where you went wrong.
     
  4. May 21, 2017 #3
    Thanks for the quick reply, I did think it was rather small tbh,

    I did...
    [tex]
    w^2(z)=w_0^2 \left(1+\frac{z^2 \lambda^2}{\pi^2 w_0^4} \right) \\
    w^2(z)=0.1^2 \left(1+\frac{(3.84 \times 10^8)^2 (532 \times 10^{-9})^2}{(\pi)^2 (0.1)^4} \right) = 65 \textrm{m}^2\\
    w = \sqrt{65} = 8.1 \textrm{m}
    [/tex]

    If that is correct, then I must have entered it in my calculator wrong, although I did do it twice,

    EDIT: Yeah I entered it in wrong as I just redid it and got w=650.3m !! I must have forgot to square something in my calculation of ##w^2(z)##
     
  5. May 21, 2017 #4

    phyzguy

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    You're still making a mistake. Just estimating, the term on the right inside the parentheses is about 4x10^7 m^2. How can adding one to it and multiplying by 0.01 give 65?
     
  6. May 21, 2017 #5
    Did you notice my edit in time? I'm now getting ##w^2(z)=422849.6##m^2 and ##w=650##m , but I wanted to double check is w definitely the diameter and not the radius? As I have a sketch of beam divergence in my notes and w seems to be the radius, obviously if so then the diametre is 1.3km.
     
  7. May 21, 2017 #6

    phyzguy

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    Science Advisor

    Yes, since the initial radius is given, then w is the radius, not the diameter like I said earlier. So I think you are doing it right now and 1.3 km is the right answer.
     
  8. May 21, 2017 #7
    Ok thank you for your help. I would have normally caught something so simple r.e. the calculation, but I entered it twice in a row and got 65, so I must have forgot to square one of the terms and made the same omission twice in a row.

    Thanks.
     
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