Homework Help: Laser Physics - Diameter of a Spot on the Moon from a Laser

1. May 21, 2017

ChrisJ

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. May 21, 2017

phyzguy

(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.

3. May 21, 2017

ChrisJ

Thanks for the quick reply, I did think it was rather small tbh,

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

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 $w^2(z)$

4. May 21, 2017

phyzguy

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?

5. May 21, 2017

ChrisJ

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.

6. May 21, 2017

phyzguy

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.

7. May 21, 2017

ChrisJ

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.