- #1

ChrisJ

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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).

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

##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##

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 :)

1. Homework Statement1. Homework Statement

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

## Homework 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##

## 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 :)

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