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

• ChrisJ
In summary: 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. In summary, the homework statement is that 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.
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. 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 :)

Last edited:
(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.

ChrisJ
phyzguy said:
(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.

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)##

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?

phyzguy said:
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?

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.

ChrisJ said:
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.

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.

ChrisJ
phyzguy said:
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.

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.

## What is the diameter of a spot created by a laser on the moon?

The diameter of a spot created by a laser on the moon varies depending on the power of the laser and the distance from the laser to the moon. However, on average, the diameter can range from a few kilometers to tens of kilometers.

## What factors affect the diameter of the spot on the moon from a laser?

The main factors that affect the diameter of the spot on the moon from a laser are the power of the laser, the atmospheric conditions, and the distance between the laser and the moon. The higher the power of the laser and the clearer the atmospheric conditions, the smaller the diameter of the spot will be. Additionally, the closer the laser is to the moon, the smaller the diameter of the spot will be.

## Can a laser spot on the moon be seen from Earth?

Yes, under the right conditions, a laser spot on the moon can be seen from Earth. However, it would require a very powerful laser and clear atmospheric conditions to be visible to the naked eye. Most likely, a telescope or other astronomical equipment would be needed to see the spot.

## How accurate is the measurement of the diameter of a spot on the moon from a laser?

The accuracy of the measurement of the spot diameter on the moon can vary depending on the equipment and methods used. However, with advanced technology, it is possible to measure the diameter with an accuracy of a few meters.

## What are the practical applications of using lasers to create a spot on the moon?

The creation of a spot on the moon using a laser has various practical applications, including lunar ranging (measuring the distance between the Earth and the moon), testing theories of gravity, and conducting experiments on the moon's surface. It can also serve as a way to communicate with astronauts or spacecraft on the moon.

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