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Relation between RMS Spot Radius and Beam Size

zuzhen

Hi everyone,

 

 

I got the following, most simple setup consisting of only the gaussian beam source and a rectangular detector.

With a beam size of 0.15mm I expected to see the same as RMS Spot Radius on the detector viewer. But the detector displays RMS Spot Radius 0,10606mm which is 1/sqrt(2) of the given beam size.

 

 

The beam size radius is where the residual intensity account for 13.5% of the peak intensity. It should therefore be identical to the RMS Spot Radius.

I just can’t wrap my head around as why the RMS Spot Radius is 1/sqrt(2) of the beam size, can anyone explain?

Is there a relation between RMS Spot Radius and Gaussian beam size that I am missing?

Sincerely,

 

Zu

Best answer by Kevin Scales

It sounds like the question here is how the RMS X or Y relates to the RMS R value. The NSDD help page gives us a useful starting point: the definition of the X RMS as the square root of the variance, which is all presented like this:

The Y values are defined analogously but the R values cannot be. The relation between X or Y and R is not linear and there is a circular distribution of spots (in this case). Consider a simple layout with several points in a half-circle, all at radius 1:

It is easy to calculate that R* in that case would be (1+1+1+1+1)/5 = 1, and the variance would be (0+0+0+0+0)/5 = 0, but this is not the actual variance and the RMS comes out to .876 in this example.

The formulas for 1/e^2 and similar comes from RMS standard values for Gaussian distributions, where it turns out that for W as the 1/e^2 beam waist and sigma being the RMS beam size, W=2sigma. In the original example of this thread, the beam size is .15 and a cross section of a detector shows that at .15 the intensity has dropped by that amount. The RMS of R is, in fact, going to be equal to sqrt(RMSX^2 + RMSY^2).

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zuzhen
  • zuzhen
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  • 4 replies
  • August 2, 2023

Could it be that what is displayed at “Beam Info”  as RMS Spot Radius is the 1/e radius rather than 1/e² radius?

Benjamin.N
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  • Benjamin.N
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  • August 3, 2023

Hi zuzhen,

the 1/e^2 radius of a Gaussian distribution exp(-.5*(x/RMS)^2) is at 2 * RMS. Therefore I would expect a factor of 2 instead of sqrt(2): beam size = 2 * RMS

Interestingly, this is consistent with the RMS spot sizes along the x- and y-direction, which are approximately half the beam size. I don’t understand why the RMS spot radius is different from the radii along the individual axes, as the distribution has circular symmetry.

Best regards
Benjamin

O
  • Oran
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  • 22 replies
  • August 3, 2023

You have RMS X = RMS Y≈  0.075 = 0.15 / 2

RMS R = sqrt((RMS X)^2 + (RMS Y)^2) ≈ sqrt(2 * 0.0749) = 0.106

I don’t see the issue, I think that here RMS R is different than what you expected?

Also, as a method to calculate spot size, the RMS given by OpticStudio is the standard deviation (sigma). thus the spot diameter in each axis should be calculated as 4sigma = 4*RMS. this method is standard and will apply to different beam shapes gaussian beams.

 

As for the definition of “Beam size” of your source- it may get confusing and non-consistent so I recommend having a look at the Help interface and lookup your desired object and see detailed explanations for each parameter field.

 

 

Benjamin.N
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  • Benjamin.N
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  • August 3, 2023

Hi Oran,

I understand how you connect the RMS radii along the x- and y-direction to the RMS spot radius. Does Zemax always define the radial spot size as the square root of the quadratic sum of the spot sizes along x and y?


In particular for a Gaussian intensity distribution, I find this definition to be confusing. In this case, the intensity distribution is proportional to exp(-.5*(x^2+y^2)/s^2). Hence, you get the same RMS radius = s along both the x- and y-direction. Radially, i.e. for r = sqrt(x^2 + y^2), you would also get RMS = s, because the distribution can be expressed as exp(-.5*r^2/s^2). This makes sense because the distribution has circular symmetry and therefore the spot size along any axis should be the same.

According to the Zemax manual, at least the NSDD operator seems to agree with this definition:

Best regards
Benjamin

O
  • Oran
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  • 22 replies
  • August 3, 2023
Benjamin.N wrote:

Hi Oran,

I understand how you connect the RMS radii along the x- and y-direction to the RMS spot radius. Does Zemax always define the radial spot size as the square root of the quadratic sum of the spot sizes along x and y?


In particular for a Gaussian intensity distribution, I find this definition to be confusing. In this case, the intensity distribution is proportional to exp(-.5*(x^2+y^2)/s^2). Hence, you get the same RMS radius = s along both the x- and y-direction. Radially, i.e. for r = sqrt(x^2 + y^2), you would also get RMS = s, because the distribution can be expressed as exp(-.5*r^2/s^2). This makes sense because the distribution has circular symmetry and therefore the spot size along any axis should be the same.

According to the Zemax manual, at least the NSDD operator seems to agree with this definition:

Best regards
Benjamin

Hi Benjamin,

 

Yes, this is always how OpticStudio calculates the RMS Radius, at least to my knowledge in NSC mode. (maybe other users/ Zemax staff would provide a more accurate definition)

 

What your saying is you’d expect that:

(RMS R)^2 = (RMS X)^2 + (RMS Y )^2. let’s say RMS = A

→ A^2 = 2A^2. this equation has only a trivial solution.

I hope that this example along with the explanation of how it is calculated will help you understand the definitions.

 

 

Benjamin.N
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  • Benjamin.N
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  • August 3, 2023

Hi Oran,

I think there’s a misunderstanding. I am not claiming that “(RMS R)^2=(RMX X)^2+(RMS Y)^2” but rather that I would expect RMS R = RMS X = RMS Y for a Gaussian intensity distribution if the RMS values are calculated according to the definition provided with the NSDD operator.

Best regards
Benjamin

O
  • Oran
  • Visible
  • 22 replies
  • August 3, 2023
Benjamin.N wrote:

Hi Oran,

I think there’s a misunderstanding. I am not claiming that “(RMS R)^2=(RMX X)^2+(RMS Y)^2” but rather that I would expect RMS R = RMS X = RMS Y for a Gaussian intensity distribution if the RMS values are calculated according to the definition provided with the NSDD operator.

Best regards
Benjamin

Hi Benjamin,

 

Actually the truth is that (RMS R)^2=(RMX X)^2+(RMS Y)^2 holds, and if you expect RMS R = RMS X = RMS Y you’ll see in the equation in my previous message that if both terms are true then RMS=0.

 

Oran 

zuzhen
  • zuzhen
  • Author
  • Student
  • 4 replies
  • August 3, 2023
Oran wrote:

You have RMS X = RMS Y≈  0.075 = 0.15 / 2

RMS R = sqrt((RMS X)^2 + (RMS Y)^2) ≈ sqrt(2 * 0.0749) = 0.106

I don’t see the issue, I think that here RMS R is different than what you expected?

Also, as a method to calculate spot size, the RMS given by OpticStudio is the standard deviation (sigma). thus the spot diameter in each axis should be calculated as 4sigma = 4*RMS. this method is standard and will apply to different beam shapes gaussian beams.

 

As for the definition of “Beam size” of your source- it may get confusing and non-consistent so I recommend having a look at the Help interface and lookup your desired object and see detailed explanations for each parameter field.

 

 

Hi Oran, 

can you explain why rms_x = 1/e2-beam size(0.15)/2?

 

I totally get why rms_r = sqrt(2) * rms_x,y as rms r represents the radial distribution using the Pythagoras.

 

So my question remains why rms_r =/= 1/e2-beam size (0.15) = w_0 because as Benjamin correctly stated, the rms is by definition the standard deviation s. The intensity of the gaussian beam is I(r=w_0) =0,135. Also the beam size parameter in Zemax OpticStudio defined as the radius where there is still 13,5% of the peak intensity - so basically rms_r. 

O
  • Oran
  • Visible
  • 22 replies
  • August 3, 2023
zuzhen wrote:
Oran wrote:

You have RMS X = RMS Y≈  0.075 = 0.15 / 2

RMS R = sqrt((RMS X)^2 + (RMS Y)^2) ≈ sqrt(2 * 0.0749) = 0.106

I don’t see the issue, I think that here RMS R is different than what you expected?

Also, as a method to calculate spot size, the RMS given by OpticStudio is the standard deviation (sigma). thus the spot diameter in each axis should be calculated as 4sigma = 4*RMS. this method is standard and will apply to different beam shapes gaussian beams.

 

As for the definition of “Beam size” of your source- it may get confusing and non-consistent so I recommend having a look at the Help interface and lookup your desired object and see detailed explanations for each parameter field.

 

 

Hi Oran, 

can you explain why rms_x = 1/e2-beam size(0.15)/2?

 

I totally get why rms_r = sqrt(2) * rms_x,y as rms r represents the radial distribution using the Pythagoras.

 

So my question remains why rms_r =/= 1/e2-beam size (0.15) = w_0 because as Benjamin correctly stated, the rms is by definition the standard deviation s. The intensity of the gaussian beam is I(r=w_0) =0,135. Also the beam size parameter in Zemax OpticStudio defined as the radius where there is still 13,5% of the peak intensity - so basically rms_r. 

 

I think that the root of your confusions is that Zemax defines RMS Spot Radius differently than what you’d expect.

Maybe some Zemax staff could shed more light to solve your confusion.

Kevin Scales
En-Lightened
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  • Kevin Scales
  • En-Lightened
  • 185 replies
  • Answer
  • August 21, 2023

It sounds like the question here is how the RMS X or Y relates to the RMS R value. The NSDD help page gives us a useful starting point: the definition of the X RMS as the square root of the variance, which is all presented like this:

The Y values are defined analogously but the R values cannot be. The relation between X or Y and R is not linear and there is a circular distribution of spots (in this case). Consider a simple layout with several points in a half-circle, all at radius 1:

It is easy to calculate that R* in that case would be (1+1+1+1+1)/5 = 1, and the variance would be (0+0+0+0+0)/5 = 0, but this is not the actual variance and the RMS comes out to .876 in this example.

The formulas for 1/e^2 and similar comes from RMS standard values for Gaussian distributions, where it turns out that for W as the 1/e^2 beam waist and sigma being the RMS beam size, W=2sigma. In the original example of this thread, the beam size is .15 and a cross section of a detector shows that at .15 the intensity has dropped by that amount. The RMS of R is, in fact, going to be equal to sqrt(RMSX^2 + RMSY^2).

Kevin Scales

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