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Square source - how to get rays to spread


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  • Infrared
  • 7 replies

Hi

I have a defined a source rectangle in OpticStudio, and am trying to analyse the rays that it will produce.

When I analyse it, the rays come out in a straight line orthogonal to the plane of the source, which is not what I expected, as I expected to see rays emanating out to the sides as well.  It is not a laser light, and I expected some diffusion to occur.

What do I need to do to see these rays?

Many thanks

Best Regards

Jo Ling

 

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Best answer by David

Hi Jo Ling,

The Help documentation for Source Rectangle references Source Ellipse for defining the angular distribution. Three methods are offered: 1) rays appear to emit from a point , 2) a Cosine^n distribution, 3)  a distribution which is Gaussian in the x and y direction cosines which are l and m. If you follow the directions given, any of these distributions will produce rays which are not collimated.

I often use a Gaussian distribution. From the given formula, I = I0 Exp[-(Gx l^2+Gy m^2)] it is possible to show that  if Gx = Gy = G then the distribution is radially symmetric and that if G = 2 / Sin(theta)^2 then the intensity falls to 1/E^2 at an angle theta from the z axis. Further, the distribution is approximately Gaussian in theta for small angles.

Kind regards,

David

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David
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  • Luminary
  • 336 replies
  • Answer
  • May 16, 2021

Hi Jo Ling,

The Help documentation for Source Rectangle references Source Ellipse for defining the angular distribution. Three methods are offered: 1) rays appear to emit from a point , 2) a Cosine^n distribution, 3)  a distribution which is Gaussian in the x and y direction cosines which are l and m. If you follow the directions given, any of these distributions will produce rays which are not collimated.

I often use a Gaussian distribution. From the given formula, I = I0 Exp[-(Gx l^2+Gy m^2)] it is possible to show that  if Gx = Gy = G then the distribution is radially symmetric and that if G = 2 / Sin(theta)^2 then the intensity falls to 1/E^2 at an angle theta from the z axis. Further, the distribution is approximately Gaussian in theta for small angles.

Kind regards,

David


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  • Author
  • Infrared
  • 7 replies
  • May 19, 2021

Hi David.

Thank you for this; I will try the Gaussian distribution for my analysis.

Best Regards

Jo


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