Question

Modeling night sky radiance in OpticStudio?

  • 6 June 2023
  • 3 replies
  • 58 views
Mike.Jones
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I could use a little radiometry help here.  Night sky luminance can be quantified in terms of visual magnitude/arcsecond².  Pristinely dark skies with no light pollution or airborne dust are in the range of 22.0 mag/arcsec² (~172 μcd/m²).  The Milky Way in a clear moonless sky has a luminance of around 19.6 mag/arcsec² (~1561 μcd/m²).  Light-polluted skies can have a luminance of 18 mag/arcsec² (~6814 µcd/m²) or worse. The formula to convert mag/arcsec² to µcd/m² is 

value in µcd/m² = (1.08E+11) × 10 ^ [-0.4*(value in mag/arcsec²) ]

1 cd = 1/683 watt/ster at 555nm.  In non-sequential mode, how would one model a patch of the night sky based on its mag/arcsec²?  I’m assuming one would use a Source Ellipse of some diameter at a negative distance from an optical aperture to define the solid angle, and with the Cosine Exponent set to 1.0 to give a Lambertian luminance pattern.  Say I set the number of analysis rays to 1E7.  What formula gives me the power in watts for a given number of analysis rays to correctly simulate a given sky luminance in mag/arcsec²?

 

 

3 replies

Jeff.Wilde
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Can’t you just use a Lambertian source with uniform irradiance that has the same shape and size as your entrance aperture (say a Source Ellipse), located at the aperture, shooting rays directly into your optical system?  Following Smith, if the source has a uniform radiance of N (watts per unit area per steradian) and an area of A, then the power associated with the source is simply pi*N*A (watts).  You can start with the night sky radiance in whatever units you like (or luminance, converted into radiance), convert this to say watts per cm^2 per steradian, then take your entrance aperture area in cm^2, and easily find the required source power.

Here’s a snippet from Smith, Modern Optical Engineering:

Regards,

Jeff

Mike.Jones
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Hi and thanks, Jeff.  My question involves the relationship between the number of Lambertian analysis rays and the power collected at the focal plane. If I assign 1 watt to 1E6 rays, shouldn't each ray carry 1 microwatt of power? That will deliver P watts of Lambertian power to the focal plane. If I increase the number of rays to 1E7 but keep the power at 1 watt, shouldn't each ray now carry 0.1 microwatt of power?  And shouldn't there still only be approximately P watts collected at the sensor due to the Lambertian source?  I'm not seeing that happen; the collected power seems to be connected to the number of rays.  Have you observed this as well?

Jeff.Wilde
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Userlevel 7
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The total detected power shouldn’t depend on the number of source rays unless you happen to have the Minimum Absolute Ray Intensity parameter set to a value that causes weaker rays to terminate before reaching the detector.  A value of 0 turns this intensity threshold feature off.

 

If that’s not the problem, then I would need to see an example file to take a closer look.

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