Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A How does Radiance change with a parabolic reflector?

  1. Apr 18, 2016 #1
    Lets say I have a coil that is 1cmx1cm in area that gets heat up. I am told that I get 2W/cm^2/sr output. So if I had 100 of them in an array, the total radiant intensity would be (2W/cm^2/sr x 100 x 1 sqcm =) 200W/sr.

    Now, let's say I add a parabolic reflector around each coil. How would I estimate the increase in radiance and radiant intensity (in the direction of the reflector? Do I now use the aperture area of each reflector instead of the area of the coil? Do I multiply by 2pi since now I can able to see the radiance from the back of the coil? This is where I am getting confused.
  2. jcsd
  3. Apr 18, 2016 #2

    Charles Link

    User Avatar
    Homework Helper

    The intensity(watts/sr) ## I=L*A ## where ## L ## is the radiance and ## A ## is the area of the source or reflector (i.e. illuminated/projected area) assuming 100% reflector reflectivity. The total power ## P ## will be the same (approximately) with reflector as without the reflector. The bare source will have a solid angle of as much as ## 4\pi ## steradians depending upon the intensity pattern that it has. (If it is spherical, it would radiate equally in all directions). The parabolic reflector will have a greatly reduced effective solid angle of beam coverage. Ideally (with 100% reflectivity and no light returning to the source from the reflector), power ## P=I*\Omega ## (where ## \Omega ## is the effective solid angle of the radiated pattern) will be the same for the bare source as for the source plus reflector. The radiance ## L ## is a constant in all of these calculations, so that in the case of conserved power ## P ##, one has ## A_1 * \Omega_1=A_2 * \Omega_2 ## where the subscript "1" is the bare source and subscript "2" is the source plus reflector. The parabolic reflector will make a narrow pattern( effective solid angle ## \Omega_2 ##), but the intensity normally would not be constant throughout the region of illumination. The last equation allows you to get a quick estimate of the size of the beam pattern (in steradians) that a reflector will provide. One can also write ## I_1*\Omega_1=I_2*\Omega_2 ## as a useful equation to see the tradeoff between intensity that the reflector generates versus the size of the beam pattern (solid angle). A larger reflector will give a higher intensity, but will necessarily generate a narrower beam.
    Last edited: Apr 18, 2016
  4. Jun 23, 2016 #3
    So if I understand correctly, the area represents the projected aperture area?
    So in my example, I said I have a coil that is 1cmx1cm in area. The coil radiates 2 W/cm^2/sr. Lets assume spherically.
    With no reflector, the radiant intensity would be 2 W/sr?
    If I have a 10 sqcm (100% reflective) reflector, the radiant intensity would be 20 W/sr?
    If I have a 100 sqcm (100% reflective) reflector, the radiant intensity would be 200 W/sr?

    Now how would I compare this with a very narrow beam, like from a laser? That is usually given in W. I am usually given an approx beam pattern where I can estimate the divergence angle. Based on that I can estimate a solid angle. However then the radiant intensity is an extremely high number since the angle is so small.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted