Intensity at detector for NT58-840

This caculation doesn't account for transmission through the glass of the photons. Some will be absrobed, but information, such as refractive indices of glass are not forth coming from Edmund. Also I not sure how to do it if I had them, as it may be necassary to integrate Fresnel's equations over a range of angles.

Knowing the size of the lens, the flux of optical radiation at a 500km altitude, and the image circle of the lens at its imaging place, the intensity of the optical flux on the detector can be determined.

Firstly, the optical flux can be taken from the evaluation done previoiusly:

Optical flux = 99.57 W.m-2.
Fphoton = 2.55453 x 1020 photons.s-1.m-2.

Basic assumptions that have to be made for this model is that dimension A in the Edmund catalogue specs is the same as the lens diameter. Obviously, this will be an overestimate, so the value obtained at the end of this working will be an upper limit for optical photon intensity at the imaging plane. Also, given the small area of the lens compared to the surface area over which the Earth is emitting radiation, the rays are assumed to be parallel.

The first stage of the calculation was to find the rate of photons through the lens:

Ratephoton = FPhoton x Alens,
Ratephoton = 2.55453 x 1020 x pi x (21.0 x 10-3 / 2)2,
Ratephoton = 88478851812156448.098985989830826,
Ratephoton=8.85 x 1016 photons.s-1.

This is the total amount of photons entering the lens per second. Assuming none are lost through the lens (a fact which shall be checked), the intensity on the imaging plane can be found as shown:

Intensityimage plane = Ratephoton / Areaimage,
Intensityimage plane = 8.85 x 1016 / (pi x (3 x 10-3)2),
Intensityimage plane = 3129299249999999999999.9999999998,
Intensityimage plane = 3.13 x 1021 photons.s-1.m-2.

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