Optical Photon Flux

To model the light levels entering a pinhole camera (to a first order approximation), the following calculations must be made:

First the intensity of solar flux at Earth is roughly 1370W/m2.

The Earth presents a circular face to the solar radiation. Some of it is reflected directly (the fraction being determined by the Earth's albedo; A = 0.367) the rest being absorbed by the Earth,and then re-emitted as a black body spectrum of a different temperature.

Soalr radiation peaks in the optical band, and in the Earth's in the microwave. It can be assumed that the contribution to the optical component of the spectrum from the Earth's blackbody spectrum will be negligable (Dr Steward). The light reflected by the atmosphere (albedo) takes the form of the Sun's Planck function.

The proportion of stellar flux that is optical is 46%.

So, firstly, the total flux reflected by the Earth's albedo is calculated:

Optical Flux = (Flux in x Area reflecting x Albedo x Optical fraction) / (Area at of surface at satellite altitude)

The satellite is assumed to be in LEO (500km) and the reflected flux is radiated hemispherically.

Optical flux = (1372 x pi x (6370 x 103)2 x 0.367 x 0.46) / (2 x pi x ((6370+500) x 103)2)

Optical flux = 99.57 W/m2

As the optical spectrum peaks in the green region, and green light has an average wavelength of 510nm, the photon flux can now be found:

Ephoton = hc / lamda
Ephoton = (6.626 x 10-34 x 3 x 108) / (510 x 10-9)
Ephoton = 3.898 x 10-19 Joules

By dividing the total energy flux, by the energy per photon, the approximate optical photon flux can be shown to be:

Fphoton = Fenergy/Ephoton
Fphoton = 99.57/(3.898 x 10-19)
Fphoton = 2.55453 x 1020 photons/s/m2.


The pinhole size in this page is modelled on the assumption that the CMOS sensor used in this model is STAR1000BRK.

The pixel array has 1024 x 1024 image pixels. From the technical specifications in the documentation supplied from Cypress each pixel has dimensions of 15 x 15 microns.

From this information, the size of the optical array on the CMOS chip is 15.36mm x 15.36mm.

Assuming that the light enters the aperture at a maximum of 45o from the normal to the aperture plane (Danny Brandt). The focal length of the chip can be found.

tan(theta) = opposite/adjacent
tan(45) = (half maximum pixel array length) / focal length
Focal length = (15.36 x 10-3) / (2 x tan(45))
Focal length = 7.68 x 10-3m

Using the equation for the pinhole diameter as derived by Lord Rayleigh;

d = 1.9 x (f x lamda)0.5
d = 1.9 x (7.68 x 10-3 x 510 x 10-9)0.5
d = 1.9 x (9.3534 x 10-10)0.5
d = 1.9 x 6.258 x 10-5
d = 1.189 x 10-4m

Photons into the Optical Array

Given the optical photon flux and the dimensions of the aperture, the total number of optical photons into the array can be found:

Aaperture = pi x r2
Aaperture = pi x ((1.189 x 10-4) / 2)2
Aaperture = pi x 3.535 x 10-9
Aaperture = 1.111 x 10-8m2

Photons per second = Aaperture x Fphoton
Photons per second = 2.6520 x 10-9 x 2.55453 x 1020
Photons per second = 2.837 x 1012 photons/second

Photons per pixel

Now knowing the total photons and the number of optical pixels, the photons per pixel per second can be shown to be:

Photons per pixel per second = Photons per pixel / pixels
Photons per pixel per second = 2.837 x 1012 / (1024 x 1024)
Photons per pixel per second = 2.705 x 106 photons/s/pixel


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