In the last two posts, we set out to find the relationship between four radiometric quantities:

*Exitance*(M): the power density emitted by a surface, measured in Watts / m^{2}. We will reserve*B*for the per-micron un-integrated quantity.*Radiance*(L): the power density emitted by a surface into each solid angle, measured in Watts / m^{}^{2}/ steradian.*Intensity*(I): the power emitted by a point source into each solid angle, measured in Watts / steradian.*Irradiance*(E): the power incident upon a receiving surface, measured in Watts / m^{2}.

We discovered that the relationship between exitance and radiance followed the simple formula for a lambertian surface: M = πL. We will now look at the relationship between these and irradiance. If anything, properly calculating this last is the most critical; as its name implies, *radiometry* is about using the power incident on our measuring device to infer the power being emitted from the surface of interest.

Let us assume that a spherical object of radius *R _{obj} *is sufficiently far away from our sensor that we can treat it as a point source of intensity

*I*. The total power Φ it emits is equal to the product of its exitance

*M*and surface area

*A*; its intensity

*I*is that power divided by the total surrounding solid angle, 4π. Thus:

_{}

You will remember that intensity is in units of Watts / steradian. Thus, if we assume that the sensor is looking directly at the emitting object, we can find the irradiance at the aperture of our receiver by multiplying the intensity by the solid angle Ω subtended by the aperture and dividing by the area A_{rcvr} of the aperture. Remembering that the definition of solid angle is the ratio of the area of a sphere subtended by that angle divided by the square of the radius gives us:

_{}

where R_{path} is the distance from the emitting object to our sensor.

Alternatively, we can treat our emitting object as a disk of radius *R _{obj}*. It’s radiance

*L*is the exitance

*M*divided by π as found previously. The irradiance on a sensor at distance is the product of the radiance, the solid angle Ω subtended by the sensor’s aperture, and the disk area A

_{obj}= πR

_{obj}

^{2}, divided by the area of the sensor aperture:

_{}

Thus, we can see that whether we approximate our emitting object of radius R_{obj} as a point source or a disk, we get the same result for the amount of radiation incident on a sensor at distance R_{path}. (The textbook irradiance of, for instance, the sun at the top of the earth’s atmosphere is 1480W/m^{2}, for instance.)

*However*, we must point out the obvious: an emitting sphere is not a disk, and it may not appear as a point source. For instance, the sun’s dimensions are apparent to an observer on the earth, 93M miles away.

In our next post, we will calculate the *exact* irradiance on the aperture of a sensor from a spherical emitting object.

(Note: I apologize for the crappy looking equations. I have a procedure for cleaning them up, but it’s tedious, and I’m tired.)

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