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Assuming on the below image the source is a point source, I can understand that since the light flux falls on dA is the same within the solid angle and since the intensity I = Φ/dω and E = Φ/dA it can be deduced that E = I / d^2(I am not clear about d here. I can understand d to be the distance from the point source to the dA and not to the horizontal surface A).

enter image description here

But I don't seem to understand and give the same reasoning for the illumination E in the horizontal surface plane A. Does illumination on A equals E = Φ/A? That is, equal to that of dA? If yes it then seems to be possible to conclude E = I*cosθ/d2 for surface A(and again here I am not quite clear about d; is it the distance to the surface A as it's shown on the image? or is it the distance on the definition of the solid angle dω which is from the source perpendicular to dA?).

My confusion has risen from this sentence I read in Image Acquisition book on describing irradiance(would be the same concept in photometry):

Another very important distinction is that irradiance describes the flux per unit area that is perpendicularly incident(normal) to a surface. Hence, while radiant incidence has the same units as exitance, it is of a different nature. Existance ignores the direction taken by the exiting flux, whereas irradiance implicitly takes account of the flux component, which is normal to surface. If the flux direction is not normal, then the value of the radiant incidence must be modified to be the perpendicular component of the angularly incident flux density.

It clearly says "the value of the radiant incidence must be modified to be the perpendicular component of the angularly incident flux density." Does this mean illuminance is only defined on dA and not on the (horizontal) surface A? Or in another words is illuminance of A equals to that of dA? And does that possibly imply illuminance is only defined for a surface on a sphere?

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  • $\begingroup$ I think your quote means you should report the** irradiance** for that beam over a perpendicular cross-section of the light cone. You can then calculate the total power (or whatever) in that cone and "spread" it over the total surface it covers. What was that quote's "distinction" comparing the term "irradiance" with? $\endgroup$ – Carl Witthoft Jan 29 '16 at 15:08
  • $\begingroup$ @CarlWitthoft it was comparing irradiance vs radiant exitance. I have updated the quote. $\endgroup$ – Ali Fatourechi Jan 29 '16 at 23:42
  • $\begingroup$ @CarlWitthoft "you should report the** irradiance** for that beam over a perpendicular cross-section of the light cone." I didn't quite get what you mean here by perpendicular cross-section of light cone. Does it mean the dA surface? and that means the irradiance is only defined on dA? $\endgroup$ – Ali Fatourechi Jan 29 '16 at 23:51
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This stuff is confusing, but you have it about right. Irradiance is $\Phi / A$ if the flux is constant over the area, which usually means that $A$ is small compared to $d^2$. Note that the total flux within the cone is the same wherever you choose to measure it. The irradiance at the surface $dA$ is $\Phi / dA$, and the irradiance at the surface $A_0$ is $\Phi / A$, and $A = dA / \cos\theta$.

It's assumed in all this that the source is far enough from the surface that the distance $d$ can be taken as either the distance to $dA$ or the distance to $A_0$. The actual difference in the distance between the two is assumed to be negligible. If the source is close to the area, then things get a little more complicated because different rays from the source his the surface at different angles.

I don't understand what that quote is getting at. At best it's confusing, and at worst it's wrong. From Wikipedia:

irradiance is the radiant flux (power) received by a surface per unit area

Clear enough. No mention of perpendicularity.

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  • $\begingroup$ " The actual difference in the distance between the two is assumed to be negligible". The area that you said is negligible is ASinθ. and if Asinθ is negligible, can one not argue why not ACosθ then? $\endgroup$ – Ali Fatourechi Jan 30 '16 at 0:00
  • $\begingroup$ Also, if irradiance at A, as you said, is E1 = Φ/A and at dA is E2 = Φ/dA; since Φ is same for both one can write E1 = E2 cosθ. But this, I imagine, is not the case. As E1 should equal =Em cosθ, where Em is the maximum illuminance when the ray is perpendicular to surface A(θ=0, from top) not at angle. $\endgroup$ – Ali Fatourechi Jan 30 '16 at 0:42
  • $\begingroup$ I think I wasn't clear enough. There are two things to clarify, they are related, but different. 1.) For the simple approximations to be valid, $A$ has to be small compared to $d^2$. We're talking about order-of-magnitude here, so what I say about $A$ also applies to $A\cos\theta$ and $A\sin\theta$. 2.) I think you were asking if $d$ is the distance from the source to $dA$ or from the source to $A$. If $A$ is small compared to $d^2$, then those two distances are very nearly the same. You can choose either. Any calculations you do with one will be very nearly equal to the other. $\endgroup$ – garyp Jan 30 '16 at 0:53
  • $\begingroup$ @AliFatourechi But the light incident on $dA$ is perpendicular to the surface. $\theta = 0$ for that surface. $\endgroup$ – garyp Jan 30 '16 at 0:56
  • $\begingroup$ I think you are right about illuminance on A and dA and this answers my questions. But by assuming those two distances the same, as you say, are you basically not eliminating the effect of cosθ(since it is this distance, however small, that makes θ angle) which is the whole point and making the illuminance same on both surfaces? $\endgroup$ – Ali Fatourechi Jan 30 '16 at 18:45

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