Emission spectra of different materials

[Charles Link]

For the case of uniform brightness $L$ radiating from any shape, the irradiance that reaches the receiver is simply $E=L \, \Omega$, where $\Omega$ is the solid angle measured from the receiver. There is basically a factor of $cos(\theta)$ in intensity fall-off per unit area with tilt angle $\theta$, but this is exactly compensated for because the radiating area in an incremental solid angle $d \Omega$ increases by a factor of $\frac{1}{cos(\theta) }$. The distance factor does not affect things as your diagram of post #277 explains quite clearly. $\\$ And I was glad to see you were able to follow the explanation that the moon needs to have some kind of jagged surface in order to have the nearly uniform brightness that it has.

 

[JohnnyGui]

And I was glad to see you were able to follow the explanation that the moon needs to have some kind of jagged surface in order to have the nearly uniform brightness that it has.

I was reading about this and it indeed makes sense. There's this effect also called Seeliger effect that explains why a full moon has much more than twice the brightness compared to a half moon, even though the illuminated surface ratio is 2:1. Part of it is explained by the fact that because the moon surface is jagged, the sun illuminating the moon from the side w.r.t. us would cast shadows along the illuminated surface (because of the mountains, etc.) while if the sun is directly behind us illuminating the moon in the same direction as our viewing direction, there are no shadows (or very short ones) from the mountains, which makes the moon look brighter.

Another factor that causes a higher brightness is called "coherent backscattering". This is when two (or more) light rays travel different distances beneath the moon's surface while they're being scattered, but when they come out in your direction (backscattering) the total scatter distance difference traveled among those light rays is a whole (or $n$) wavelength and therefore giving each other a constructive interference right back at your eyes.

For the case of uniform brightness LL L radiating from any shape, the irradiance that reaches the receiver is simply E=LΩE=LΩ E=L \, \Omega , where ΩΩ \Omega is the solid angle measured from the receiver. There is basically a factor of cos(θ)cos(θ) cos(\theta) in intensity fall-off per unit area with tilt angle θθ \theta , but this is exactly compensated for because the radiating area in an incremental solid angle dΩdΩ d \Omega increases by a factor of 1cos(θ)1cos(θ) \frac{1}{cos(\theta) } . The distance factor does not affect things as your diagram of post #277 explains quite clearly.

So, the distance doesn't even matter if the spherical emitter is small and near like this?

So this scenario still won't give dimmer edges?

 

[Charles Link]

So, the distance doesn't even matter if the spherical emitter is small and near like this?

So this scenario still won't give dimmer edges?

The distance isn't the important factor here. If the red surface (which is assumed to be a smooth diffuse scattering surface) is illuminated by parallel rays from the left, then it will show some dimming from the center to the edges. $\\$ And your additional reading of the moon's appearance was quite interesting.

 

[JohnnyGui]

The distance isn't the important factor here. If the red surface (which is assumed to be a smooth diffuse scattering surface) is illuminated by parallel rays from the left, then it will show some dimming from the center to the edges.

Apologies, I was talking about a diffuse black body emitter, not an illuminated sphere. Will in that case the edges in the picture still look equally bright as the center?

And your additional reading of the moon's appearance was quite interesting.

I'm glad I finally gave something in return after all your efforts XD.

 

[Charles Link]

Apologies, I was talking about a diffuse black body emitter, not an illuminated sphere. Will in that case the edges in the picture still look equally bright as the center?

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

 

[JohnnyGui]

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

Yes, this is what I indeed expected. So if distance doesn't matter regarding the received energy in a fixed solid angle, can I also say the following:

If the two $d\Omega$'s in the picture are equal in size, the power received from each solid angle would be the same?

 

[sophiecentaur]

The answer is yes, it will look equally bright everywhere. You might find it of interest that you also wouldn't be able to tell whether it is flat in shape or concave inward or outward.

This would need the emission to be taking place at the surface. Observing the Sun, you see it is a bit dark at the edges so the flux per solid angle is not actually the same in that (practical) case. (Danger - do no observe it directly!)

 

[Charles Link]

Yes, this is what I indeed expected. So if distance doesn't matter regarding the received energy in a fixed solid angle, can I also say the following:
View attachment 207264
If the two $d\Omega$'s in the picture are equal in size, the power received from each solid angle would be the same?

Yes, you have it correct. One thing to note that the irradiance $E=L \, d \Omega$ is the same for both. The power received per unit area by a flat detector could differ by a $cos(\theta)$ factor because the irradiance $E$ is measured in a plane normal to the direction of the small solid angle increment $d \Omega$.

 

[JohnnyGui]

Yes, you have it correct. One thing to note that the irradiance $E=L \, d \Omega$ is the same for both. The power received per unit area by a flat detector could differ by a $cos(\theta)$ factor because the irradiance $E$ is measured in a plane normal to the direction of the small solid angle increment $d \Omega$.

If you say a flat detector could receive less $E$ from the left sided $d\Omega$ (in the picture) by a factor of $cos(\theta)$ since $E$ is measured in the normal plane of $d\Omega$, doesn't this imply that if I don't look at the edges but straight normal to the flat source, I'd notice dark edges from the sides of my viewing angle because my (more or less) flat retinas are not in the normal plane of the left sided $d\Omega$. And when I move my eyes to the edges so that my retinas are normal to the left sided $d\Omega$, I'd be receiving the same $E$ from that $d\Omega$ as the normal $d\Omega$ that I was looking at previously?

This would need the emission to be taking place at the surface. Observing the Sun, you see it is a bit dark at the edges so the flux per solid angle is not actually the same in that (practical) case

I was thinking the same thing. Do you mean that because when looking at the center of the sun, there are way more emitting particles beneath the surface as compared to the edges?

(Danger - do no observe it directly!)

Too late, I have no idea what I'm typing right now.

 

[Charles Link]

If you say a flat detector could receive less $E$ from the left sided $d\Omega$ (in the picture) by a factor of $cos(\theta)$ since $E$ is measured in the normal plane of $d\Omega$, doesn't this imply that if I don't look at the edges but straight normal to the flat source, I'd notice dark edges from the sides of my viewing angle because my (more or less) flat retinas are not in the normal plane of the left sided $d\Omega$. And when I move my eyes to the edges so that my retinas are normal to the left sided $d\Omega$, I'd be receiving the same $E$ from that $d\Omega$ as the normal $d\Omega$ that I was looking at previously?

Very good. You have it correct. The eye is slightly more complicated having a lens, and the precise details of the brightness of the image at points far off-axis is beyond the scope of what we need here. In any case, if you look in the direction that is off-axis, you see the same $E$ as you stated in the last sentence.

 

[sophiecentaur]

Very good. You have it correct. The eye is slightly more complicated having a lens, and the precise details of the brightness of the image at points far off-axis is beyond the scope of what we need here. In any case, if you look in the direction that is off-axis, you see the same $E$ as you stated in the last sentence.

The varying luminosity can be seen on solar photographs too -wherever the Sun is placed on the sensor. So it's not a visual anomaly

Too late, I have no idea what I'm typing right now.

Haha.
It's worth treating fairly seriously though. Too much time looking for solar features without the right precautions actually makes people blind. Before the sextant was invented, ship's navigators always ended up blind in one eye, apparently from using the Cross Staff.

 

[JohnnyGui]

The varying luminosity can be seen on solar photographs too -wherever the Sun is placed on the sensor. So it's not a visual anomaly

You mean since the edges are "thin" compared to the center, w.r.t. your viewing direction, and therefore you'd still see darker edges?

Yes, you have it correct. One thing to note that the irradiance E=LdΩE=LdΩ E=L \, d \Omega is the same for both. The power received per unit area by a flat detector could differ by a cos(θ)cos(θ) cos(\theta) factor because the irradiance EE E is measured in a plane normal to the direction of the small solid angle increment dΩdΩ d \Omega .

I think I've finally got how brightness is independent from distance! I managed this by reasoning $L$ itself and using the following scenario (please bear with me):

The 2 factors that would make brightness differ with the angle are the observed source surface $dA$ and the eye's solid angle. Therefore, to objectify how one sees the brightness of a radiating surface $dA$ so that it can be compared to what others see, it is necessary to calculate the received energy per fixed solid angle (1 steradian) per fixed source area( 1 m²).

Therefore, if the observing eye has a surface of $dA_o$, then receiver $A$ would measure a total received power of $I_0\cdot \frac{dA_o}{R^2} = P_A$. Receiver $A$ would thus say that if $dA$ is emitting at distance $R$ an energy of $I_0\cdot \frac{dA_o}{R^2} = P_A$ on a surface of $dA_o$, that the energy per steradian per $1$ m² radiating source would be
$$I_0 \cdot \frac{dA_o}{R^2} \cdot \frac{R^2}{dA_o} \cdot \frac{1}{dA} = \frac{I_0}{dA} = L$$
Receiver $B$ would receive a total power of $I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 = P_B$ because of Lambert's cosine law and the increase in distance. B would reason that per steradian, this power would be $I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 \cdot \frac{R^2}{dA_o \cdot cos(\theta)^2} = P / Sr$ (intensity). B would furthermore see that this energy is coming from a surface of $dA \cdot cos(\theta)$. Per $1$ m² this would give :
$$I_0 \cdot \frac{dA_o}{R^2} \cdot cos(\theta)^3 \cdot \frac{R^2}{dA_o \cdot cos(\theta)^2} \cdot \frac{1}{dA \cdot cos(\theta)} = \frac{I_0}{dA} = L$$
Therefore, the brightness is the same for both A and B. Apologies for taking so long to finally understand this. I tend to take things step by step to guarantee I fully understand something.

 

[JohnnyGui]

@Charles Link : Is there actually another intrinsic physical explanation, other than "looking at a smaller projected source area and therefore the same radiance", that explains why we would see the same brightness? In other words;

Why would our retinas register the same brightness even though the total received energy as well as the photon density that bombards the retinas are both less when they're coming from a $dA$ at an angle? (regardless of the observed source area being smaller by a factor of $cos(\theta)$). I mean, isn't photon density the intrinsic factor that determines how the retina measures brightness instead of observing a smaller projected source area?

 

[Charles Link]

@Charles Link : Is there actually another intrinsic physical explanation, other than "looking at a smaller projected source area and therefore the same radiance", that explains why we would see the same brightness? In other words;

Why would our retinas register the same brightness even though the total received energy as well as the photon density that bombards the retinas are both less when they're coming from a $dA$ at an angle? (regardless of the observed source area being smaller by a factor of $cos(\theta)$). I mean, isn't photon density the intrinsic factor that determines how the retina measures brightness instead of observing a smaller projected source area?

A very good question...You need to keep the solid angle of all sources the same for this kind of comparison. The area on the retina $dA_r$ that gets imaged is $dA_r=f^2 \, d \Omega$, where $d \Omega$ is the solid angle subtended by the source as measured from the eye, and $f$ is the focal length of the lens of the eye. $\\$ (In the focal plane, for small angles $d \theta$ , we can write $dx=f \, d \theta$ where $dx$ is the image size on the retina (which is in the focal plane) and $d \theta$ is the range of angles covered by the incoming rays. From the center of the object, essentially the rays are parallel and on-axis and image at the center of the focal plane, and if the rays come in from the edge of the object all parallel but at some angle $d \theta$ , they will come to focus in the focal plane (retina) at distance $dx =f \, d \theta$ from the center. Thereby $dx=f \, d \theta$ and $dA_r=f^2 \, d \Omega$). $\\$ If you look at an area $dA_r$ of the retina, for sources of equal brightness, it receives the same number of photons per second. $\\$ Additional item, in case you didn't know it: Whether it is a camera lens or the lens of your eye, it creates a focused image of the scene in the image plane, which for a camera is the focal plane array of pixels, and for your eye, it is the retina. The focusing is such that $\frac{1}{f}=\frac{1}{b}+\frac{1}{m}$ where $b$ is the object distance and $m$ is the image distance. For faraway objects=large $b$, $m \approx f$, and the focused image occurs in the focal plane. For objects at closer distances, a refocusing is necessary. $\\$ Additional item: Oftentimes, there is too much stray light and you don't see the image, but try seeing the image of a light bulb with a magnifying glass: Place the (lit up) light bulb about 4 ft. from the magnifying glass, and look for the focused image using a sheet of paper as the screen. Your focused image should appear on the screen about 4" from the magnifying glass for a typical magnifying glass with $f=4"$.

 

[JohnnyGui]

...You need to keep the solid angle of all sources the same for this kind of comparison.

Thanks for the detailed explanation! One question regarding keeping the solid angle the same. Is it correct to say the following:

If one has a fixed viewing angle, for example the human eye (150 degrees). And that eye first looks perpendicularly at a radiating surface, and after that looks at it from an angle, doesn't this mean that the solid angle $\frac{dA_{projected}}{s^2}$ will always be different, regardless of the viewing angle being fixed? So in other words, one can not keep the solid angle fixed unless you change the viewing angle?

 

[Charles Link]

I don't understand your question. Basically, a given amount of solid viewing angle $d \Omega$ corresponds to a given amount of area on the retina $dA_r$ with $d \Omega= \frac{dA_r}{f^2}$. If you mean that it will use a different portion of the area on the retina to see it if you view it at an angle, the answer is yes, and the response of those receptors could be different. Basically the retina is like the array of pixels of a digital camera that are in the focal plane of the lens. I think different parts of the retina even have fewer pixels per unit area, but for the eye, you aren't generating quantitative brightness information.

 

[JohnnyGui]

I don't understand your question. Basically, a given amount of solid viewing angle $d \Omega$ corresponds to a given amount of area on the retina $dA_r$ with $d \Omega= \frac{dA_r}{f^2}$. If you mean that it will use a different portion of the area on the retina to see it if you view it at an angle, the answer is yes, and the response of those receptors could be different. Basically the retina is like the array of pixels of a digital camera that are in the focal plane of the lens. I think different parts of the retina even have fewer pixels per unit area, but for the eye, you aren't generating quantitative brightness information.

Apologies, I meant it more generally. So in the following picture:

If viewing angle $\alpha_A$ = $\alpha_B$ (same angle, in degrees), then this means that the projected area (dashed line) that $B$ sees is the same as the area that A sees. I'm taught that $B$ would see this projected area at a larger distance than $A$ would see his surface by $\frac{s_A}{cos(\theta)} = s_B$. However, from the drawing and from geometry, the distance from $B$ to the dashed line looks to be the same as the distance from $A$ to his own surface, $s_A = s_B$.

If the drawing is incorrect and $B$ is seeing the projected area at a distance of $\frac{s_A}{cos(\theta)} = s_B$, then it should be drawn like this instead:

If this is correct, then that means that the solid angle of $A$ subtended by his own surface is different from the solid angle of B subtended by the projected surface, regardless of the viewing angle $\alpha$ being the same for both.

Is this right?

 

[Charles Link]

B in some ways thinks he is seeing the projected area, but he is actually seeing the surface which is farther away and with an increased area. He really can't tell the difference between the two if there is nothing of contrast to focus on. And remember, solid angle $\Omega=\int \frac{dA_{projected}}{s^2}$. $s$ is different for the different parts of $dA_{prrojected}$. You can't compute the solid angle properly in your diagram if $s$ changes throughout the diagram. Also, your diagram is unclear on what you want to call $s$ and $A_{projected}$. In the formula that I just wrote with the integral, the $dA_{projected}$ and $s$ are measured right to that portion of the radiating surface. You can of course even measure $\Omega$ from your eye or from any surface that intercepts that $\Omega$, but you need to be precise, or you will get inconsistencies in the results. The formula are all very consistent.

 

[JohnnyGui]

B in some ways thinks he is seeing the projected area, but he is actually seeing the surface which is farther away and with an increased area. He really can't tell the difference between the two if there is nothing of contrast to focus on.

Ah ok, but if one wants to calculate B's radiance using his solid angle $d\Omega_B$ he should use $$d\Omega_B = \frac{A_{projected}}{(\frac{s_A}{cos(\theta)})^2}$$
Where $s_A$ is the distance of A to his own surface, right?

 

[Charles Link]

Ah ok, but if one wants to calculate B's radiance using his solid angle $d\Omega_B$ he should use $$d\Omega_B = \frac{A_{projected}}{(\frac{s_A}{cos(\theta)})^2}$$
Where $s_A$ is the distance of A to his own surface, right?

The angle $\theta$ here doesn't stay constant throughout the calculation. $s_A$ is the perpendicular distance to the second surface, but since you have an extended angle of viewing, $\theta$ will not be constant. Also, $A_{projected}$, if you are using this formula, which has an increased distance $s=s_A/cos(\theta)$, needs to come from the location of that point on the surface. (The $A_{projected}$ will be larger if you measure it at a location which is farther away, as is necessary to keep the ratio $\Omega=\frac{A_{projected}}{s^2}$ constant for constant $\Omega$).

 

[JohnnyGui]

Also, AprojectedAprojected A_{projected} , if you are using this formula, which has an increased distance s=sA/cos(θ)s=sA/cos(θ) s=s_A/cos(\theta) , needs to come from the location of that point on the surface. (The AprojectedAprojected A_{projected} will be larger if you measure it at a location which is farther away, as is necessary to keep the ratio Ω=Aprojecteds2Ω=Aprojecteds2 \Omega=\frac{A_{projected}}{s^2} constant for constant ΩΩ \Omega ).

This is exactly what I was wondering since $dA_{projected}$ indeed changes depending on where you draw it. So from the 2 pictures here:

The picture on the right is the correct one if you're using the increased distance $s = s_A / cos (\theta)$?

 

[Charles Link]

This is exactly what I was wondering since $dA_{projected}$ indeed changes depending on where you draw it. So from the 2 pictures here:
View attachment 207845 View attachment 207846
The picture on the right is the correct one if you're using the increased distance $s = s_A / cos (\theta)$?

I think you have the basic idea. The $dA_{projected}$ can be measured anywhere, just so long as the $s$ that is used is the distance at which $dA_{projected }$ is measured, in order to properly determine $d \Omega=d A_{projected}/s^2$.

 

[JohnnyGui]

I think you have the basic idea. The dAprojecteddAprojected dA_{projected} can be measured anywhere, just so long as the ss s that is used is the distance at which dAprojecteddAprojected dA_{projected } is measured, in order to properly determine dΩ=dAprojected/s2dΩ=dAprojected/s2 d \Omega=d A_{projected}/s^2 .

That's indeed what I thought. But I have a feeling that this is merely done just for the sake to keep $d\Omega$ constant. Which $dA_{projected}$ does $B$ truly observe when the radiating surface covered by B's viewing angle is the only surface that is radiating?

 

[Charles Link]

That's indeed what I thought. But I have a feeling that this is merely done just for the sake to keep $d\Omega$ constant. Which $dA_{projected}$ does $B$ truly observe when the radiating surface covered by B's viewing angle is the only surface that is radiating?

Your eye is not an accurate measure of brightness regardless because (if I have this part of the anatomy correct), it contains an adjustable iris, so that when we are in bright light, we only use a part of the lens of our eye. Cameras also have this same feature, which is the f-stop #. In any case, a given area on the retina maps into a given solid angle, and if the area of the lens stays constant, then what we see is a good measure of the surface brightness. $\\$ Editing... To answer your question of where doe the radiating surface originate? , it really doesn't matter. When we see a solid angle of our field of vision illuminated with a certain brightness, the actual location of the surface (i.e. the distance that the surface is from us), is irrelevant. $\\$ Additional editing.. An important part of a previous discussion was the action of the lens of your eye. If you could crawl inside a person's eye and get a view of the retina, you would see the scene the person is looking at: It would appear like a photograph does across the surface of the retina. The focused scene is projected by the lens onto the surface of the retina like it were a projection screen. If there is a surface of brightness $L$ (e.g. bright orange) in the field of view, it would show up in the scene on the retina basically as an orange spot (shaped like the surface) with a brightness proportional to $L$. (If you can control and/or account for the iris mentioned in the first sentence above, the eye would actually be a good measure of surface brightness.)

 

[Charles Link]

@JohnnyGui I edited the last post, so be sure and read the updated version.

 

[JohnnyGui]

Editing... To answer your question of where doe the radiating surface originate? , it really doesn't matter. When we see a solid angle of our field of vision illuminated with a certain brightness, the actual location of the surface (i.e. the distance that the surface is from us), is irrelevant.

Ah, I forgot about the fact that anything that fully covers the solid angle of our field of vision would seem to have the same size, regardless of distance (sort of like the moon and the sun).

Additional editing.. An important part of a previous discussion was the action of the lens of your eye. If you could crawl inside a person's eye and get a view of the retina, you would see the scene the person is looking at: It would appear like a photograph does across the surface of the retina. The focused scene is projected by the lens onto the surface of the retina like it were a projection screen. If there is a surface of brightness LL L (e.g. bright orange) in the field of view, it would show up in the scene on the retina basically as an orange spot (shaped like the surface) with a brightness proportional to LL L . (If you can control and/or account for the iris mentioned in the first sentence above, the eye would actually be a good measure of surface brightness.)

Regarding the projection of the radiating surface on the retina, can I say the following explanation about brightness?:

Watching a radiating surface from an angle would indeed make an observer receive less power, i.e. less photons, but the projected surface area on the retina is also smaller because of the angle. So the lesser amount of photons are concentrated in a smaller retina surface so that photon density stays the same.
If this causes the equal brightness, then this must mean that the optical nerve doesn't measure brightness by the absolute number of light receptors that are stimulated but by the photon density per unit retina surface.

 

[Charles Link]

The eye is not an ideal measure of surface brightness, especially if you try to use it as a standard for peripheral viewing. If you consider the case of a lens (such as your eye or a magnifying glass or a camera lens) being used to put an image onto a screen, (in the case of a camera lens onto the pixels that generate the image), it is necessary to have everywhere in the the scene be incident on the same projected area of the lens. The projected area of the lens only decreases by $cos(\theta)$ so for small angles not to far off axis, this remains relatively constant. For a perfect image onto the screen, (or retina), for perfect mapping of the scene, the image brightness (on the screen) at any location should be proportional to the brightness of the scene. For angles not too far off axis, this will be the case with most lenses, including the eye. Far off axis, this generally will not be the case, either for your eye or any other lens system. $\\$ To quantify the previous discussion, for irradiance $E_{screen}$ from the other side of the lens onto a perfectly diffuse white projector screen, the image brightness $L_i$ will be such that $E_{screen}=L_i \pi$. (The incident irradiance is perfectly reflected in a Lambertian pattern that has an effective solid angle of $\pi$ steradians). $\\$ And a couple additional calculations: If the scene is in the far field and has brightness $L_s$ and occupies a solid angle $\Omega_s$ as measured from the lens, it will image with an area $A_i=f^2 \Omega_s$ and total power collected by the lens will be $P=L_s \Omega_s A_{Lens}$. This gives $E_{screen}=\frac{P}{A_i}=\frac{L_s A_{Lens}}{f^2}$ and image brightness (on a perfectly diffuse white screen) $L_i=\frac{E_{screen}}{\pi }$ which in general will be proportional to but less than the brightness $L_s$.

 

[JohnnyGui]

The eye is not an ideal measure of surface brightness, especially if you try to use it as a standard for peripheral viewing. If you consider the case of a lens (such as your eye or a magnifying glass or a camera lens) being used to put an image onto a screen, (in the case of a camera lens onto the pixels that generate the image), it is necessary to have everywhere in the the scene be incident on the same projected area of the lens. The projected area of the lens only decreases by $cos(\theta)$ so for small angles not to far off axis, this remains relatively constant. For a perfect image onto the screen, (or retina), for perfect mapping of the scene, the image brightness (on the screen) at any location should be proportional to the brightness of the scene. For angles not too far off axis, this will be the case with most lenses, including the eye. Far off axis, this generally will not be the case, either for your eye or any other lens system. $\\$ To quantify the previous discussion, for irradiance $E_{screen}$ from the other side of the lens onto a perfectly diffuse white projector screen, the image brightness $L_i$ will be such that $E_{screen}=L_i \pi$. (The incident irradiance is perfectly reflected in a Lambertian pattern that has an effective solid angle of $\pi$ steradians). $\\$ And a couple additional calculations: If the scene is in the far field and has brightness $L_s$ and occupies a solid angle $\Omega_s$ as measured from the lens, it will image with an area $A_i=f^2 \Omega_s$ and total power collected by the lens will be $P=L_s \Omega_s A_{Lens}$. This gives $E_{screen}=\frac{P}{A_i}=\frac{L_s A_{Lens}}{f^2}$ and image brightness (on a perfectly diffuse white screen) $L_i=\frac{E_{screen}}{\pi }$ which in general will be proportional to but less than the brightness $L_s$.

But if you compare an $L_i$ from a not too far-off axis angle of a radiating source to an $L_i$ from a perpendicular part of that radiating source, the angled $L_i$ must be less than the perpendicular $L_i$ right? If this is correct, then in order for the retina to measure the angled $L_i$ to be the same as the perpendicular $L_i$, this must mean that the retina measures brightness based on the photon density per unit projected retina image surface instead of the absolute amount of retina receptors being activated. (the latter being more for the perpendicular $L_i$ than for the angled $L_i$)
This is all while the retina surface is directed perpendicularly to the radiation, whether it's coming from an angle or not.

It this statement then correct?

 

[Charles Link]

The retina is in the focal plane of the lens of the eye when the eye is focused on a distant scene. Each location on the retina corresponds to a given angle in the far-field. If the scene in the far-field is uniformly illuminated (like a blue sky), the retina will receive an approximately uniform image. It won't be a perfect mapping, but we normally can not discern any lack of uniformity that might result.

 

[JohnnyGui]

The retina is in the focal plane of the lens of the eye when the eye is focused on a distant scene. Each location on the retina corresponds to a given angle in the far-field. If the scene in the far-field is uniformly illuminated (like a blue sky), the retina will receive an approximately uniform image. It won't be a perfect mapping, but we normally can not discern any lack of uniformity that might result.

Ah, so if I understand correctly, for the following 2 scenarios, in which $\alpha_A=\alpha_B$,

The projected surface size on the retina is in both cases the same (if it's not too far off-axis)?

 

[Charles Link]

Ah, so if I understand correctly, for the following 2 scenarios, in which $\alpha_A=\alpha_B$,
View attachment 208665
The projected surface size on the retina is in both cases the same (if it's not too far off-axis)?

The size on the retina will be $\Delta x=f \alpha$ where $f$ is the focal length of the lens of the eye, and $\alpha$ is the angle measured in radians. $\\$ To supply some extra detail on how this is computed, parallel rays incident at some angle come to a focus in the focal plane, and the ray that is incident on the center of the lens, (even it comes in at angle $\theta$), passes through as a straight line to the focal plane, while the other rays incident at angle $\theta$ at other locations on the lens will be focused by the lens and meet the central ray in the focal plane. The position of these incident parallel rays that come to a focus in the focal plane is thereby given by $x=f \theta$, for small angles $\theta$.

 

[JohnnyGui]

The size on the retina will be $\Delta x=f \alpha$ where $f$ is the focal length of the lens of the eye, and $\alpha$ is the angle measured in radians. $\\$ To supply some extra detail on how this is computed, parallel rays incident at some angle come to a focus in the focal plane, and the ray that is incident on the center of the lens, (even it comes in at angle $\theta$), passes through as a straight line to the focal plane, while the other rays incident at angle $\theta$ at other locations on the lens will be focused by the lens and meet the central ray in the focal plane. The position of these incident parallel rays that come to a focus in the focal plane is thereby given by $x=f \theta$, for small angles $\theta$.

Got it, and this calculated $x$ would be the same for both $A$ and $B$?

And it needs to be not too far-off axis because otherwise the same $\alpha$ would cover a too large surface for which an integration would be needed to calculate the projected surface area instead of just a $cos (\theta)$ factor?

 

[Charles Link]

One of the reasons the formula works for small angles $\theta$ is because for the distance on the retina $x$, we have $tan(\theta)=x/f$, which means $x= f \, \theta$ for small angles where $tan(\theta) \approx \theta$. There are probably other reasons as well, but in any case, here we are only interested in angles that are within a few degrees of normal incidence=otherwise the area of the lens also appears to fall off as $cos(\theta)$. In the simple equations we are presenting here, we want to be able to ignore effects such as these.

 

[JohnnyGui]

One of the reasons the formula works for small angles $\theta$ is because for the distance on the retina $x$, we have $tan(\theta)=x/f$, which means $x= f \, \theta$ for small angles where $tan(\theta) \approx \theta$. There are probably other reasons as well, but in any case, here we are only interested in angles that are within a few degrees of normal incidence=otherwise the area of the lens also appears to fall off as $cos(\theta)$. In the simple equations we are presenting here, we want to be able to ignore effects such as these.

Got it.

Sorry if I'm blacking out again but is it correct that, even though the projected retina image is the same for $B$ and $A$, the total energy that $B$ receives on his retina is less than $A$'s received energy because the distance from $B$'s lens to his radiating source part (that $B$ is looking at) is larger than the distance from $A$'s lens to the radiating surface part that $A$ is looking at? ($B$'s radiating surface is larger by a factor of $\frac{1}{cos(\theta)}$ but that is already compensated by Lambert 's cosine law, so there's a net less energy for $B$ by a factor of $cos(\theta)^2$)

 

[Charles Link]

Got it.

Sorry if I'm blacking out again but is it correct that, even though the projected retina image is the same for $B$ and $A$, the total energy that $B$ receives on his retina is less than $A$'s received energy because the distance from $B$'s lens to his radiating source part (that $B$ is looking at) is larger than the distance from $A$' s lens to the radiating surface part that $A$ is looking at? ($B$'s radiating surface is larger by a factor of $\frac{1}{cos(\theta)}$ but that is already compensated by Lambert 's cosine law, so there's a net less energy for $B$ by a factor of $cos(\theta)^2$)

Almost, but no cigar. B's radiating surface is larger by $\frac{1}{cos^2(\theta)}$ precisely accounting for the inverse square extra distance. There is no accounting for the first $cos(\theta)$ factor though, which is the projected area of the lens. For small angles $cos(\theta) \approx 1-\frac{\theta^2}{2} \approx 1$ so that we like to work with small angles $\theta$. Otherwise, it gets overly complicated mathematically, and really offers little additional instructional value.