<p>It is easiest to understand if you fix the incident angle and explain why the emergent intensity is largest at highly oblique angles.  Since the light comes from the outside, it only penetrates so far, and this in turn causes the atmosphere to act like a source of scattered light.  But the source of scattered light is brightest near the top of the atmosphere, since that's where more of the external light penetrates.  Whenever the sources are brighter near the top, it produces what is called "limb brightening," where if you look from highly oblique angles, you mostly probe those higher, brighter regions.  Looking down the normal is where you see the deepest into the atmosphere, where the external light does not penetrate as well. </p>

<p>You can also think about what the individual photons are doing, and ask what is their distribution over emergent angle.  If cos(i)=0, all the photons scatter right at the surface, so it is tantamount to introducing an isotropic radiation field right at the surface.  The photons that go outward will of course have an isotropic distribution, and an isotropic incident radiation field must scatter isotropically (that is a consequence of the principle of reciprocity).  So the emergent distribution is isotropic, but intensity is also per solid angle, so accounts for the foreshortening, and that's where the 1/cos(e) in the intensity comes from.  If cos(i)=1, on the other hand, the incident photons tend to penetrate more, and must diffuse their way out, which gives less of an advantage to low cos(e) after the foreshortening is included.</p>

<p>As for fixing the angle at which you are looking and altering i, here your result says that the intensity always peaks as cos(i) rises.  Your expression claims that is true at every e, so to me this suggests a normalization error.  You want to keep the incident F fixed, but that requires you must get the same outgoing F if you integrate over all e.  Your result says the I is higher at all e if cos(i) is higher, but that contradicts the idea that you are keeping the incident F the same.  Maybe your result is actually comparing the incident and emergent intensities, not the emergent intensity to incident flux.  Then at higher cos(i), for the same incident I, the incident F is falling, explaining your rise in I/F.</p>

CC BY-SA 3.0
2018-04-24T01:08:48.023Z
Your answer for the first part, with emission angle really helped!

However, I found the inverse about the incident angle. $I/F$ decreased with increased $cos(i)$.
Spuds
CC BY-SA 3.0
2018-04-24T01:23:20.447Z
The expression you give for I/F certainly increases as cos(i) increases, at all e.
Ken G
CC BY-SA 3.0
2018-04-24T01:25:27.583Z
Woops I meant to say that it decreases with increased $i$, not $cos(i)$.
Spuds
CC BY-SA 3.0
2018-04-24T02:53:46.750Z
Right, but that's what I said, and that cannot be true for I/F because it would not conserve F, but it could be true for I_out / I_in .
Ken G
CC BY-SA 3.0
2018-04-22T15:42:33.680Z
2018-04-22T15:42:33.680Z
Ken G
<p>So I just derived an expression for the scattered intensity of light from an atmosphere with ideal isotopic scattering: $I/F=\frac{cos(i)}{4(cos(e)+cos(i))}$, where $I$ is intensity, $F$ is flux, $i$ is the angle off incidence, and $e$ is the angle of emission ($i,\ e=0$ are normal to atmosphere).</p>

<p>When I plot the scattered intensity as a function of $i$ with $e$ constant, it decreases while the angle increases. Why is this? My theory is that at a large incidence angle, the light encounters a larger surface area, so reflected light will have a higher variation of emission angle, meaning that scattered light at a given emission angle will be lower. Does this make sense? Or is there another reason for this?</p>

<p>Next, I plotted scattered intensity as a function of $e$ with $i$ constant. This time, the intensity increases with increases emission angle. I don't have a theory as to why this is the case, nor an intuition to back it up. Why would this be the case?</p>

CC BY-SA 3.0
2018-04-22T14:26:44.017Z
How did you arrive at that result?  It has some surprising elements-- first of all, it is simple, and few things are with scattering, and second of all, it doesn't have anything in it about the thickness of the atmosphere, yet scattering should depend on that.
Ken G
CC BY-SA 3.0
2018-04-22T15:04:52.203Z
The full expression I derived was $I/F=\frac{\varpi_0\,\mu_0}{4(\mu+\mu_0)}P(\alpha)(1-exp[-\tau(1/\mu+1/\mu_0)])$, where $\varpi_0$ is the single-scattering albedo, $\mu_0 = cos(i)$, $\mu = cos(e)$, $P(\alpha)$ is the normalized phase function, and $tau$ is the optical depth. I got this expression by integrating over layers of an atmosphere. In my question, I'm using a bunch of simplifying assumptions to look at how scattered intensity depends on incidence and emission angles.
Spuds
CC BY-SA 3.0
2018-04-22T15:11:21.223Z
OK, so it's a semi-infinite atmosphere with some simplifying assumptions attached to get a nice result.  Now I think I can answer.
Ken G
<light><planetary-atmosphere><radiation><atmospheric-effects><radiative-transfer>
Scattered Intensity of Light vs Angles of Incidence and Emission
Spuds
# Scattered Intensity of Light vs Angles of Incidence and Emission

(Asked by: username_0 on 2018-04-22)

So I just derived an expression for the scattered intensity of light from an atmosphere with ideal isotopic scattering: $I/F=\frac{cos(i)}{4(cos(e)+cos(i))}$, where $I$ is intensity, $F$ is flux, $i$ is the angle off incidence, and $e$ is the angle of emission ($i,\ e=0$ are normal to atmosphere).

When I plot the scattered intensity as a function of $i$ with $e$ constant, it decreases while the angle increases. Why is this? My theory is that at a large incidence angle, the light encounters a larger surface area, so reflected light will have a higher variation of emission angle, meaning that scattered light at a given emission angle will be lower. Does this make sense? Or is there another reason for this?

Next, I plotted scattered intensity as a function of $e$ with $i$ constant. This time, the intensity increases with increases emission angle. I don't have a theory as to why this is the case, nor an intuition to back it up. Why would this be the case?

**Tags:** light, planetary-atmosphere, radiation, atmospheric-effects, radiative-transfer

> **Comments:**
> **username_1**: How did you arrive at that result?  It has some surprising elements-- first of all, it is simple, and few things are with scattering, and second of all, it doesn't have anything in it about the thickness of the atmosphere, yet scattering should depend on that.  
> **username_0**: The full expression I derived was $I/F=\frac{\varpi_0\,\mu_0}{4(\mu+\mu_0)}P(\alpha)(1-exp[-\tau(1/\mu+1/\mu_0)])$, where $\varpi_0$ is the single-scattering albedo, $\mu_0 = cos(i)$, $\mu = cos(e)$, $P(\alpha)$ is the normalized phase function, and $tau$ is the optical depth. I got this expression by integrating over layers of an atmosphere. In my question, I'm using a bunch of simplifying assumptions to look at how scattered intensity depends on incidence and emission angles.  
> **username_1**: OK, so it's a semi-infinite atmosphere with some simplifying assumptions attached to get a nice result.  Now I think I can answer.  

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## 1 answers

### Answer by username_1 on 2018-04-22. Score: 3 | Accepted answer

It is easiest to understand if you fix the incident angle and explain why the emergent intensity is largest at highly oblique angles. Since the light comes from the outside, it only penetrates so far, and this in turn causes the atmosphere to act like a source of scattered light. But the source of scattered light is brightest near the top of the atmosphere, since that's where more of the external light penetrates. Whenever the sources are brighter near the top, it produces what is called "limb brightening," where if you look from highly oblique angles, you mostly probe those higher, brighter regions. Looking down the normal is where you see the deepest into the atmosphere, where the external light does not penetrate as well.

You can also think about what the individual photons are doing, and ask what is their distribution over emergent angle. If cos(i)=0, all the photons scatter right at the surface, so it is tantamount to introducing an isotropic radiation field right at the surface. The photons that go outward will of course have an isotropic distribution, and an isotropic incident radiation field must scatter isotropically (that is a consequence of the principle of reciprocity). So the emergent distribution is isotropic, but intensity is also per solid angle, so accounts for the foreshortening, and that's where the 1/cos(e) in the intensity comes from. If cos(i)=1, on the other hand, the incident photons tend to penetrate more, and must diffuse their way out, which gives less of an advantage to low cos(e) after the foreshortening is included.

As for fixing the angle at which you are looking and altering i, here your result says that the intensity always peaks as cos(i) rises. Your expression claims that is true at every e, so to me this suggests a normalization error. You want to keep the incident F fixed, but that requires you must get the same outgoing F if you integrate over all e. Your result says the I is higher at all e if cos(i) is higher, but that contradicts the idea that you are keeping the incident F the same. Maybe your result is actually comparing the incident and emergent intensities, not the emergent intensity to incident flux. Then at higher cos(i), for the same incident I, the incident F is falling, explaining your rise in I/F.

> **Comments:**
> **username_0**: Your answer for the first part, with emission angle really helped!  However, I found the inverse about the incident angle. $I/F$ decreased with increased $cos(i)$.  
> **username_1**: The expression you give for I/F certainly increases as cos(i) increases, at all e.  
> **username_0**: Woops I meant to say that it decreases with increased $i$, not $cos(i)$.  
> **username_1**: Right, but that's what I said, and that cannot be true for I/F because it would not conserve F, but it could be true for I_out / I_in .  

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