<p>Remember radiance corresponds to a particular direction. When the equation is considering directions quasi parallel to the surface normal, the differential flux (the numerator) will typically be a much greater quantity than for grazing angles. The numerator is not remaining constant.</p>

CC BY-SA 3.0
2017-02-22T18:45:00.967Z
2017-02-22T18:45:00.967Z
PeteUK
<p>It's only more intense if $d\phi$ remains constant. However, for light source with constant luminance the flux is a function of $cos(\theta)$, i.e. depends on the light's angle of incidence with the surface and cancels out the term.</p>

<p>Edit: I feel the need to clarify this a bit because you say that:</p>

<blockquote>
  <p>thus light is somehow more "intense"</p>
</blockquote>

<p>It's not that the "light is somehow more intense", but rather that the light <strong>has to be</strong> more intense to have same flux in grazing angles as when it's parallel to the surface. If you think of flux as particles hitting to this differential surface, the number of particles hitting the surface is smaller more perpendicular the light gets to the surface. So you have to throw more particles at the surface (i.e. increase the radiance) for the same flux until when the surface is completely perpendicular to the light, none of the particles hit the surface, thus the radiance goes to infinity.</p>

CC BY-SA 3.0
2017-02-22T17:53:08.380Z
I know calculating irradiance, for example, cancels out the term, but this doesn't answer my question regarding radiance itself.  More precisely, how can it approach infinity when theta is approaching 90.
Ganea Dan Andrei
CC BY-SA 3.0
2017-02-22T18:05:59.653Z
Because if the flux stays constant, you need infinite radiance that there's any flux through a surface that's perpendicular to the light source
JarkkoL
CC BY-SA 3.0
2017-02-22T21:25:36.877Z
It makes sense that the light "has" to be more intense in this context, indeed.
Ganea Dan Andrei
CC BY-SA 3.0
2017-02-22T17:46:46.373Z
2017-02-22T21:18:16.620Z
2017-02-22T21:18:16.620Z
JarkkoL
<p>As far as I understand, radiance is a measure of the "intensity" of light in a direction defined as a differential cone . It's measure is Flux per solid angle per projected area.  <span class="math-container">$L = \frac{d\Phi}{d\omega dA^\perp} = \frac{d\Phi}{d\omega dAcos(\theta)} $</span>. Using this definition, it seems like as the <span class="math-container">$\theta$</span> angle increases, radiance increases, thus light is somehow more "intense" in a direction nearly perpendicular to the normal at point p. This contradicts my intuition.<br>
Where am I interpreting this wrong?</p>

CC BY-SA 3.0
2017-02-22T13:54:47.150Z
It should be $L = \frac{d\phi}{d\omega dAcos(\theta)}$
JarkkoL
CC BY-SA 3.0
2017-02-22T14:14:19.107Z
Yeah, sorry, fixed that now.
Ganea Dan Andrei
CC BY-SA 3.0
2017-02-22T20:11:00.083Z
I've posted an answer to this, but have been thinking about it and think I can see what you might be getting at. If there was some differential flux (even a teeny weeny bit) at a grazing angle and $\theta$ was **very close** to $90°$, then the radiance would be very large because of the very small denominator. Is this what you're getting at? If so, I'm not sure what the explanation would be and wonder "is there any surfaces that would  have a significant $d\phi$ at a grazing angles?"
PeteUK
<raytracing><lighting><radiosity>
Radiance at a nearly perpendicular angle is almost infinite?
Ganea Dan Andrei
# Radiance at a nearly perpendicular angle is almost infinite?

(Asked by: username_0 on 2017-02-22)

As far as I understand, radiance is a measure of the "intensity" of light in a direction defined as a differential cone . It's measure is Flux per solid angle per projected area. $L = \frac{d\Phi}{d\omega dA^\perp} = \frac{d\Phi}{d\omega dAcos(\theta)} $. Using this definition, it seems like as the $\theta$ angle increases, radiance increases, thus light is somehow more "intense" in a direction nearly perpendicular to the normal at point p. This contradicts my intuition.  
Where am I interpreting this wrong?

**Tags:** raytracing, lighting, radiosity

> **Comments:**
> **username_1**: It should be $L = \frac{d\phi}{d\omega dAcos(\theta)}$  
> **username_0**: Yeah, sorry, fixed that now.  
> **username_2**: I've posted an answer to this, but have been thinking about it and think I can see what you might be getting at. If there was some differential flux (even a teeny weeny bit) at a grazing angle and $\theta$ was **very close** to $90°$, then the radiance would be very large because of the very small denominator. Is this what you're getting at? If so, I'm not sure what the explanation would be and wonder "is there any surfaces that would  have a significant $d\phi$ at a grazing angles?"  

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## 2 answers

### Answer by username_1 on 2017-02-22. Score: 4 | Accepted answer

It's only more intense if $d\phi$ remains constant. However, for light source with constant luminance the flux is a function of $cos(\theta)$, i.e. depends on the light's angle of incidence with the surface and cancels out the term.

Edit: I feel the need to clarify this a bit because you say that:

> thus light is somehow more "intense"

It's not that the "light is somehow more intense", but rather that the light **has to be** more intense to have same flux in grazing angles as when it's parallel to the surface. If you think of flux as particles hitting to this differential surface, the number of particles hitting the surface is smaller more perpendicular the light gets to the surface. So you have to throw more particles at the surface (i.e. increase the radiance) for the same flux until when the surface is completely perpendicular to the light, none of the particles hit the surface, thus the radiance goes to infinity.

> **Comments:**
> **username_0**: I know calculating irradiance, for example, cancels out the term, but this doesn't answer my question regarding radiance itself.  More precisely, how can it approach infinity when theta is approaching 90.  
> **username_1**: Because if the flux stays constant, you need infinite radiance that there's any flux through a surface that's perpendicular to the light source  
> **username_0**: It makes sense that the light "has" to be more intense in this context, indeed.  

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### Answer by username_2 on 2017-02-22. Score: 1

Remember radiance corresponds to a particular direction. When the equation is considering directions quasi parallel to the surface normal, the differential flux (the numerator) will typically be a much greater quantity than for grazing angles. The numerator is not remaining constant.


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