<blockquote>
<p>Why 2 is divided by 10^-19?</p>
</blockquote>
<p>A <a href="https://en.wikipedia.org/wiki/Photon" rel="nofollow noreferrer">photon</a> of a specific wavelength has a specific energy; no more, no less — light is <a href="https://en.wikipedia.org/wiki/Quantum" rel="nofollow noreferrer">quantized</a>, and photons are the quanta of light, the minimum-size piece it comes in. Whenever there is enough light to see by, that light is made up of <em>very many</em> photons, such that the energy of all the photons summed together makes up the amount of energy emitted. Therefore, we can use the energy of one photon as the conversion factor to convert between energy and photon count, or between power and photon rate.</p>
<p>When we divide <span class="math-container">$2\,\mathrm{W}$</span>, the total power, by <span class="math-container">$3 \cdot 10^{-19}\,\mathrm{J}$</span>, the energy of one photon of this wavelength, we get the rate at which those photons being emitted, <span class="math-container">$0.66 \cdot 10^{19}\,\mathrm{/s} = 6.6 \cdot 10^{18}\,\mathrm{/s}$</span>, such that all their energies together make up <span class="math-container">$2\,\mathrm{W}$</span> of power.</p>
<blockquote>
<p>Suddenly, 10^12 comes out but I have no idea how did the author calculated...</p>
</blockquote>
<p>You're not missing something obvious; they didn't explain that well. The key thing to think about is that given a light source in a room, all the light hits objects in the room. They're <em>comparing areas</em> as a way to compute approximately how much light hits one particular piece of surface in the room. They're saying that there are <span class="math-container">$10^{6}\,\mathrm{cm}^2$</span> of wall/object surface area in the room, so we can divide the number of photons being emitted by the light source in all directions, around <span class="math-container">$10^{18}$</span>, by the <span class="math-container">$10^6\,\mathrm{cm}^2$</span> to obtain <span class="math-container">$10^{12}$</span>, the approximate number of photons (per second) falling on an individual <span class="math-container">$\mathrm{cm}^2$</span> of the room.</p>
<p>This calculation is very approximate and not especially relevant to actually making computer graphics; it's meant to give you an idea of the rates/quantities involved in the real physics, to contrast with how the simulations that we build for graphics work.</p>

CC BY-SA 4.0
2022-12-11T00:55:36.247Z
I appreciate your detailed explanation!
Winter Endless
CC BY-SA 4.0
2022-12-10T15:37:46.520Z
2022-12-11T02:53:11.010Z
2022-12-11T02:53:11.010Z
Kevin Reid
<p>Recently I'm reading the book computer grahpics : principle and practice and I encountered first weird things. Ordinary bulb consumes 100W per second, and 2% is used to emit the light. Then how do I know the photon of the light bulb? The textbook said that 2W is divided by 10^-19.. What I tried to understand is followed : 2W is the energy consumed, and typical photon is 10^-19. Why 2 is divided by 10^-19?</p>
<p>The example is followed; area is 100m^2 -&gt; 10^6, and it consumes 100W.=10^2. Suddenly, 10^12 comes out but I have no idea how did the author calculated...</p>
<p>It looks like very simple, but I'm currently struggling.. It would be appreciated if you give any single hints. Thank you<a href="https://i.sstatic.net/PPENT.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PPENT.jpg" alt="enter image description here" /></a><a href="https://i.sstatic.net/WDwSs.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/WDwSs.jpg" alt="enter image description here" /></a></p>

<photon-mapping>
Calculation of the photon of the object
Winter Endless
# Calculation of the photon of the object

(Asked by: username_0 on 2022-12-10)

Recently I'm reading the book computer grahpics : principle and practice and I encountered first weird things. Ordinary bulb consumes 100W per second, and 2% is used to emit the light. Then how do I know the photon of the light bulb? The textbook said that 2W is divided by 10^-19.. What I tried to understand is followed : 2W is the energy consumed, and typical photon is 10^-19. Why 2 is divided by 10^-19?

The example is followed; area is 100m^2 -> 10^6, and it consumes 100W.=10^2. Suddenly, 10^12 comes out but I have no idea how did the author calculated...

It looks like very simple, but I'm currently struggling.. It would be appreciated if you give any single hints. Thank you[![enter image description here](https://i.sstatic.net/PPENT.jpg)](https://i.sstatic.net/PPENT.jpg)[![enter image description here](https://i.sstatic.net/WDwSs.jpg)](https://i.sstatic.net/WDwSs.jpg)

**Tags:** photon-mapping


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## 1 answers

### Answer by username_1 on 2022-12-10. Score: 2

> Why 2 is divided by 10^-19?

A [photon](https://en.wikipedia.org/wiki/Photon) of a specific wavelength has a specific energy; no more, no less — light is [quantized](https://en.wikipedia.org/wiki/Quantum), and photons are the quanta of light, the minimum-size piece it comes in. Whenever there is enough light to see by, that light is made up of *very many* photons, such that the energy of all the photons summed together makes up the amount of energy emitted. Therefore, we can use the energy of one photon as the conversion factor to convert between energy and photon count, or between power and photon rate.

When we divide $2\,\mathrm{W}$, the total power, by $3 \cdot 10^{-19}\,\mathrm{J}$, the energy of one photon of this wavelength, we get the rate at which those photons being emitted, $0.66 \cdot 10^{19}\,\mathrm{/s} = 6.6 \cdot 10^{18}\,\mathrm{/s}$, such that all their energies together make up $2\,\mathrm{W}$ of power.

> Suddenly, 10^12 comes out but I have no idea how did the author calculated...

You're not missing something obvious; they didn't explain that well. The key thing to think about is that given a light source in a room, all the light hits objects in the room. They're *comparing areas* as a way to compute approximately how much light hits one particular piece of surface in the room. They're saying that there are $10^{6}\,\mathrm{cm}^2$ of wall/object surface area in the room, so we can divide the number of photons being emitted by the light source in all directions, around $10^{18}$, by the $10^6\,\mathrm{cm}^2$ to obtain $10^{12}$, the approximate number of photons (per second) falling on an individual $\mathrm{cm}^2$ of the room.

This calculation is very approximate and not especially relevant to actually making computer graphics; it's meant to give you an idea of the rates/quantities involved in the real physics, to contrast with how the simulations that we build for graphics work.

> **Comments:**
> **username_0**: I appreciate your detailed explanation!  

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