Photon Arrival Rate, Chapter 1.5 from Computer Graphics by Folly

[CGandC]

Homework Statement

( Not a homework but something I wanted to understand: ) I haven't understood how he got to the result in the following calculation, that if there exists a single 100W light bulb, emitting an order of $10^{12}$ photons per second and illuminating an office of 4m x 4m x 2.5 m, then each photon of the $10^{12}$ photons emitted during the duration of a second, arrives to a square centimeter of surface each second.

Relevant Equations

Don't know, probably equation relating number of photons emitted from a source per second to how much distance they had traveled ( probably photon flux )

I've been reading Computer Graphics: Principles and Practice by Folly and I've encountered the following paragraph in chapter 1.5.

A single photon (the indivisible unit of light) has an energy $E$ that varies with the wavelength $\lambda$ according to $E=h c / \lambda$
where $h \approx 6.6 \times 10^{-34} \mathrm{~J}$ sec is Planck's constant and $c \approx 3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ is the speed of light; multiplying, we get $E \approx \frac{1.98 \times 10^{-25} \mathrm{Jm}}{\lambda}$ Using $650 \mathrm{~nm}$ as a typical photon wavelength, we get $E \approx \frac{1.98 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{650 \times 10^{-9} \mathrm{~m}} \approx 3 \times 10^{-19} \mathrm{~J}$ as the energy of a typical photon. An ordinary $100 \mathrm{Watt}$ incandescent bulb consumes $100 \mathrm{~W}$, or $100 \mathrm{~J} / \mathrm{sec}$, but only a small fraction of that-perhaps $2 \%$ to $4 \%$ for the least efficient bulbs-is converted to visible light. Dividing $2 \mathrm{~J} / \mathrm{sec}$ by $3 \times 10^{-19} \mathrm{~J}$, we see that such a bulb emits about $6.6 \times 10^{18}$ visible photons per second.

An office-say, $4 \mathrm{~m} \times 4 \mathrm{~m} \times$ $2.5 \mathrm{~m}$-together with some furniture has a surface area of very roughly $100 \mathrm{~m}^{2}=$ $1 \times 10^{6} \mathrm{~cm}^{2}$; thus, in such an office illuminated by a single $100 \mathrm{~W}$ bulb on the order of $10^{12}$ photons we arrive at a typical square centimeter of surface each second.

By contrast, direct sunlight provides roughly 1000 times this arrival rate; a bedroom illuminated by a small night-light has perhaps $1 / 100$ the arrival rate. Thus, the range of energies that reach the eye varies over many orders of magnitude. There is some evidence that the dark-adapted eye can detect a single photon (or perhaps a few photons). At any rate, the ratio between the daytime and nighttime energies of the light reaching the eye may approach $10^{10}$.

I haven't understood how he got to the result in the following calculation, that if there exists a single 100W light bulb, emitting an order of $10^{12}$ photons per second and illuminating an office of 4m x 4m x 2.5 m, then each photon of the $10^{12}$ photons emitted during the duration of a second, arrives to a square centimeter of surface each second.

I attempted on my own as follows, by trying to look at how much distance the light will travel in a second in a straight line:
Using the known formula for distance under constant velocity and zero acceleration, $\Delta X = v \cdot t$, we have $t = 1 s$ and $v = 3 \cdot 10^9 m/s$, thus $\Delta X = 3 \cdot 10^9 m/s \cdot s = 3 \cdot 10^9 m$. I aimed to get a result of $\Delta X = 1 cm$ but obviously my result is blatantly false, but why? ( I'd say the photon arriving from the orthogonal Y axis also makes a distance of $\Delta Y = 1 cm$ , hence we'd multiply then and have that the photon travels $1 (cm)^2$ in a second )

 

[bob012345]

I believe the author means that ~$10^{12}$ photons hit each square cm every second.

 

[kuruman]

And farther above, the author says that "$\dots$ we see that such a bulb emits about $6.6\times 10^{18}$ visible photons per second." The author then divides that number by the total surface area of the room in cm² to find the average number of photons per second per cm². Although the distance from the source to a part of the wall determines the time of flight of each photon, the fact remains that all the photons per second emitted hit the wall, so on average the same amount of photons that are emitted the source are intercepted by the wall.

and have that the photon travels $1(cm)^2$ in a second )

Actually, a photon travels about $3\times 10^{10}$ cm in one second.

 

[Orodruin]

The text clearly states that the estimated emission is around $7\cdot 10^{19}$ photons per second, not $10^{12}$.

 

[Orodruin]

It should also be mentioned for completeness that the number of photons registered by an eye in a given time will depend on the reflectance of the walls. If wall reflectance is high, this will lead to more photons. If it is low and you are not looking directly at the light, there will be less photons. If you are looking directly at the light then the flux is more accurately described by the inverse square law.

 

[CGandC]

And farther above, the author says that "$\dots$ we see that such a bulb emits about $6.6\times 10^{18}$ visible photons per second." The author then divides that number by the total surface area of the room in cm² to find the average number of photons per second per cm². Although the distance from the source to a part of the wall determines the time of flight of each photon, the fact remains that all the photons per second emitted hit the wall, so on average the same amount of photons that are emitted the source are intercepted by the wall.

Actually, a photon travels about $3\times 10^{10}$ cm in one second.

Thanks. Indeed, I got the result the author got, following what you said: $\frac{6.6 \times 10^{18} \cdot \frac{photons}{second} }{ 10^6 \cdot cm^2 } = 6.6 \cdot 10^{12} \cdot \frac{photons}{second \cdot cm^2 } \approx 10^{12} \frac{photons}{second \cdot cm^2 }$.

In general, if I want to find it in books, how is the equation called which relates a point-source emitting photons to the number of photons from the source that swift through an arbitrary surface at some distance $\vec{r}$ from the source? is it called photon-flux equation?

 

[hutchphd]

is it called photon-flux equation?

No it is called doing geometry. The flux of photons emitted isotropically get spread over the area of the sphere of radius r . What are the units (dimensions) of a flux of objects? Some care must be taken to specify flux per steradian and total flux (per total sphere): there are some "funny" units

 

[SammyS]

Actually, a photon travels about $3\times 10^{10}$ cm in one second.

which is about 1 foot / nanosecond .
FWIW .

 

[kuruman]

which is about 1 foot / nanosecond .
FWIW .

or about 1.8 terafurlongs per fortnight .

 

[CGandC]

No it is called doing geometry. The flux of photons emitted isotropically get spread over the area of the sphere of radius r . What are the units (dimensions) of a flux of objects? Some care must be taken to specify flux per steradian and total flux (per total sphere): there are some "funny" units

So under the assumption that my photons are emitted isotropically from a point source, the flux of the photons over a surface will be the number of photons per second divided by the area of the surface they hit, regardless of the shape of the surface? ( Because from what I understand, the calculation of flux under the assumptions I wrote just depends on the area of the surface, not on its shape )

 

[hutchphd]

What matters is the area perpendicular to the the direction of flow. For a point source and a sphere that will always be true. For a general surface it will involve a vector "dot" product with the local surface normal. This is the usual notation for a projected surface area.

 

[Orodruin]

or about 1.8 terafurlongs per fortnight .

Or 1.

 

[bob012345]

Or 1.

1 what?

 

[kuruman]

1 what?

Just 1 as in "Using E = mc², the mass of a proton is 938.272 088 16 MeV."

 

[bob012345]

Just 1 as in "Using E = mc², the mass of a proton is 938.272 088 16 MeV."

It's cheating I tell you! Cheating! :)

 

[Orodruin]

It's cheating I tell you! Cheating! :)

I’ll reserve my constitutional rights to use a reasonable system of units where energy has dimensions of inverse length and time and length are the same dimensionally tyvm