# Intensity vs. Photon Flux

1. May 17, 2015

### lulzury

1. The problem statement, all variables and given/known data
A sodium lamp emits light at the power P = 130 W and at the wavelength λ = 570 nm, and the emission is uniformly in all directions. (b) At what distance from the lamp will a totally absorbing screen absorb photons at the rate of 1.00 photon /cm^2s?

2. Relevant equations
1. (Rate of emission/absorption) $R= \frac {P\lambda} {hc}$
2. (Intensity) $I=\frac {P}{A} = \frac {P}{4\pi r^2}$
3. (Photon Flux) $I= \frac {R} {A} =\frac {R}{4\pi r^2}$

3. The attempt at a solution
My confusion here comes from the units of Intensity vs. Photon Flux
From the equations above, I get that
P = R, but this is not the case right?
Here is how I had set up my solution:
Since they give us I = 1.0/cm^2, this means I = 1e4 photons/m^2s

IF, I use the second equation
$I=\frac {P}{A}$
$I=\frac {P}{4\pi r^2}$
$r=\sqrt{\frac{P}{4\pi I}}$
$r = \sqrt{\frac{130}{4\pi 1e4}}$

this yields 0.0321 m, which is NOT the answer.
I get the correct answer using the third equation, but I don't understand why the first equation doesn't work in this case.

2. May 17, 2015

### TSny

The intensity is the amount of energy that strikes a unit area during a unit of time. In the SI system, this is the number of Joules of energy that strike a square meter of area each second. This is the same as Watts per square meter. (The area is assumed to be perpendicular to the direction of propagation).

The photon flux is the number of photons that strike a unit area during a unit of time. So the SI units would be number of photons per square meter per second. (Again, the area is assumed to be perpendicular to the direction of propagation.)

Intensity and photon flux are different quantities with different units.

The confusion might be due to using the same symbol $I$ for these two different quantities. This is not a good idea.

$I$ is usually used for intensity. For photon flux, people sometimes use $\Phi$.

You can convert from $\Phi$ to $I$ by multiplying $\Phi$ by the energy of each photon ($hc/\lambda$).

3. May 18, 2015

### lulzury

Thanks TSny, I checked the units of both photon flux and intensity and got different units.

It's so weird that my book decided to use the same symbol for different quantities.