# B Light intensity independant of frequency and E = hf

1. Nov 2, 2016

### DoobleD

Isn't it weird that light intensity (which is time average of Poynting vector) doesn't depend of the light frequency, while the energy of a photon does ?

From E = hf it seeems that frequency would have an impact of light energy flux (even time averaged). But intensity, which is a (time averaged) measure of energy flux, doesn't depend of frequency ? What am I missing ?

I found a close debate but not really answering here.

2. Nov 2, 2016

### Jonathan Scott

Light intensity is the total amount of energy per time, regardless of the number of photons arriving. If the photons are a higher frequency, then you'd need less of them arriving per unit time to achieve the same intensity.

3. Nov 2, 2016

### DoobleD

So somehow intensity is dependant of the number of photons. Doesn't this implies that intensity indirectly depends of frequency (since the energy of each photon does) ? But frequency (or number of photons) doesn't show up in <S>.

4. Nov 2, 2016

### Jonathan Scott

This is true of any sort of "stuff" which comes in packages. The amount of stuff is simply the number of packages times the amount in each package.

5. Nov 2, 2016

### DoobleD

Right, but what puzzles me is that frequency doesn't show up in the formula for intensity. Yet, it seems to depend of it.

6. Nov 2, 2016

### Jonathan Scott

Consider two electromagnetic waves of the same amplitude but different frequency. The time-averaged Poynting vector relates to the mean square value, which is the same for both for any whole number of cycles.

7. Nov 2, 2016

### Staff: Mentor

I believe this is true of many formulas in physics. A property can depend on another property without having it appear directly in one of the equations. Usually one of the variables already contains that property and it isn't shown for convenience and simplification.

8. Nov 2, 2016

### pixel

If we start with an expression for the plane wave solution for the electric field, E = Em sin(kx-ωt), and similarly for B, we end up with an expression for the average intensity that depends on Em2. So I think you are asking why the frequency doesn't appear in this result. But Em, the amplitude of the electric field is proportional to the number of photons and the energy of each photon (mixing metaphors here - wave and particles descriptions). For a given number of photons, higher frequency ones will have more energy and therefore larger values of Em. Just another way of saying what Jonathan and Drakkith have already pointed out.

9. Nov 3, 2016

### DoobleD

Oh ok, I didn't thought to that.

So in this case I suppose :
- the photons of the wave with higher fequency each have a higher amount of energy (E = hf),
- but the wave with lower frequency has more photons,
- so that their respective maximum amplitudes $E_m$ (thus their intensities $E_m^2$) are the same values.

Is this correct ?

Last edited: Nov 3, 2016
10. Nov 6, 2016

### Staff: Mentor

Pretty much, yes.

There are some subtleties about what a photon is (almost certainly not what you're thinking) that you'll find discussed in some other threads here, but the basic arithmetic of the same amount of energy being delivered by a larger number of photons with less energy in each one is OK.

11. Nov 6, 2016

Thank you !