# B Spectral Intensity as a Function of Wavelength in Blackbody Radiation

Tags:
1. Mar 7, 2017

### Samama Fahim

A blackbody is also a perfect emitter giving off electromagnetic waves at all frequencies. A detector could measure the intensity of the radiation it receives through the prism. By moving the detector to different positions, you could measure the intensity of light as a function of color or wavelength. (https://goo.gl/bqwl7K)

So each wavelength has a finite (not an infinitesimal) intensity? If we consider wavelengths in the interval [λ, λ + dλ] and each wavelength in this interval has a finite intensity, then the total intensity for this interval would be infinite because this interval has infinitely many wavelengths. Where am I going wrong? Are there not infinitely many wavelengths within this interval?

Between the red and green beams there are infinitely many beams with a finite intensity, so the total for this interval would be infinite?

2. Mar 7, 2017

### phyzguy

You need to review the concept of integration. By your argument, any integral would be infinite. The point you are missing is that as the number of intervals [λ,λ+dλ] increases to infinity, the size of the interval dλ decreases to become infinitesimally small in such a way that the total sum remains finite.

3. Mar 7, 2017

### Samama Fahim

I don't get it. For example the intensity of a red beam is x units. The neighboring beams would have an intensity between x + dx. If you add all the intensities and each of them has an x in them since they are greater than x, then for all the beams the sum of x's is infinite. How is it similar to finding distance which is an integral?

4. Mar 7, 2017

### phyzguy

I think I see what you are missing. The spectral intensity at a given wavelength is given by:
$$B(\lambda, T) = \frac{2 h c^2}{\lambda^5}\frac{1}{e^{\frac{h c}{\lambda k T}-1}}$$ ,
but if you want to know the total flux between λ and λ+dλ, you need to multiply this expression by dλ, giving:
$$B(\lambda, T) d \lambda = \frac{2 h c^2}{\lambda^5}\frac{1}{e^{\frac{h c}{\lambda k T}-1}} d \lambda$$

5. Mar 7, 2017

### Samama Fahim

Intensity or intensity per wavelength band? I don't know why it is so confusing. I know how integral works, I don't understand the physics.

6. Mar 7, 2017

### pixel

From a practical point of view, we can never measure the intensity of a single wavelength. Any measuring device has a wavelength range Δλ that it measures. Let's say we measure the intensity at λ0 with one instrument having a range Δλ1 and get a value of I1. Next we measure the intensity at λ0 with another instrument that has a smaller range, Δλ2, and get I2, where I2<I1. If we continue doing this with more and more accurate instruments, the resultant intensity measurement would continue to get smaller and smaller. In the limit as Δλ → 0, the measured intensity also → 0.

An analogy would be a probability density function, P(x). The probability of a value of region dx about x is given by P(x)dx. The probability of one specific value of x is 0 (dx = 0).

7. Mar 7, 2017

### pixel

According to the Wikipedia article on Black-body radiation, the first equation in phyzguy's post is (noting that c = ν λ):

the power (the energy per unit time) of frequency ν radiation emitted per unit area of emitting surface in the normal direction per unit solid angle per unit frequency by a black body at temperature T

Note all the instances of "per." It is not power, but rather a power density type of function, similar to the probability density function.

8. Mar 7, 2017

### Samama Fahim

So the measured intensity will go to zero as we shrink the wavelength interval? Is this like at any instant the distance traveled in that instant is zero? Δs → 0, as Δt → 0?

9. Mar 7, 2017

### pixel

I would say yes. Δs = v Δt (for constant velocity). I wouldn't make sense to ask how far a particle travels at a specific, single time t, but rather how far it moves in a time interval Δt.

10. Mar 7, 2017

### Samama Fahim

Apart from blackbody radiation case, any monochromatic light wave should have a zero intensity? Is that true?

11. Mar 7, 2017

### Samama Fahim

The intensity corresponding to any one wavelength or frequency should then be zero. Like some wave EM wave with a single frequency has zero intensity?

12. Mar 7, 2017

### Ron19932017

I think you can say " The intensity of a single frequency in a blackbody radiation is zero" but I would like to talk about an example.
Consider a class of students. We can then plot a histograms of weights. That means we are distributing the total number into spectrum of the students' individual weights. You can say that the number of student with exactly 50kg = 0. However if we do the sum/ integral we can get the number of students to be some number.
We need to keep in mind that the y axis on the histogram does not really mean number of students, but specific number of students. Just like the case in blackbody radiation we talked about specific intensity at a particular wavelength.

13. Mar 7, 2017

### phyzguy

Maybe it would be easier to think of it as counting photons. In your drawing in post #1, the red emitter is emitting a certain (finite) number of photons per second. The prism is splitting them up into wavelength bands, As you make the wavelength band dλ narrow and narrower, the number of photons per second arriving in that band gets smaller and smaller. In the limit of dλ=0, the number of photons per second arriving in that bandwidth goes to zero.

14. Mar 7, 2017

### pixel

A plane wave, with a single frequency, is an idealization since it would have to fill all space.

15. Mar 10, 2017

### Samama Fahim

What's the frequency/wavelength of these photons?

16. Mar 10, 2017

### phyzguy

The source is emitting a large number of photons, each with a different wavelength, distributed according to the Planck law. Is that your question?

17. Mar 10, 2017

### OmCheeto

I think I found it so confusing, as there are WAY TOO MANY WAYS to describe the measurement of light, IMHO.

36, by my count.

It took me a weeks worth of googling, a lot of studying, and one nap, before I finally figured out why the y-axis on the "Planck's Law" graph was labeled in so many different ways.

wiki: kW/sr/m^2/nm