What is the purpose for the blackbody radiation graph to be graphed using the below parameters?

Question

If you observe the above graph, for y axis, "intensity per wavelength" is used as the parameter. I am aware we use "per wavelength" because it is hard to measure and graph isolated wavelengths, thus a small range is measured.

But, my question is, why divide it by the wavelength range. Now we have different parameter from intensity. We already had the intensity of the range we measured, so we could have just assumed "because it was a narrow range all the wavelengths in that range had same intensity."

the reason I have this problem is because, the graph above doesn't give us a clear representation of how intense a certain wavelength is. For example, my textbook gives this this graph and tells us to observe of lower wavelengths have higher intensity. How could we say that since, in order to get the intensity we have to multiply the height of the graph (intensity per wavelength value) at a certain wavelength from its wavelength. Then we will have a different graph.

I don't know if the graph above (made by myself) is correct but, it signifies my logic. Observe how I made the slope on the right hand side less steeper while making the left hand side more steep. I did this because to get intensity of a certain wavelength we need to multiply x value with y value like I said earlier (in my opinion. please correct me if I am wrong). So, for lower wavelengths the multiplied product is lower than before and for higher wavelengths it is high because we multiply a big value with a small value ( I am trying to say is it doesn't vary much in the right hand side due to that reason). Please help me understand the graph and correct my confusions.

Answer 1

I think you're misinterpreting the "per wavelength" part of the description of the quantity. For example, if $B(\lambda)$ is the spectral density with respect to wavelength, then the total intensity of radiation between 600 and 601 nm would be approximately $B(600 \text{ nm}) \times (1 \text{ nm})$; the total intensity of radiation between 1200 and 1201 nm would be approximately $B(1200 \text{ nm}) \times (1 \text{ nm})$; and so on. The process of finding the amount of radiation in a small wavelength band between $\lambda$ and $\lambda + \Delta \lambda$ only involves multiplying by $\Delta \lambda$, and does not involve multiplying by $\lambda$ itself.

I do say "approximately" above because the statement above is only valid when $\Delta \lambda$ is "sufficiently small". More generally, the intensity between two wavelengths $\lambda_1$ and $\lambda_2$ will be $$ \int_{\lambda_1}^{\lambda_2} B(\lambda) \, d \lambda $$ which has a natural interpretation as the area under the curve of $B(\lambda)$ between the two wavelength bounds. It's also not hard to see that my "approximate" statements above for narrow wavelength bands are in fact good approximations to this integral so long as $B(\lambda)$ doesn't change significantly over the wavelength band.

Answer 2

Spectral radiance is a function of wavelength, i.e., it is changing continuously with wavelength (or frequency, which is related via $\lambda f=c$), following the Planck's formula. We can approximately speak of intensity in a region of width $\Delta\lambda$ about wave length chosen wave length $\lambda$, but it is an approximation, since we are dealing with a continuous function, i.e., with calculus of infinitesimals.

The regions of infrared, visible light and ultraviolet are chosen for illustration - indeed, the range of visible spectrum is not clearly defined and may somewhat vary from person to person, although not significantly. See electromagnetic spectrum.