# Number of photons in range of frequencies

I was trying to calculate the number of photons emitted by a light of constant power $$P$$ between frequencies $$\nu_1$$ and $$\nu_2$$. I have already checked this question but the reply marked as correct asserts that $$\dfrac{dP}{d\nu}$$ is a constant and has a value of:

$$\dfrac{dP}{d\nu}=\dfrac{P}{\nu_2-\nu_1}$$

However, I cannot see where this result comes from. Shouldn't $$\dfrac{dP}{d\nu}=0$$ since $$P$$ is constant throughout $$[\nu_1,\nu_2]$$.

First of all, $$P$$ is the total power. When they say, $$dP$$, they mean a portion of the power belonging to a frequency range $$d\nu$$. From $$E=h\nu$$, photons of different frequency would radiate with different power.

Now, just to explain what that formula in the question is:

$$\frac{dP}{d\nu} = \frac{P}{\nu_2 -\nu_1}$$Assume $$P(\nu)$$ or the power emitted along one frequency alone (i.e. power is a function of frequency, as we know it is). In fact, if you plotted a graph with the power on the y-axis and frequency on the x-axis, you get a straight line (as expected due to $$E=hf$$).

$$\frac{dP}{d\nu}=\frac{P(\nu_2)-P(\nu_1)}{\nu_2-\nu_1}$$ And you can see that $$\int dP=P$$. If you put the limits on this particular integral, you can see that the result is $$P=P(\nu_2)-P(\nu_1)$$.

• I think I understand your reasoning, but I believe it does not agree with the fact that $P$ is constant throughout $[\nu_1,\nu_2]$. The exercise I am attempting to solve states that for any wave length between $[\lambda_2,\lambda_1$], power $P$ is constant. Moreover it includes a $P$ vs. $\lambda$ graph where you can see a horizontal line – Elementarium Feb 9 at 11:11
• If that were the case, then $dP/d\nu = 0$. The question you have linked and your is very different then – KV18 Feb 9 at 11:15
• But I don't understand how it can be so: then the only possible explanation for such a power $P$ for every single frequency is if the intensity varies. But how is that really possible? – KV18 Feb 9 at 13:21
• The question asked was the same as yours. But the answer given that was accepted, assumes $P$ to be the total power of the beam across the range of wavelengths $[\lambda_1,\lambda_2]$ – KV18 Feb 9 at 13:23
• Actually I cannot understand the situation either. That is why I decided to search some information here and found the answer I linked, and that answer only contributed more to my confusion – Elementarium Feb 9 at 13:24

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