Relation between frequency and intensity of light I was going through a question in photoelectric effect and it was a true/false which says that the intensity of the incident light gets doubled on doubling the frequency. The answer is given as true and the explanation is that $I=nh\nu$ where $n$ is the number of photons. So if frequency is doubled intensity also gets doubled. I really don't think this is right because if frequency is doubled keeping $I$ constant. Then the only thing possible is that $n$ is reduced to half due to which the photocurrent should also be reduced but that doesn't happen experimentally. 
So does the intensity of light depend upon the frequency?
 A: Yes, the intensity depends, in part, on the frequency.
Intensity is power per unit area. Power is energy per time. For a photon, the energy is $h\nu$. So, the intensity will be $$I=Nh\nu / A$$ if $N$ is the monochromatic photon emission rate (photons per second), $\nu$ is the frequency of the photons,  and $A$ is the area these photons are hitting.
If the only thing one changes is the frequency of the photons, then doubling the frequency will double the intensity.  Alternately, doubling only the emission rate, or focusing the photons to hit half the area will also double the intensity.
In the explanation you saw, maybe $n$ is the photons per time per area so that $n=\frac{N}{A}$?
A: It sounds as though the true/false question you read was poorly worded. If the frequency gets doubled, AND the number of incident photons per unit area does not change, then the intensity doubles.  So the question would only be certainly "true" if something in the wording indicated that the number of incident photons per unit area does not change.
