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?
[Physics] Relation between frequency and intensity of light
frequencyphotoelectric-effect
Related Solutions
Well it may help to first simply consider the famous the photoelectric effect, and see how photon frequency $\nu$ is related to the photoelectron's kinetic energy, once you are sure you understand up to this point, then we can move on to your question on photoelectric current.
The picture is a typical diagram used to elaborate on the photoelectric effect, where the photon's energy $h\nu$ has to be larger than the electron's binding energy to its atom usually denoted as work function $E_0$, only then a photoelectron is created. As for its kinetic energy, just make use of the Energy Conservation in photoelectric effect: $$h\nu = E_0 + \frac{\hbar^2 k^2}{2m} $$ where the second term in the right hand side is the kinetic energy, $h$ Planck's constant, $m$ mass of the electron and $k$ its wavenumber (momentum $p=\hbar k$)
Now back to our photoelectric current:
When a photosensitive surface is subject to incident light (x-ray for example), photoelectrons can be ejected from the surface (metallic surface) if the photons' frequency is high enough to reach the necessary work function of the metal, then electrons can be ejected, and now you should already know how to define their kinetic energy. It is important to note again that the maximum kinetic energy depends on the frequency of the photons and not the intensity of the ray.
Next step: if the electrons now reach a collecting plate, a current can be detected. Furthermore if an external retarding potential is placed between the metallic surface and the collecting electrode, the current can be reduced, because at high enough potentials, even the fastest electrons will be prevented from reaching the collector. With the potential $U=qV$ (charge $q$, the voltage $V$), the work-energy theorem is written simply $W = \Delta KE = -\Delta U$. The electron starting from rest, strikes the plate at zero potential relative to its first plate: $\Delta KE = \frac{1}{2}mv^2$ and $-\Delta U = qV$, so an electron failing to reach the plate, must have had a kinetic energy of $KE = eV$, where $eV$ is the work done on charge moving through the retarding potential V. The kinetic energy of the fastest electrons can then be obtained by finding the critical retarding potential necessary to reduce the current flow to zero: $$eV_{crit}=h\nu - W $$
Finally as for the intensity of the ray and current saturation, if the intensity is high enough (very high number of incident photons) then all the electrons get the chance to be ejected (assuming $\nu$ high enough) and contribute to the current, once all possible electrons are ejected from the plate, saturation current can be reached.
This overview should give you the necessary tools to mull over your next questions on your own, but feel free to ask if you face new "understanding" problems.
Best Answer
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 $n=\frac{N}{A}$?