Is the intensity of light ONLY dependent on the number of photons, and nothing else? Recently, my teacher just told us that intensity is not linearly dependent on temperature and that it's ONLY dependent on photons. But then, what about Boltzmann's law? Isn't intensity dependent on the fourth power of absolute temperature?
Even if you only consider temperature, we do observe that even though intensity is non-uniformly changed, increasing temperature does increase the MAXIMUM intensity. So.. technically, isn't intensity also non-linearly dependent upon temperature?
UPDATE: I asked my teacher this question and he explained that Stefan's law applies when we're restricted to Classical Physics, not Quantum Physics. So, when you're talking about the macro-level, AKA, a body emitting light as some of the answers here explain, Stefan's law applies. But when you're solely talking about a photon, Stefan's law doesn't apply and so.. intensity of a photon isn't dependent on temperature IN QUANTUM PHYSICS. Don't know how... that works.
UPDATE: I might have, sort of, figured it out.. fact of the matter is, when we're discussing BLACK BODY RADIATIONS, specifically, the intensity of light does increase with temperature, regardless of whether you're considering classical or quantum physics. Classical physics is correct in predicting the behavior of radiation emitted when the body was heated, but only to specific wavelengths. And, finally, Plank's Quantum Theory is correct for all wavelengths. So, yeah.. intensity of light coming from a BLACK BODY does increase, regardless of whether we're studying classical theories or quantum theories. Cheers, everyone :D! Thank you so much for giving me your time!
 A: Yes, but your instructer was refering to the intensity as measured by a detector independent of the source.  In that case the intensity detected is just the number of photons.  The temperature of a star will dictate the number of photons emitted.  But the number detected gives the intensity of photons on a detector.
A: The problem is that the statement is a little imprecise, so it has some truth and some falsity.
A photon is, by definition, a quantum of energy exchanged between the elecromagnetic field (more precisely: a mode of the electromagnetic field) and something else. So by definition the energy exchanged is proportional to the number of photons, and therefore the light intensity is proportional to the number of photons exchanged through a surface, divided by the surface area and the time of the exchange. We speak of "quanta" of energy because the electromagnetic field cannot give or receive energy continuously, but only in multiples of a minimum amount – the quantum, the photon.
But this isn't the full story. The energy of a photon depends on the frequency $\nu$ of the electromagnetic field exchanging it: the energy is $h\nu$, where $h$ is Planck's constant. So if the same number of photons is exchanged in two different situations, through equal areas and in equal amounts of time, the intensity can still be different in the two situations.
How does the temperature enter in all this? Roughly speaking, in many situations we're actually uncertain about the number of photons exchanged by the electromagnetic field. We only have a probability that a particular number is exchanged. Typically this probability depends on the temperature $T$. For example (thermal state of light) we can have that the probability that $n$ photons of frequency $\nu$ are exchanged is
$$P(n)=\frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ ,$$
where $k$ is Boltzmann's constant.
This means that the precise value of the intensity is also uncertain, because the number of photons is. We can calculate the average or expected value of the intensity: by definition this average is proportional to
$$\sum_n n\ P(n) \equiv \sum_n n\ \frac{\exp\bigl(-n\,\frac{h\nu}{kT}\bigr)}{1-\exp\bigl(-\frac{h\nu}{kT}\bigr)} \ .$$
To summarize, the intensity is proportional to the number of photons, but the probability of the exact number of photons depends on the temperature. Therefore the average intensity depends, indirectly, on the temperature. It's also important to specify whether we're speaking about the temperature at one specific frequency (or of one specific electromagnetic mode), or through the whole spectrum, and so on.
It's good to emphasize that notion of temperature in electromagnetism is tricky. From a classical thermodynamic point of view the electromagnetic field does not have any temperature. Rather, we speak of the temperature of the matter in which it resides or with which it interacts. From a statistical-mechanical point of view we can speak of the (statistical) temperature $T$ of the electromagnetic field, in the sense that if we're uncertain about its state we then give each possible state a probability proportional to $\exp[E/(kT)]$ (or some other Gibbsian formula), where $E$ is the state's energy. This is almost a definition of the (statistical) temperature of the field. This definition goes over to the quantum description, and that's where the formula for the probability given above comes from. The two notions of temperature get connected only through the presence of matter: in equilibrium the statistical temperatures of the electromagnetic field and of the matter it interacts with must be the same; but the statistical temperature of matter is in turn connected with its thermodynamic temperature (in ways that are still under debate).
Good references about all this are Leonhardt's Measuring the Quantum State of Light (Cambridge U. Press 1997); Jackson's Classical Electrodynamics (Wiley 1999); Hutter, van de Ven, Ursescu's Electromagnetic Field Matter Interactions in Thermoelastic Solids and Viscous Fluids (Springer 2006); and also Biró's Is There a Temperature?: Conceptual Challenges at High Energy, Acceleration and Complexity (Springer 2011).
A: Checking Wikipedia:

Several measures of light are commonly known as intensity:Radiant intensity, a radiometric quantity measured in watts per steradian (W/sr)
Luminous intensity, a photometric quantity measured in lumens per steradian (lm/sr), or candela (cd)
Irradiance, a radiometric quantity, measured in watts per meter squared (W/m2)
Intensity (physics), the name for irradiance used in other branches of physics (W/m2)
Radiance, commonly called "intensity" in astronomy and astrophysics (W·sr−1·m−2)

Without any further qualification, I would take "intensity" to mean the first. Wattage is number of photons per unit of time times energy per photon, and energy per photon is proportional to frequency.

Recently, my teacher just told us that intensity is not linearly dependent on temperature and that it's ONLY dependent on photons.

If you claim that intensity is dependent on temperature, that's a rather unclear what that means. The term "temperature" doesn't apply to photons, at least not directly. One can establish a correspondence between frequency and temperature, for instance by taking the energy and dividing by Boltzmann's constant. But that's the temperature that corresponds to the photon, not the temperature of the photon. Given that interpretation, the intensity is proportional to the temperature.
Another interpretation of the statement is that it's referring to the temperature of an object emitting light. And one can further interpret that as meaning the intensity of the light emitted by an object emitting black-body radiation at that temperature. Note that when an object is emitting light, it's not necessarily emitting it as black body radiation. If it is being emitting as black body radiation, then Boltzmann's law applies.
A: You are both at least somewhat right.
Photons are more or less particles of light. So the more particles, the more light. This is true at any temperature.
But making an object hot makes it emit more photons. Furthermore, it makes it emit photons of higher frequency and thus higher energy. This makes the energy go up faster than the number of photons.
There are ways of making photons other than heating an object. For example, a laser makes photons all of the same frequency. So there intensity is proportional to number of photons.
A: It depends how you defined intensity and what units you are measuring that in.
Lux, lumens, candela, watts/m^^2,  cycles per second, photons per steradian, or what?
