Intensity is an objectively measurable attribute of light. It is the rate at which energy is delivered to a surface. Intensity is energy delivered per unit time per unit area. The intensity of light is a measurement of photon irradiance, which is the number of photons delivered per square meter per second. You can measure intensity with a photoelement, such as a solar cell or a photomultiplier, which converts light to electric current. As the electric current varies strictly with the intensity of the light, an objective measurement is possible. The more energetic the photon (the shorter its wavelength), the fewer will be the number of photons required for a given intensity. See this link: http://www.pveducation.org/pvcdrom/properties-of-sunlight/photon-flux.
Brightness is a "subjective" quality of light. It depends on the perception of whoever is viewing the light. It can't be objectively measured, but it can be scaled, so that the same viewer (or viewers with similar perceptions) can agree that certain light is more or less bright. For instance a less bright surface may be deemed 50% of the brightness of a brighter surface. In astronomy, for example, stars may be graded according to their apparent magnitude, which is their brightness in comparison to a very bright benchmark star. Brightness may also be called luminous flux. Here is a list of units in which various luminous standards of brightness are scaled: https://en.wikipedia.org/wiki/Luminous_flux
Photon is a term used to describe the particle attribute of light. A photon may be considered the smallest packet of energy into which light can be separated.
You could deliver greater intensity by emitting light of a shorter wavelength, or by increasing the surface area emitting the light (greater surface area means more electrons emitting more photons).
The energy of each photon is measured in joules, and depends on the wavelength of the light according to this formula: Q = h*c / lambda, where Q is energy in joules, h is Planck's constant, c is the speed of light in a vacuum, and lambda is the wavelength of the light in meters. Photons are considered to be "massless particles", but since a photon has energy, it must have mass. This conundrum is solved by saying that photons have relativistic mass when they are traveling, but possess zero rest mass because they are never at rest. Here is a better explanation: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html.
The wavelength of a photon determines its energy, and its energy determines its wavelength. The shorter the wavelength, the shorter the cycle of the wave, the greater its frequency, and the more periods of the wavelength can be crammed into a unit time. Short wavelengths are more energetic than long wavelengths. As photons of light possess both particle and wave natures, a photon when traveling has velocity (c), momentum (relativistic mass * c), and wavelength, but when a photon is emitted or absorbed (at the beginning and the end of its journey), it is considered a particle.
The power of a light source can be measured in watts. A watt is the rate of energy of one joule per second.
The surface brightness of the Sun is -10.6 mag per square arcsecond.
The full moon on the other hand is about 14.5 (astronomical) magnitudes fainter than the Sun, has a similar apparent angular size and is just visible in a bright daytime sky.
The flux from the daylight sky incident upon the eye is therefore around $10^{14.5/2.5}$ times less than the solar constant. i.e. About $2\times 10^{-3}$ W/m$^2$.
The pupils of the eye might have a 2mm diameter in bright light, so receive around $6.2\times 10^{-9}$ W.
Let's assume that the average blue sky photon is at 400 nm with an energy of 3.1 eV, then you receive about $10^{10}$ per second (in each eye).
Ah, but this would be correct for a small patch of blue sky with the same angular extent as the full moon (about $7\times 10^{-5}$ steradians). The eye actually collects light from $\sim \pi$ steradians, but then the projected area of the pupil is reduced by a small factor (I think 0.75) because of the $\cos \theta$ term.
So the final result is $3\times 10^{14}$ photons per eye.
Best Answer
You don't use EM or QED to explain most of this, because its more biology than anything. Hue saturation and brightness describe perceptive color. It's based on how the human mind perceives color. Thus, the one bit of EM you need is this chart:
This is a chart of the sensitivity of the three color receptive pigments in the eye (plus the one used for low light vision). When a wide spectrum of light hits your eye, each of the three color receptors responds to that light proportional to the integral of the number of photons of a given wavelength multiplied by this sensitivity curve at that wavelength.
After that step has occurred, the signal is now a data signal rather than a stream of photons, and our eyes process it as such.
Saturation - Saturation describes how much the signal appears to be a single wavelength. If it is a single wavelength, then there will be a very particular set of RGB signals that can come out of it. Our brain perceives this as saturation. On the other hand, if the light does not have one wavelength that stands out above the rest, we will perceive it as desaturated.
Hue - If the color is saturated, then hue represents what wavelength the eye perceives as being the only wavelength. This, of course, can be fooled by having 3 wavelengths carefully tuned to the colors our eyes are most sensitive to, which is how monitors work.
Brightness - The strength of the overall response. The more photons that hit the eye, the stronger the signal from the cones in our eyes, so the brighter we perceive it.
This one actually is an EM effect. Dry surfaces have a lot of rough edges which scatter light in all directions. Wet surfaces have a smooth surface of water on them. If you've ever been startled by the bright reflection of the sun off of a still pool of water, you have your answer. Most of the energy that hits a wet surface reflects in one direction. Thus, the object appears very bright in that one direction, and darker in all other directions.