Many of the things you write sound OK. But I wouldn't say that the other directions in the graph are mere strengths, they indicate the actual value (strength/magnitude and direction as well) of the electric and magnetic fields.
First note that technically the electric field is a vector and it is a field so it should have a vector (possibly zero at every point). Each vector has a head and a tail. You can think of the location of the tail as telling you the place where the field has a particular value. Then you can think if the difference between the head and the tail as telling you both the magnitude and the direction of the electric field at that point, in some specified unit system. So imagine a bunch of arrows all the same color the location of each is telling you where it is telling you the field and how the arrow points from there tells you the value.
Then draw the magnetic fields in a different color. And in both cases you can't draw a vector at every single point because it would just be too much to see them all.
For a plane wave traveling through vacuum there is a great deal of regularity.
If the wave travels in the x direction then the electric and magnetic fields all point in the y-z plane. And their values only depend on time and on x, so you can draw just one longitudinal line and it will tell you about all of the values at all the locations. And since the electric fields and magnetic fields all point in the y-z plane you can reimagine the y-z plane as like the independent axis of a graph.
And for a classical plane wave travelling through a vacuum your picture should have the electric and magnetic fields both be strong together at one plane of a fixed value in the longitudinal direction and then at a different plane corresponding to a different fixed value of the longitudinal direction. Strong together and weak together and then strong again but pointing in the opposite direction.
If so, you get a picture much like the first one you drew (though that is just for one the electric and magnetic fields, a wave has both). That is all fine for a classical wave travelling through a vacuum.
For your other questions I think you should look at existing questions about the different topics.
From a quantum perspective it is much more complicated. From a quantum view there isn't an electric or a magnetic field, there is a photon field. When you have a large number of photons all in phase with each other it can look or act like an electromagnetic wave, but it is still different, and if you have a small number or they aren't in phase then it just truly different.
The electric and magnetic fields classically are stand-ins for saying how charges interact (though they do have their own energy and momentum, pressure and stress, etc.) And in quantum mechanics different interactions are possible and so the photon field is a stand in for saying how those different interactions go.
Since in quantum mechanics the charges don't have a location and momentum saying that their momentum changes based on the field where the charge is located just isn't going to be possible because you have none of those things. And if you can't verify the field's values at any location it becomes difficult to say it is there in that way.
We make the objects we need to predict results, for quantum mechanics we need different objects because we are predicting different interactions. Only in some limits (that only hold sometimes) do we expect quantum mechanical effects to start to approximately look like classical effects.
As for a transverse spacing in a Faraday cage stopping a wave that varies only in the longitudinal direction remember that you were drawing only the simplest case, the case of a plane wave.
In a plane wave the entire plane parallel to the y-z plane has the exact same electromagnetic field. This very simple (to describe) solution allows the wave to travel entirely in the x direction, over time each part of the wave simply slides over, the new value at distance $x=x_0+\Delta x$ over is the old value at $x_0$ a time $\Delta x/c$ earlier.
But no real wave is ever that perfect. They might expand in a spherical front instead of a plane front. They might have a central region where it is strong and it gets weaker farther out (like a Gaussian beam) some people even make beams where the energy and momentum travels along a cone so there is a central strong region but if you place a small object there the region beyond it still gets light because the light from farther out came from a base of the cone that started out farther from the central region.
Sort of like if you had some racers all lined up for a race but instead if running in their own tracks they ran towards the same point that is in front on the center runner. The center runner gets there first, but if parts of the center of the track are damaged the runners that left later will eventually close in and it won't matter that that the runners originally closer to the center got affected.
So a real wave has a more complicated wave front. And in fact if you are trying to avoid the wires of the Faraday cage you need your beam to be focused to not extend out too far in the direction transverse to the direction of propagation. And this isn't restricting the amplitude, this is saying that the actual wave shouldn't have field values spread out in the yz plane, it should be focused to a small portion of the yz plane.
Imagine you shot narrow beams all towards the same hole in the Faraday cage, you need very precision aiming each point has to has "its x" aimed very precisely to get in that same hole.
In general if you aim your initial propagation directions with accuracy on the order of the wavelength then you gave changed the wave.
Thus is because the direction of propagation isn't a magic thing you can freely associate with a point. It depends on how the electromagnetic field varied in space. When you field didn't vary in the y and z directions then the field travelled in the x direction. Now you are trying to get each region to aim slightly differently so you try to adjust the electric and magnetic field here and there to have each part aim just right.
Waves tend to spread, so if you aim some at that one whole, lots will be lost in other directions and what aimed well for one whole is nor going to aim for the other holes, so the Faraday cage succeeds at blocking almost all fields coming at it assuming the cage is big enough to have many many holes.
I think many of these question can be (and have been) asked as separate questions.
The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by $E=h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy.
This is where the confusion comes. The wavelength and frequency characterize the emergent electromagnetic wave from very many photons. How the classical wave emerges can be seen here although it needs a quantum field theory background to understand it. The photon, as a quantum mechanical entity, has a quantum mechanical wavefunction. This wavefunction complex conjugate squared gives the probability density for the specific photon to be at $(x,y,z,t)$. The frequency in the wavefunction is the frequency of the possible emergent classical wave, but for the individual photon it is only connected with probability of manifestation, as for example in the single photon double slit experiments.
single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames
you ask:
"How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?
It does not. It takes zillions of photons to build up the classical electromagnetic wave. In the photos above each individual photon gives a little dot. The build up gives the probability density distribution for photons, and lo, there is a frequency associated with the interference pattern, even though the photon manifests individually as a dot at the $(x,y)$ of the screen.
That is why we need quantum mechanics.
Edit after this question became the main duplicate of another one, where I have a long answer/comment that might be of interest to readers.
Does a single photon have a wavelength or not? [duplicate]
Best Answer
The classical electromagnetic waves may have many forms. The simplest examples are "monochromatic" waves with a well-defined frequency $f$ – the number of periods per second. The wavelength – the distance between two maxima of the wave – is $\lambda=c/f$.
When you change something about the electromagnetic field at one point, i.e. when you try to modify an electromagnetic wave, you can't instantly modify anything at a distant place. The information and influence can never propagate faster than $c$, the speed of light in the vacuum. This is a completely general fact about Nature that follows from the special theory of relativity and it applies to quantum theory, too. If you start with a wave of a huge wavelength $L$ and modify quickly something about the wave in a small region and in a timescale much shorter than the period of the electromagnetic wave, in a hope that this allows you to "instantly" affect distant points of space, you will fail. Instead of making a long electromagnetic wave collaborate on your project, you will create some electromagnetic waves of a much higher frequency and shorter wavelength. Too long electromagnetic waves just don't allow you to make things "really quickly" – in times shorter than $T$, their period – and too locally – in regions much shorter than the wavelength. Whenever the resolution of time or space is much better, it just proves that some electromagnetic waves of much higher frequency and shorter wavelength are present.
To describe the waves in the microwave oven, it is enough to consider classical physics i.e. avoid the notion of photons. The microwaves – electromagnetic waves whose wavelength is just a little bit shorter than the size of the oven – have a negligible chance to get through the holes because these holes are much smaller than the wavelength. The mechanism is sometimes called electromagnetic shielding. How does it work?
Well, if you want to discuss waves of wavelength $\lambda$ only, their dependence on space must always have the form $A\cdot \cos(2\pi x / \lambda)$: waves with the right distance between the minima. However, the metallic cage surrounding the interior of the oven imposes the potential $\phi=0$ at a very dense network of points. When you try to write down the potential as $A\cdot \cos(2\pi x / \lambda)$, while making sure that it's zero at all points where there is a conductor, you will find out that there is no solution except for $A=0$. The wave of the given wavelength just can't get through at all. Alternatively, you could calculate the reflection from the metallic points of the mesh (not counting the holes). They would interfere with each other and guarantee that the probability of reflection is nearly 100 percent.
In other words, the microwave-frequency photon is "really large", at least as large as the wavelength, and it just "doesn't fit in". It could fit in if it "pretended" it was a shorter-wavelength photon but that would be a different one. The microwave oven prescribes a frequency $f$ and the corresponding wavelength is always $c/f$ and cannot change. You may also look at the situation with a "poor resolution" so that you "neglect" distances shorter than the wavelength. From this viewpoint, the metallic caging of the oven looks solid despite the small holes. That's ultimately why it looks solid to the electromagnetic waves.
All these things may also be phrased in terms of photons – which is needed in quantum theory – even though, as I said, it is not necessary for microwaves because they contain a huge number of coherent photons so that this large group of photons behaves classically.
Individual photons are described by wave functions that mathematically look like an electromagnetic wave. It may be monochromatic but it may be a mixture of different frequencies, too. Individual photons' wave functions propagate through the oven or anything else pretty much just like the classical electromagnetic waves. That's why photons of microwave frequencies can't escape from the microwave, either. But the interpretation is different: the wave function isn't directly measurable like the electric fields. Instead, it encodes (after squaring) the probability density that the photon is here or there.
If you try to catch a photon whose wave function is spread over a large region, you have a certain calculable chance that you will catch it "somewhere here", in a particular nearby region $V$. But when you do so, the probability is zero that the photon is simultaneously at some other, distance place $P$. So whatever you do to the photon here – if you are lucky at all and the photon "shows up" here – will not affect what is happening very far from you. Effectively, if you see the photon here, the "wave function collapses" and you're sure that the photon isn't "over there" so it can't do anything there. The photon was never over there; it only had a chance to be there but your measurement showed that the chance didn't materialize. These points are often misunderstood because people are trying to imagine that the wave function of the photon is a real wave that must leave some traces even if the photon is ultimately seen elsewhere. But it can't and it doesn't leave any traces whatsoever: the nonzero wave function only quantified a "potential" that something may be seen somewhere. When it's seen elsewhere, it's a proof that the potential was never "real".
At any rate, you will see that photons can't be used to send superluminal signals, either.