How can you draw the wavelength of a wave to show that when a slit is of the order of the wavelength diffraction is at a maximum? I was looking for a diagram to show how when the wavelength of the wave is of the order of the slit width, then diffraction is maximised.
The diagram is supposed to be an intuitive explanation for this, because I just don't know how to visualise wavelength and how diffraction is affected by it.
 A: I don't know if this will hold as an "intuitive explanation", but I will try my best.
You have to remember that light is a wave and that it obeys the principle of superposition: at any point in space, the total electric field is the sum of all electric fields coming from different directions.
Now consider what happens if you come with a plane wave (i.e many individual waves that oscillate together) on a small aperture (with a size on the order of $\lambda$). According to the Huygens-Fresnel principle, you can guess what will happen after the aperture by "drawing" a lot of waves going in every direction at every point of the interface, and trying to see at which condition these waves will stay in phase. If you look at the forward direction, you can convince yourself that even after a long distance, all waves will still be oscillating at the same time, and thus the resulting amplitude will be important. See the relevant drawing below:

However, you also have to consider waves going in different directions. A simple criterion to know wether or not the total intensity will be zero is to only consider one wave on the top of the aperture, and one wave in the middle of the aperture. If these two add up destructively, you can show that all waves from the top part of the aperture will add up destructively with one wave of the bottom part of the aperture, and thus the resulting intensity will be zero (otherwise, it is more complicated, because you would have a add up an infinite number of waves, one for each point of the aperture, in order to find the total intensity). Consequently, this criterion will only be useful to find the directions in which the light is not transmitted.
The idea behind this is that, for the two waves to add up destructively, the red wave (for instance) has to "lag behind" the blue wave until it has a delay of half a wavelength (see picture below). This is possible because as you can see, if you consider an angle $\theta$ in the upward direction, the red wave will have to travel more than the blue wave to reach the same point. The greater $\theta$, the more dramatic the effect. But intuitively, if the diameter of the aperture $a$ is "small" (i.e on the order of a few $\lambda$), then the starting positions of the blue and the red wave are very close one to another, so you will have to select large values of $\theta$ in order to create a significant path difference. For instance, if $a = 2 \lambda$ (shown below), you can compute the corresponding angle to be $30^{\circ}$, which is significant.
 
As you can see above, when looking far away from the aperture, the blue and red waves appear to be exactly cancelling each other. This would be the same for any pair of waves separated by $a/2$ (one in the top part of the aperture, one in the bottom part of the aperture).
Of course, you could argue that I have only shown that the waves will add up to $0$ at a given angle $\theta$, not that there was actually diffraction occuring. In fact, when working with all the waves, not just two, you can show that most of the light will be distributed between an angle $0$ and $\theta$, and that the light field will be small beyond $\theta$. This means that the typical "angle of diffraction" is indeed $\theta$.
Maybe you have figured out what happens when you increase $a$. Because we add up one wave from the bottom of the aperture with one wave from the center, if you increase $a$, you increase the separation between blue and red at the start. But that means that for a given angle, the delay accumulated by red WRT blue will be larger, or, similarly, that the angle $\theta$ for which the waves add up to $0$ will be smaller than before (you can show in fact that $\theta$ is given by the equation $\sin(\theta) \simeq \theta = \lambda/a$, which is decreasing when $a$ increases). You can see it more clearly on the next picture, where $a$ was increased compared to the previous case.

I have not drawn the waves "on top" of each other for convenience, but you can see at the end that the two waves have the position of their maxima and minima inverted.

Last but not least, you could ask me "If $\theta$ is increasing for $a$ decreasing, does that not mean that we will have even better diffraction for $a \ll \lambda$?". This is in fact not the case. In the above formula, for $\theta$, i.e $\sin(\theta) = \lambda/a$, you might notice that there is no solution for $a < \lambda$, because otherwise $\sin(\theta)$ would be large than $1$. What happens for $a < \lambda$, which is not described by the simple arguments I have developed above, is that there will be light in most directions, BUT a significant fraction of the power will be lost to so-called evanescent waves that are absorbed in the transverse direction of the aperture. See the corresponding picture below (I have left out the "many waves" picture, because I don't think it can explain this phenomenon on its own):
 
With everything added, it seems quite reasonable that the maximum of diffraction (not accounting for the evanescent waves), is reached for $\lambda \sim a$. Hope this helps.
