If wavelength is just the distance between 2 consecutive waves, then shouldn’t waveheight be the determining factor on if it can reflect or pass through an object.

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For eg, in the wikipedia for radar, it says that they were only able to use radar with radio waves with a wavelength smaller than a plane’s length. But won’t wavelength be perpendicular to it’s direction of travel? But waveheigth is parallel, so won’t that be the deciding factor?

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Wavelength determines the frequency of the wave, whereas waveheight determines the amplitude of the wave. Waveheight is an important factor in the wave having enough energy to reach the object and be reflected back.

The ability of a material to absorb light is based mostly on the exact arrangement of that material’s electrons. When the light hits the material, the photons hit the electrons and are absorbed^(1), pushing the electron temporarily into a higher energy state. Electrons are able to ‘recover’ from this higher energy state more or less quickly based on their arrangement inside their atom – different atoms absorb photons at different rates, in other words. If a photon hits while an electron isn’t in position to absorb it, it’ll just pass that atom by without being absorbed^(2). The frequency of a light wave (one over the wavelength) determines the time interval between different photons hitting the material. If this frequency lines up well with the ‘recovery’ period of the electrons, the electrons will do a really good job of absorbing all of the photons and basically none of them will get through. If it doesn’t like up as well, only the occasional photon will get reflected/absorbed and almost all of the light will pass through, making the object transparent to that wavelength. This is also why (for example) glass is opaque to UV rays and infrared rays, but transparent to the visible rays between them – it happens to sync up for UV and infrared, but it’s just slightly off for the visible spectrum.

1. Or reflected, but it has to get absorbed before it gets reflected, so I’m just going to say absorbed. Reflection is also a bit more complicated, but in terms of this particular question that extra complexity mostly doesn’t matter.

2. Well, kinda. Photons aren’t quite distinct particles like that, because of quantum physics. And electrons don’t really jump between fixed positions based on energy level either. But you can treat it as working basically like that for the purposes of this explanation.

There is no waveheight, you’re picturing this as an ocean wave which is incorrect. There is no perpendicular up and down movement or height to a radiowave. Any diagram of a radiowave you’ve seen depicting this is using that axis to depict something else other than spatial extent. This is not a mechanical wave where something is actually moving up and down.

What is perpendicular to the wave direction of travel is the direction of the electric and magnetic fields. They point perpendicular to the wave, but they are directly in line with the wave. Imagine a compass in the middle of a very low frequency (as most are too fast) radiowave. The needle will simply swap north and south with the wave. A higher amplitude (not height) wave simply means a stronger electric and magnetic field. You could strap some more weight to the compass needle and it would still be able to turn it. However, if you move slightly off the wave, the compass won’t move. The wave does not exist outside of its path. Increasing the amplitude will not make the wave wider, just stronger. Amplitude is height for an ocean wave, it’s not height for a radiowave.

So waveheight doesn’t exist. But why does wavelength matter? You’re right, this is spatial and parallel to the wave travel. But the waves aren’t balls traveling in straight lines bouncing off, they’re waves. They don’t actually travel straight, not unless they have adjacent waves or a barrier keeping them straight. [They can diffract.](https://upload.wikimedia.org/wikipedia/commons/thumb/c/c2/Wave_diffraction_at_the_Blue_Lagoon%2C_Abereiddy.jpg/1280px-Wave_diffraction_at_the_Blue_Lagoon%2C_Abereiddy.jpg). This is water, and dealing with a small slit. But a small obstacle sees the same behaviour. And radiowaves do the same thing, but in 3D and invisible to us. You would not see the circular pattern in the photo here if the wavelengths were way smaller than that gap in the rocks they are passing through.

The radiowaves of radar are a continuous wavefront, they can simply morph around something smaller than their wavelength by their tendancy to try and spreadout. If the object is much larger than their wavelength, they can’t just ignore it, some reflects and it leaves a shadow behind it. Imagine ocean waves hitting a pole, like the one holding a dock up. They will just completely ignore it and go around it. Not through it, but they will warp by around it and look like it wasn’t even there. Small ripples from a stone thrown in a pond would hit the pole and bounce off though, leaving a shadow behind it. Wavelength dictates diffraction with different scale objects.

Should note, the radar is absolutely not going through a small plane. Radiowaves do not go through metal. See your other answers about why certain types of light can go through some things, but not others. I think photons being brought into the answer is not a good way at all to go about answering your question as this is a classical problem with a classical answer, it’s not really about photons and electron energy levels. But it does answer why glass is transparent and stuff like that. In general, no light from the longest radiowaves right up to visible light can go through a metal. Them being conductive allows them to repel any external electric field, like that from a radiowave.

RF engineer here.

The answers you’re given in this theead kind of miss the point. I was in your shoes for a very long time, even after I finished my degree, since this is a very unintuitive concept. However, an answer on Physics Stack Exchange really helped my understanding:

https://physics.stackexchange.com/questions/125903/why-does-wavelength-affect-diffraction

Hope this helps! I’ll be glad to explain further if you have anything to ask about this link.