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Wavelength Question

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MrJammin

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Why is it that if one wanted to measure a particle by scattering a light wave off it, the light wave would have to have a wavelength shorter than the particle.
 

dknguyen

Well-Known Member
Most Helpful Member
Hmmm. GOod question. I always just though it as the wavelength must be "finite" and must complete full period. So if the thing was smaller than the wavelength or something like that you couldn't see it. Really really hand wavey, but when I think hard about it I'm not sure I know why. Could it be something to do with diffraction?
 

3iMaJ

New Member
MrJammin said:
Why is it that if one wanted to measure a particle by scattering a light wave off it, the light wave would have to have a wavelength shorter than the particle.

Rather than thinking about this in terms of wavelengths stay with me and lets talk about this in the time domain. Consider the following problem:

We have a transmitter positioned at x = 0, and we have two targets a and b positioned at xa and xb respectively. What you've asked above is how long of a pulse can my transmitter have in order to resolve my two targets in range?

Lets do some algebra to figure out the 2nd question and then relate it to the first.

Lets say our transmitter has a pulse width of T. So the amount of time that it takes the front of our pulse to get to target a is xa/c seconds. Similarly the amount of time it takes for our pulse to get to target b is xb/c seconds. However if our pulse covers both xa and xb at the same time we cannot resolve targets a and b because we will not see individual echos from the targets, there will be just 1 echo. So that means we need our pulse width to be smaller than the distance between the two closest targets. In other words:

dr is the distance between target a and target b. if xa > xb dr = xa - xb.

dr > cT/2 The factor of 2 is due to the round trip time. Now, we're going to make a hand waving arguement and say that the bandwidth of a square pulse in time is B (its not, but we'll pretend it is because many people pretend in a similar fashion).

dr = c/(2*B). So how does this relate to wavelength?

Lets make a semi-sketchy arguement about wavelength here. It's sketchy here because wavelength isn't really determined by bandwidth (in most cases). So the resolution of particles isn't generally determined by wavelength either, but you can see how its related.

lambda = c/f: So

dr = lambda/2 or in inequality form: dr > lambda/2.

Thats how "wavelength" determines target resolution. Although I hope I showed here its really not that simple and there is some pretending that takes place. What you really should take away from this is that BANDWIDTH determines resolution, and a pulse isn't the only transmit signal that radar systems use. A popular type is a linear frequency modulated signal (which in reality isn't all that good) however it has a well defined bandwidth which makes determination of resolution quite simple.

You can see though if you're going to try and resolve something as small as a particle the bandwidth will have to be quite high which in turn drive the wavelength to be quite small.
 
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