I have read that to measure the position of a particle to an accuracy $\Delta x$, we need light of wavelength $\lambda < \Delta x$. Is it true? Why is it so?


4 Answers 4


It is because the wavelength is being used as a yardstick to measure the object ( actually the uncertainty in its position). You can measure a 10 meter object with a meter stick, but you can't measure a 10 cm object with the meter stick (assuming no markings on the stick)

  • $\begingroup$ Is the accuracy in position a function of the wavelength of light used, i.e., does $\Delta x$ depend on $\lambda$?? $\endgroup$
    – user155325
    Jan 11, 2018 at 16:36
  • $\begingroup$ It depends on what you mean by accuracy. If you mean experimental accuracy, then yes. If you mean quantum uncertainty, then no. $\endgroup$ Jan 11, 2018 at 16:39
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    $\begingroup$ It seems to me that this answer basically restates the question without explaining anything. $\endgroup$
    – Javier
    Jan 11, 2018 at 16:40
  • $\begingroup$ So is it fair to say that if you were measuring a distance with a specific wavelength, you could only define that distance in whole units of that wavelength? Using the meter stick example, you could only tell whether an object is 5 meters away, not 5.1? It essentially provides a maximum "resolution"? $\endgroup$
    – thanby
    Jan 11, 2018 at 19:34
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    $\begingroup$ @thanby Roughly speaking that is correct. $\endgroup$ Jan 11, 2018 at 20:30

When you want measure the location of an object by throwing light to it and watching what is coming back to you, you are actually measuring the intensity of light coming from different positions (our eyes are an example of an apparatus that can measure this intensity pattern, though in this case the eyes measure the intensity of light coming from different directions and the brain interprets it in terms of positions).

A valid criteria to assert that a particle is localised in some position is to require that the intensity pattern you receive is sharply peaked around that position. Ideally, the peak would need to be really sharp for you to be extremely sure that the particle is in that position. However, imagine that the peak was not so sharp, but that it was something like the profile of a mountain (or a gaussian distribution). Then, the uncertainty of where exactly the particle is would be given by some measurement of how spread the intensity pattern was (like the width of the mountain).

Now, how can you get a sharply peaked intensity pattern? Consider throwing very long wavelength light. Each of the wave crests is almost constant through space, so the pattern you will receive from it after it has been reflected by the particle will be very wide as well. Thus, even if you receive a lot of them simultaneously, summing them will not create a sharp peak (it is like saying that summing almost constant functions can create a sharp peak function). On the contrary, if your wavelength is short, each of the wave crests fluctuates rapidly throughout space so that the intensity pattern of each of them will also fluctuate a lot. So summing them you can actually receive a sharp peak (this is actually the Fourier decomposition of a sharp peak, where mainly high frequency modes contribute).

More quantitatively, if you are using light of wavelength $\lambda$, you can show mathematically (using Fourier mode decomposition) that the width of the sharpest peak you can construct is actually $\lambda$.

All this explanation is without mentioning quantum mechanics, where light is quantised in photons. Einstein discovered that they have momentum $p \sim \lambda^{-1}$, so by the uncertainty principle you cannot by any means locate a photon with an uncertainty lower than $\lambda$.

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    $\begingroup$ I feel like the real question is why the wavelength which is measured in the direction of travel has any bearing on the width of the detected peak which is measure perpendicularly to the direction of travel? $\endgroup$ Jan 11, 2018 at 19:18
  • $\begingroup$ Good point. This is because the particle will not actually reflect the ingoing plane wave as an outgoing plane wave in the oposite direction, but more as a spherical wave. In the case of an extended object it is similar. This is why this phenomenon is studied as the diffraction of light rather than mere reflection. Actually, if the object is a crystal, you can use the diffraction pattern to see its atomic structure, which is pretty amazing. $\endgroup$
    – amrgoloh
    Jan 11, 2018 at 21:47
  • $\begingroup$ Is that because of Huygens' principle? $\endgroup$
    – Neil G
    Jan 18, 2018 at 15:50

Light will be diffracted by the object and, say, projected on a ruler. The angular span of the diffracted light is proportional to $\frac{\lambda}{\Delta X}$. You would like this ratio to be small for higher accuracy.


This may not be the answer expected, however it is not true if you have a light source that is very narrow bandwidth and stable.

You can interpolate between fringes on an interferometer. For example, a commercial laser interferometer using a 633nm source can achieve a 1nm resolution.

Even if you simply bounce light off the object the round trip path changes 2λ for every λ you move the object.

There are additional and alternative techniques that can improve upon that considerably.


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