The electromagnetic spectrum is a continuum of wavelengths of light, and we have labels for some ranges of these and numerical measurements for many.

Question: Is the EM spectrum continuous such that between two given wavelengths (e.g. 200nm and 201nm) there is an infinite number of distincts wavelengths of light? Or is there some cut-off of precision with which light might exist (e.g. can light only have wavelengths of whole number when measured in nanometers, etc.)?


4 Answers 4


Yes, there are an uncountable infinity of possible wavelengths of light.

In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2).

Two things to consider in practice:

  1. There are situations in which the only relevant frequencies are discrete (such as the modes in a cavity).

  2. For any given experimental measurement you will always have a finite precision or bandwidth with which you can measure, and so although light at 200nm and 200.01nm is in principle different, you might not be able to tell in practice.

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    $\begingroup$ Doesn't Planck time and Planck length contradict this? $\endgroup$
    – vsz
    Mar 9, 2015 at 8:30
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    $\begingroup$ Practically, two frequencies may not be distinguishable, if they differ by less then (1/lifetime of the universe) or something like that, no? $\endgroup$ Mar 9, 2015 at 9:32
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    $\begingroup$ @vsz: They might. We really don't know, as "Planck time" is more of a concept than a physical law. Experimentally we haven't been able to find one "impossible wavelength" and there's the relativistic problem of wavelength depending on the observer. $\endgroup$
    – MSalters
    Mar 9, 2015 at 10:42
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    $\begingroup$ The 'allowed' wavelengths could be rational multiples of some fundamental wavelength with no-one being any the wiser... Food for thought: is there an experiment that can be performed in a finite time to distinguish between there being a countably or an uncountably infinite number of wavelengths? $\endgroup$
    – yatima2975
    Mar 9, 2015 at 15:39
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    $\begingroup$ @yatima2975: The problem would be that there is exactly one inertial frame in which your minimal fundamental wavelength exists. This violates the very fundaments of relativistic physics. Now I'm not dismissing the idea merely because of that, but extraordinary claims require extraordinary proof. $\endgroup$
    – MSalters
    Mar 10, 2015 at 15:21

Formally there are an infinite number of different wavelenghts. However, any given physical system can only be found in a finite number of distinct physical states. To create a light source with a wavelength $\lambda$ that is well defined up to some resolution $\delta\lambda$, requires observing it within a system of size of the order of $\lambda^2/\delta\lambda$. So, the smaller we make $\delta\lambda$, the larger the system must be before we get physically distinct states within each such smaller interval.


Sir Elderberry, Punk_Physicist and the Count Iblis have all given correct answers in principle.

There are two phenomena (really thought experiment, rather than practical, devices) that one needs to heed.

  1. A finite measuring time $T$ can only resolve frequencies to within an uncertainty of the order of $1/T$. This is the reciprocal relationship between the time spread $\Delta t$ in a pulse and the frequency spread $\Delta \omega$ in its Fourier transform, given by $|\Delta t|\,|\Delta\,\omega|\geq\frac{1}{2}$ as I show in my answer to the Physics SE question "Heisenberg Relation". This is the mathematical phenomenon underlying the Heisenberg uncertainty principle (but not the same as the latter: the latter arises because the Fourier transform relates a quantum state's expression in co-ordinates related by the Canonical Commutation Relationship). In practice, though, a finite pulse is modelled by a spread of frequencies over an interval, and all frequencies in the interval are present in the Fourier transform.

  2. If you model a finite universe as a cavity, you will indeed get only a finite number of modes per frequency. So one aspect of the quantum light field ground state energy divergence is not really a divergence at all as the energy per unit volume is finite, as I discuss here. If you were Odin, say, creating a universe, it would cost you roughly a fixed energy per unit volume to make one like ours.

  • $\begingroup$ Best answer in this thread. Can't say I'm surprised, though ;). $\endgroup$ Mar 11, 2015 at 11:45
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    $\begingroup$ I'm not entirely comfortable with using the steady-state result for a cavity universe unless the age of the universe is large enough for light to have made several crossings. We know from multi-photon processes that non-eigenstates of bound systems have (short but) finite lifetimes with observable consequences. $\endgroup$ Jun 3, 2018 at 17:28

I believe that currently, light in free space (i.e., not in a waveguide or a crystal or anything tricky) is believed to be able to have all values of frequency/wavelength/energy. As you say, it is continuous. There are highly speculative theories that perhaps this is not true "all the way down" but thus far we have no evidence that the wavelength spectrum is discrete to my knowledge.

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    $\begingroup$ Well in our experience it looks continuous, but the spectrum is more like the natural numbers or integers in its uncountability based off the smallest quantum of light having a minimum step of plancks constant $\endgroup$
    – Triatticus
    Mar 9, 2015 at 8:34
  • $\begingroup$ @Dan: Consider two observers moving relative to each other. If one sees a step size of 1.00... Planck constant and the other observer sees a step size of 1.1 Planck constant, relativity breaks down. This might be real, but it's not at all backed by experiments. $\endgroup$
    – MSalters
    Mar 9, 2015 at 10:48
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    $\begingroup$ @Dan: Planck's constant sets the smallest energy that a photon of a given wavelength can have. It does nothing to restrict which wavelengths are available. $\endgroup$
    – zeldredge
    Mar 9, 2015 at 12:19
  • $\begingroup$ @Dan also natural numbers and integers are countable not uncountable, you are mixing your terms up $\endgroup$
    – jk.
    Mar 9, 2015 at 17:45
  • $\begingroup$ Sorry i had intended to say countably infinite rather than uncountable $\endgroup$
    – Triatticus
    Mar 9, 2015 at 18:08

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