# If the energy of light is determined by its wavelength... **URL:** \ **Category:** Factual Questions\ **Created:** [March 19, 2015, 8:09pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420 "2015-03-19T20:09:45Z")\ **Posts on this page:** 14\ **Page:** 1
**Author:** ![Schnitte](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/schnitte/32/9033_2.png) [@Schnitte](https://boards.straightdope.com/u/Schnitte)\ **Post date:** [March 19, 2015, 8:09pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/1 "2015-03-19T20:09:45Z")
…then what’s the difference between two light sources of same wavelength but different output power? I would assume that, from the particle point of view of electromagnetic radiation, the wavelength, i.e. frequency, determines the energy content per photon, and the power output of the light source determines the number of photons emitted by it per time unit. Correct so far? If so, what would be the corresponding view seen from the wave interpretation of electromagnetic radiation? Would the power output of the light source determine the amplitude of the electromagnetic wave while leaving the frequency/wavelength unchanged? ---
**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\ **Post date:** [March 19, 2015, 8:19pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/2 "2015-03-19T20:19:48Z")
Well power is simply the time derivative of energy. So you can have 2 objects emitting the same kind of photons (of wavelength X) but one simply emits more per second. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [March 19, 2015, 8:20pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/3 "2015-03-19T20:20:21Z")
> [@Schnitte](#): > > I would assume that, from the particle point of view of electromagnetic radiation, the wavelength, i.e. frequency, determines the energy content per photon, and the power output of the light source determines the number of photons emitted by it per time unit. Correct so far? If so, what would be the corresponding view seen from the wave interpretation of electromagnetic radiation? Would the power output of the light source determine the amplitude of the electromagnetic wave while leaving the frequency/wavelength unchanged? Everything you’ve said is completely correct. The power delivered by an electromagnetic wave is dependent only on its amplitude (specifically, the square of its amplitude), and frequency doesn’t enter into it. ---
**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\ **Post date:** [March 19, 2015, 8:23pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/4 "2015-03-19T20:23:58Z")
One possible complication in this analogy is that most light sources emit photons across a whole spectrum of light (visible and not). As you increase temperature, you shift the average wavelength, but you still have a distribution. Anyway, we can always idealize some light source that emits only photons of a particular frequency. In that kind of light source total power would be equal to the energy per photon multiplied by the number of photons. So if you keep the wavelength the same, but one light source has a lower total power output, the only way to achieve that is for the weaker light to emit fewer photons. So you’re right so far. Where I think you get lost is that we have just described the wave-like behavior of light because we’re talking about light having a frequency/wavelength. When you ask how the wave-like behavior is different, the question doesn’t make any sense. What you’d want to ask is how the particle-like behavior of a photon would be different. I think light has been understand as a wave for so long that we don’t really think that way. If we did, though, there’d be no wavelength, frequency or amplitude. There would just be a little billiard ball containing a certain amount of energy as it flies along in a straight line through space. ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [March 19, 2015, 8:31pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/5 "2015-03-19T20:31:22Z")
> [@dracoi](#): > > Anyway, we can always idealize some light source that emits only photons of a particular frequency. Use two monochromatic lasers of equal power output, and pass one of the beams through a filter to reduce its intensity. Presto: now you have two beams of light, both of same wavelength, but different intensities. One beam is passing more photons per second than the other. Is the difference between the two beams describable by referencing only the wave properties of light? ---
**Author:** ![Schnitte](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/schnitte/32/9033_2.png) [@Schnitte](https://boards.straightdope.com/u/Schnitte)\ **Post date:** [March 19, 2015, 8:35pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/6 "2015-03-19T20:35:02Z")
> [@Grey](#): > > Well power is simply the time derivative of energy. So you can have 2 objects emitting the same kind of photons (of wavelength X) but one simply emits more per second. That’s exactly what my question was getting at. Suppose you let a 10 watt and a 100 watt light source run for one second (referring to power output, not input). The first light source would emit 10 joules of energy in that second, the second 100 joules. Would the second, under the particle interpretation, simply emit ten times as many photons of the same kind as the first? ---
**Author:** ![Schnitte](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/schnitte/32/9033_2.png) [@Schnitte](https://boards.straightdope.com/u/Schnitte)\ **Post date:** [March 19, 2015, 8:39pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/7 "2015-03-19T20:39:36Z")
> [@Machine\_Elf](#): > > Use two monochromatic lasers of equal power output, and pass one of the beams through a filter to reduce its intensity. Presto: now you have two beams of light, both of same wavelength, but different intensities. One beam is passing more photons per second than the other. > > Is the difference between the two beams describable by referencing only the wave properties of light? That’s a good rephrasing of what I was getting at, thanks for that. My understanding (but I might be wrong) of the particle/wave dualism is that there are some instances where light behaves like a stream of particles, and these particles would be photons (but you wouldn’t think of these photons as packages of waves in that scenario). In other instances, light behaves like a wave, so you can think of it as a continuous, uninterrupted wave, not packaged into photons. If this is a possible interpretation of the nature of light, at least in some instances, then would the more powerful light source simply emit a wave of the same wavelength, but larger amplitude, than the less powerful source? ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [March 19, 2015, 8:47pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/8 "2015-03-19T20:47:47Z")
I am not sure about the effect of light passing through water or glass. I know frequency of light does not change when it passes through water but the wavelenght does. Assuming energy remains the same, the energy equation should work better when expressed as a function of frequency. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 19, 2015, 11:23pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/9 "2015-03-19T23:23:27Z")
> [@](#): > > Quoth **Schnitte** : > > That’s exactly what my question was getting at. Suppose you let a 10 watt and a 100 watt light source run for one second (referring to power output, not input). The first light source would emit 10 joules of energy in that second, the second 100 joules. Would the second, under the particle interpretation, simply emit ten times as many photons of the same kind as the first? Correct. ---
**Author:** ![thelurkinghorror](https://avatars.discourse-cdn.com/v4/letter/t/7c8e57/32.png) [@thelurkinghorror](https://boards.straightdope.com/u/thelurkinghorror)\ **Post date:** [March 20, 2015, 1:26am UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/10 "2015-03-20T01:26:26Z")
If you’re talking about vision, and whether a photoreceptor can tell the difference between a weaker optimal wavelength and a stronger less optimal wavelength with equal activation, the answer is that they can’t. But if you have (at least) two photoreceptor types (2 types of cones), you can compare how each responds and therefore determine the wavelength (partially) and see color. See: [univariance](http://sites.sinauer.com/wolfe3e/chap5/univarianceF.htm) (man the Wikipedia page is crappy). ---
**Author:** ![johnpost](https://avatars.discourse-cdn.com/v4/letter/j/f17d59/32.png) [@johnpost](https://boards.straightdope.com/u/johnpost)\ **Post date:** [March 20, 2015, 2:13am UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/11 "2015-03-20T02:13:22Z")
> [@am77494](#): > > I am not sure about the effect of light passing through water or glass. I know frequency of light does not change when it passes through water but the wavelenght does. Assuming energy remains the same, the energy equation should work better when expressed as a function of frequency. if the frequency doesn’t change neither does its wavelength. ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [March 20, 2015, 2:44am UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/12 "2015-03-20T02:44:30Z")
> [@am77494](#): > > I am not sure about the effect of light passing through water or glass. I know frequency of light does not change when it passes through water but the wavelenght does. Assuming energy remains the same, the energy equation should work better when expressed as a function of frequency. > [@johnpost](#): > > if the frequency doesn’t change neither does its wavelength. Well, not quite. When light enters water or glass from air, its speed slows and the wavelength decreases in direct relation to the speed. Remember, Speed=Wavelength\*Frequency. The frequency does not change. If the light then leaves the water or glass and goes back into the air, its speed, wavelength, and frequency will resume their original values. ---
**Author:** ![Carl\_Pham](https://avatars.discourse-cdn.com/v4/letter/c/e9a140/32.png) [@Carl\_Pham](https://boards.straightdope.com/u/Carl_Pham)\ **Post date:** [March 20, 2015, 9:31pm UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/13 "2015-03-20T21:31:57Z")
A photon differs from other fundamental particles in that its velocity is always known precisely ©. Uncertainty tells you therefore that its position is entirely indeterminate. From the mathematical point of view, a single photon is a quantum of amplitude, and corresponds classically to a pure sine wave with a certain (very small) amplitude that fills alls space. Two photons is a pure sine wave with twice the amplitude, et cetera. Hence the classical correspondence to “a lot of photons” is a sine wave with a large amplitude, and “a few photons” is a sine wave with a small amplitude. So far so good, except for two things: first, photons can be absorbed or emitted by point particles. So in some sense – a sense about which unfortunately quantum mechanics is silent – a photon can be “present” in one specific location, in order to be absorbed by a particle at that location, even though technically the photon also fills all space. This produces a painful contradiction of intuitions about what the beast is, that no one has really resolved satisfactorily. (There’s also a horrible apparent contradiction with relativity, since places where the photon has amplitude have to get the “message” that the photon has been absorbed instantly, no matter how far away they are. This is another instance of “quantum teleportation” or just the measurement problem in general. Fortunately it does not lead to grandfather paradoxes or Spock with a bear.) Furthermore, the behaviour of precisely one photon is rarely of practical significance. Normal light phenomena are always a superposition of many photons, usually of at least slightly varying frequencies. It’s not even possible theoretically to observe exactly one photon emitted, because the process would have to take an infinite time (there’s a derivative uncertainty relationship between knowing the energy of a change in state and knowing the time it takes). When photons combine, they of course interfere, and the classical equivalent – what you’d measure with a detector of electric fields – can take on many different extents and shapes. For the most part, people ignore all these weird complications in the extreme limit, because the differences between a theoretical ideal single photon and something that is very, very close, but still behaves semiclassicaly and intuitively like a tiny little blot of very nearly monochromatic light, as far as the dimensions and timescale of your experiment go, are usually ignorable. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [March 21, 2015, 3:01am UTC](https://boards.straightdope.com/t/if-the-energy-of-light-is-determined-by-its-wavelength/715420/14 "2015-03-21T03:01:53Z")
> [@Carl\_Pham](#): > > A photon differs from other fundamental particles in that its velocity is always known precisely (c). Uncertainty tells you therefore that its position is entirely indeterminate. Nitpick: Uncertainty tells you that if _momentum_ is known precisely (∆p = 0), then position is entirely undeterminate (∆x = ∞), and a photon’s momentum doesn’t just depend on its speed but is most directly related to its energy (p = E/c) and therefore its wavelength. Your statement would hold for a photon “prepared” in a state of definite wavelength/frequency, though. > [@](#): > > Normal light phenomena are always a superposition of many photons, usually of at least slightly varying frequencies. Moreover, a single photon could exist in a state where it was in a superposition of many different momentum states; in such a case, we would have ∆p \> 0 and so ∆x \< ∞, and we could have some partial knowledge of its position. In other words, it’s not necessarily true that a photon’s position is completely unknown.