2
$\begingroup$

This is a diagram showing the single slit interference set up:

enter image description here

The pinhole (or 'peep hole') is said to 'produce a parallel beam of coherent light, because the time interval between light passing through the pinhole and single slit is constant, producing a constant phase difference'.

However, how would this create coherent light, because the filament lamp is an incoherent source, meaning it emits light at random, so wouldn't rays emerging from the peep-hole also emerge at random?

Also is the peephole even necessary? Because the slit by itself would be narrow enough to create coherent light?

$\endgroup$
10
  • $\begingroup$ The edge of the pinhole is what creates the coherency. The same is true for the edges of a slit or even a single edge. Anything out of phase is absorbed by the edge. $\endgroup$ May 6, 2020 at 18:36
  • $\begingroup$ @BillAlsept but if the filament lamp is incoherent, it emits light rays at random, so there is no guarantee that every second a light ray will reach the edges, right? This is what I don't understand - the time intervals of light rays reaching the pinhole will never be constant, so how can coherent light emerge from it? $\endgroup$
    – XXb8
    May 6, 2020 at 18:38
  • $\begingroup$ Why does the light need to be coherent? Single photons interfere with themselves. $\endgroup$
    – Jon Custer
    May 6, 2020 at 18:54
  • 1
    $\begingroup$ @JonCuster I think the OP is interested in a method that claims to produce coherent light. I don't think they are claiming that you need coherent light for things to be interesting. $\endgroup$ May 6, 2020 at 18:56
  • 1
    $\begingroup$ What about this question? $\endgroup$ May 6, 2020 at 18:59

4 Answers 4

3
$\begingroup$

Just as it a choppy (incoherent) sea there are clear wavecrests the light emmited from the incandescent lamp has locally clear wavecrests (although on a much shorter scale). We say that the incoherent light has short range coherence. The pinhole samples only a small part of the wavefront from the light emitted from the lamp, and as long as the pinhole, is smaller than the short coherence length of the light the wavefronts that spread out from the hole will be spatially coherent and good for demonstraing interference due to different path lengths like Young's slits when both slits are at same distance from the pinole. The light from a pinhole will not have good temporal coherence because the period of waves beating on the pinhole will not be strictly periodic so neither will the light emitted from the pinhole. A Mach-Zender interferometer that interferes light against a time delayed verion of itself will not work well therefore.

There is a nice picture that illustrates this distinction at https://en.wikipedia.org/wiki/Coherence_(physics)#Temporal_coherence under the heading "Examples of spatial coherence"

$\endgroup$
1
  • $\begingroup$ This makes sense, but why is the pinhole needed at all? If the slit is already narrow, then it would create emerging coherent light too, so why is a pinhole needed before the slit? $\endgroup$
    – XXb8
    May 9, 2020 at 14:56
2
$\begingroup$

The slit does not make the light beam coherent. The beam's coherence is primarily a function of the source. You are right that light from a filament is incoherent. It is temporally incoherent (multi-wavelength) and spatially incoherent (spread out in space). A slit does increase the spatial coherence of the light: if you consider the slit to be a source, it can be much smaller than the filament. Interference is possible using light that is only spatially coherent, but if you want easily observed interference fringes, you need light that is both spatially and temporally coherent- like a laser.

$\endgroup$
9
  • 1
    $\begingroup$ Please re-phrase your comment/question to clarify what you mean by "this". E.g., "How does a slit reduce spatial coherence?" $\endgroup$
    – S. McGrew
    May 9, 2020 at 14:26
  • $\begingroup$ I don't understand how the slit (or pinhole) would increase spatial coherence. Is this because a narrower beam is created so there are less phase differences possible? And how would you answer the question about the pinhole (not the slit) creating coherent light? $\endgroup$
    – XXb8
    May 9, 2020 at 14:28
  • 1
    $\begingroup$ The only portion of light that makes it through and beyond the slit is that portion whose phase variations are negligible on a scale smaller than a wavelength. (I don't know a way to explain why to a layman.) So, the total amount of effective phase variation across the slit is roughly proportional to the ratio of the slit width to the light wavelength. That means that the phase variation drops toward zero as the slit width approaches the wavelength of the light. A pinhole is just the same as a slit, except two-dimensional instead of one-dimensional. $\endgroup$
    – S. McGrew
    May 9, 2020 at 14:50
  • $\begingroup$ This makes sense, but wouldn't the slit have the same effect if it is narrow without needing a pinhole to create emerging coherent light? $\endgroup$
    – XXb8
    May 9, 2020 at 14:52
  • 1
    $\begingroup$ Yes, that's right. A slit illuminated by an incoherent source is, for all practical purposes, itself a source whose spatial coherence is defined by its dimensions. But two such slits would not be mutually coherent, unless illuminated by a single spatially coherent source. Mutual coherence is necessary for two-slit interference. $\endgroup$
    – S. McGrew
    May 9, 2020 at 15:10
1
+50
$\begingroup$

Is the peephole even necessary? Because the slit by itself would be narrow enough to create coherent light?

To see interference effects reliably, the phase of the light should be in step everywhere across the hole. In other words, the spatial coherence length should be larger than than the width of the hole.

In general, passing light through a small hole and then letting it expand back out increases the spatial coherence length. To see why, note that for an infinitely small hole, the phase of the light is automatically the same everywhere across the hole (since it's just one point), so it automatically puts out coherent light. So if you use finite-sized holes, you increase the coherence. In fact, in principle you can repeat it, with multiple layers of pinholes, with the light getting more coherent every time. It's just not very useful in practice because this also makes the final light very dim.

The diagram is showing a situation where the slit's size and the lamp's coherence length are so that you need one layer of pinholes in between to reliably see interference effects. But this depends on the parameters. In other situations you could need zero, or two.

$\endgroup$
5
  • $\begingroup$ So the pinhole is actually used to further increase the coherence before reaching the slit and ensure almost perfectly coherent light will emerge from the slit? $\endgroup$
    – XXb8
    May 10, 2020 at 8:22
  • $\begingroup$ Also, why would a filament lamp alone be unsuitable? Is it because since light is emitted with different phases and wavelengths could cancel each other out before reaching the double slit? $\endgroup$
    – XXb8
    May 10, 2020 at 8:35
  • 1
    $\begingroup$ @XXb8 Each layer of pinholes increases coherence length. You can also increase the coherence length by just moving further away from the lamp. It's just that all of these things decrease the intensity too. $\endgroup$
    – knzhou
    May 10, 2020 at 19:25
  • 1
    $\begingroup$ @XXb8 Yeah, the light is made of parts with different phases and wavelengths. It doesn't "cancel out", but it makes the waveform complicated so that it doesn't interfere nicely. $\endgroup$
    – knzhou
    May 10, 2020 at 19:25
  • $\begingroup$ Not necessarily 'cancel our' but superposes before reaching the slit? $\endgroup$
    – XXb8
    May 11, 2020 at 7:57
0
$\begingroup$

Coherency is a very important property of waves, whether they are light, sound or water waves. It is a complex subject especially as physicists have tended to apply the wave properties of water to light ever since the time of Huygen. But there are important differences, for example water waves through a pinhole do diffract but there is no interference pattern, whereas for light we do get diffraction and interference for the single slit or pinhole. Interference is another complex area, many physicists say light interferes and we see dark spots and brights spots but this is a violation of conservation of energy, even in water 2 waves will "cancel" or "superimpose" but this is only temporary as the water waves reemerge and eventually dissipate their energy by crashing on the beach for example.

Surprisingly even sunlight and incandescent light do have some level of coherence, there is no such thing as perfectly incoherent light! Why? In order to observe coherence we scientists do need some kind of apparatus and the apparatus effects the light we measure or observe. The apparatus places geometric constraints on the possible light paths and according to Feynman (who made a determined effort to understand the double slit) we must look at many possible path and compute the ones that are most probable, eventually this was shortened to the fact that light will travel the shortest path that is a multiple of the light wavelength. A laser is a good example of this phenomenon, mis-align the mirrors and the laser ceases to lase.

So historically (and somewhat inaccurately) for light we tend to think of coherency as waves arriving in phase (to "interfere") and in many cases this model seems to work .... but for light coherency is more about light of similar wavelengths being emitting from well localized sources that are highly constrained (laser) or less constrained (slit), these light waves are only able to travel certain paths.

A certain light path or constrained path that light chooses is one where the light arrives or is absorbed at its maximum EM field (of Maxwell's plane wave), thus we can say that the light is arriving in phase, that is its most probable path. It is not that many photons are arriving at a certain point that happen to be in phase.

$\endgroup$
4
  • $\begingroup$ so by reducing the width of the beam, you are actually reducing the number of possible phase differences? $\endgroup$
    – XXb8
    May 9, 2020 at 13:55
  • $\begingroup$ Then why is the pinhole needed at all? If the slit itself is narrow, and light rays along its path will always reach it? $\endgroup$
    – XXb8
    May 9, 2020 at 14:15
  • $\begingroup$ @XXb8 Young also used a color filter on the sunlight in addition to the pinhole. There are many (infinite?) interference patterns on the screen, the pinhole and the filter let you begin to select or observe them. $\endgroup$ May 9, 2020 at 17:06
  • 1
    $\begingroup$ @XXb8 Phase is the more "traditional" method of explaining the pattern using "interference", but photons don't truly interfere so that's where the lesser known Feynman concepts are needed. Feynman paths require that only photons at maximum phase/amplitude interact with the screen/atoms (your eye), thus they are all already in phase. By using slits and pinholes we see a pattern (I don't want to use the word "interference") that shows the bright spots (many photons) and dark spots (no photons). The pattern is a result of photons needing to travel n multiples of wavelength to the screen. $\endgroup$ May 9, 2020 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.