In the double slit experiment we have something like this

enter image description here Light waves from two sources interfere at screen to give us this pattern. To reach the middle (that is a point on the perpendicular bisector) of measurement screen waves will travel equal distances (at sources they are in phase because of the same main source) and hence will be in phase at the middle of the screen and hence a constructive interference will occur.

We can write things a little more mathematically, if at time $t$ the crest of the light wave is at the sources (secondary) then at time $t+nT$ ($T$ is the time period) crests will be at the middle of the measurement screen and hence a maximum will occur. Here, I have assumed, for simplicity, that distance between the source $S_1$ and the middle point M is $n\lambda$.

Now, my question is when there was a crest at $S_1$ and $S_2$ then we had a maximum at $M$, but when there is no disturbance at sources (I mean $E(t)=0$ and $B(t)=0$) then there should be no disturbance at M too and therefore we should obtain a dark fringe (no light) for this particular interference. Why does the interference pattern remains static? The bright fringes should be turning on and off periodically.

My question comes down to normal light also, why do we see the light even when the electric field and magnetic field components reach their zero position periodically? We should see light as being turning on and off, off when $E(t)$ and $B(t)$ becomes zero and on for every other time.

P.S. :- My $M$ in the figure doesn't coincide with the maximum but by $M$ I meant the middle of the screen where the first maximum gonna occur.

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    $\begingroup$ You are quite correct, but the frequency of visible light is of order $10^{15}$ Hz, so the oscillations in brightness at M are too rapid to see. $\endgroup$ Dec 28, 2019 at 20:17
  • $\begingroup$ Keep in mind that every time you see a visible light the field strength is oscillating. Any detector system which registered the temporal oscillations of an interference fringe would also register the temporal oscillations of a plain light source. $\endgroup$ Dec 29, 2019 at 0:38

3 Answers 3


The interference of waves at M does in fact result in an oscillation which has the same frequency as the waves from S. You do not see this oscillation using light because it is around $10^{14}-10^{15}$ cycles per second, which is many orders of magnitude faster than what your brain can detect. If you did the experiment using water waves at no more than a few cycles per second you would see the oscillation.

The interference pattern is a sinusoidal variation in the amplitude (or intensity) of the oscillations of the light or sound waves. This pattern is a variation in space (from one point to another) but it is constant in time - because the amplitudes and phase differences of waves from the two slits remain constant in time. There is still a wave oscillation at each point on the screen, but there is also a sinusoidal variation in the amplitude of these waves across the screen.

You can see the oscillation in this video simulation of water wave, as the waves hit the wall/screen at 0.09 onwards. (It is a short video so I recommend re-setting the playback speed to 0.25.) Where there is constructive interference (bright interference fringe) large-amplitude waves are reaching the screen. Where there is destructive interference (dark interference fringe) the waves reaching the screen have small or zero amplitude.

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    $\begingroup$ As in these simulations: phet.colorado.edu/en/simulation/wave-interference $\endgroup$
    – user137289
    Dec 28, 2019 at 20:32
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    $\begingroup$ I wouldn’t say the interference oscillates: in a bright fringe it remains constructive at all times. But the net amplitude oscillates. $\endgroup$ Dec 28, 2019 at 20:47
  • $\begingroup$ @dmckee Thanks for the suggestion. Yes there is a spatial oscillation (variation) in the amplitude, but I am trying to point out that there is still a time-oscillation of waves resulting from interference. $\endgroup$ Dec 28, 2019 at 20:51
  • $\begingroup$ Perhaps I wasn't clear. At any given point in space: (a) net amplitude oscillates in time and (b) but the phase difference (and thus the nature of the interference (fully constructive, fully destructive, partial)) remains constant in time. $\endgroup$ Dec 29, 2019 at 0:37

Sammy gerbil already explained the fact that the field strength is, in fact, oscillating. I want to expand in a different direction, regarding how our senses work.

You are right that at the maxima of any interference pattern, the field strength is oscillating. However, you are mistaken in what that means for our senses. Our retina does not sense field strength. Neither of the electric field, nor of the magnetic field. What our retina does measure is the frequency and amplitude of a quickly oscillating electric field. If the field at the maxima didn't oscillate, we would see darkness at that point. Think about it: Have you ever seen a charged capacitor glow? Or a coil with direct current flowing through it? No, these do not glow, even though they have electric/magnetic fields, because our eyes can only see how these fields change (if they change by oscillating), not how strong they are. We need periodically changing fields to see, not fields by themselves.

By the way, it's the same with our hearing. We can't hear constant air pressures. But if pressure oscillates rapidly, we can hear that. Because our ears do not sense air pressure (except as that uncomfortable feeling in our ears we get when diving or flying, but that's basically pain receptors feeling the mechanical stress on our ear drums). Ears sense the frequency and amplitude of oscillating air pressure.


The common explanation that at any point there is an oscillation in amplitude is incorrect. If this were so we would not see bright and dark fringes but a bright fringe across the screen because of the way our eyes react to light and darkness.

The double slit interference experiment is conducted these days by holding a laser behind the two slits. The laser emits light at the same phase always. Let us assume that E and B are at their maximum amplitude at the slits. Also, assume for now that the central point on the screen is a whole number of wavelengths from the two slits. This means that E and B at the central point are always at their maximum amplitude and would be a bright point always. There is no variation in amplitude at the central point or indeed at any point on the screen.

But there is no guarantee that the central point is a whole number of wavelengths from the two slits. It only depends on the geometry of the experment. So why does it seem to be always bright? The reason is that the brightness at a point on the screen depends on two factors. One is the value of E and B which varies sinusoidally across the screen. The other is the angle which the E and B vectors make when they strike the screen; this varies across the screen. As the angle increases, the effect of E and B reduces, which is why the brightness of the fringes falls away far from the central point. So even if the central point is dark the brightness would rapidly increase on both sides and our eyes would not be able to distinguish the central dark point.

It is possible to show from simple electromagnetic considerations that there is no "interference" occurring at the screen. The pattern we observe on the screen is due to the individual effects of photons from the two slits. What is being added is not the amplitudes of photons from the two slits but their intensities. I am not sure that the rules of stackexchange allow it, but I could post a paper to show how this happens and why there is no mystery to "single photon interference".


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