Can anyone please provide an intuitive explanation of why phase shift of 180 degrees occurs in the Electric Field of a EM wave, when reflected from an optically denser medium?

I tried searching for it but everywhere the result is just used.The reason behind it is never specified.


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  • $\begingroup$ Could you check this question here, I cannot see how a brick-wall is optically denser material and which scenario is it? $\endgroup$ – hhh Sep 16 '12 at 12:11
  • $\begingroup$ It seems your question is posed the wrong way - the phase shift occurs when you go from less dense to more dense (lower to higher refractive index) not the other way around? Ron's answer is correct - it answers the opposite question. $\endgroup$ – Floris Aug 9 '15 at 5:45

This is a general property of waves. If you have waves reflecting off a clamped point (like waves running on a string that you pinch hard at one point), the waves get phase inverted. The reason is the principle of superposition and the condition that the amplitude at the clamped point is zero. The sum of the reflected and transmitted wave must be the amplitude of oscillation at all points, so that the reflected wave must be phase inverted to cancel the incoming wave.

This property is continuous with the behavior of waves going from a less massive string to a more massive string. The reflection in this case has opposite phase, because the more massive string doesn't respond as quickly to the tension force, and the amplitude of oscillation at the contact point is less than the amplitude of the incoming wave. This means (by superposition) that the reflected wave must cancel part of the incoming wave, and it is phase reflected.

When a wave goes from a more massive string to a less massive string, the less massive string responds with less force, so that the derivative at the oscillating end is flatter than it should be. This means that the reflected wave is reflected in phase with the incoming wave, so that the spatial derivative of the wave is cancelled, not the amplitude reduced.

In optical materials of high density are analogous to strings with a higher density, hence the name. If you go into a material with low speed of light, the time derivative term in the wave-equation is suppressed, so that the field responds more sluggishly, the same way that a massive material responds more sluggishly to tension pulls. Since the eletric field response in these materials is reduced, the reflected wave is phase inverted to make the sum on the surface less, as is appropriate to match with the transmitted wave.

  • $\begingroup$ How does an EM wave behave when it hits a brick wall? More here? A phasor or not i.e. a phase-change or not? $\endgroup$ – hhh Sep 16 '12 at 12:12
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    $\begingroup$ This site illustrates the concept quite well: acs.psu.edu/drussell/Demos/reflect/reflect.html $\endgroup$ – Mastergalen Apr 5 '14 at 19:46
  • $\begingroup$ Ron, can you please explain what happens if reflection happens from less dense medium ?( e.g. In air at air glass interface) $\endgroup$ – Tausif Hossain Jun 22 '17 at 16:51

Since this has just been asked again, let me attempt an intuitive explanation. The real explanation is of course to match $\vec{E}$ and $\vec{B}$ at the interface and the direction of the reflected wave drops out, but this isn't especially intuitive.

Let's calculate the ratio $E_r/E_i$ as a function of the ratio $n_t/n_i$, and let's start with the refractive indices equal, that is $n_i = n_t$, in which case there is obviously no reflection. As we decrease $n_t/n_i$, either by making $n_t$ smaller or $n_i$ bigger, the reflectivity will increase from zero so we'll get something like (this is the real calculation for the ratio, but the exact form of the graph doesn't matter):


This shows what happens when the refractive index at the incident side is equal to or greater than the refractive index at the transmitted side, but what happens when $n_i < n_t$? Obviously what happens is that we have to continue the line to the left to get something like:


This is the same as the first graph, just continued to values of $n_t/n_i \lt 1$. The point is that assuming the graph is smooth (which seems physically reasonable) the ratio $E_r/E_i$ must change sign as we pass through $n_t/n_i = 1$. In other words the phase of $E_r$ must differ by $\pi$ on the two sides of the point $n_t/n_i = 1$.

What actually happens is that $\vec{E_i}$ and $\vec{E_r}$ are in phase when $n_t/n_i < 1$ and out of phase by by $\pi$ when $n_t/n_i > 1$, and my argument doesn't prove this. However it hopefully gives you a feel for why the phase of $\vec{E_r}$ must differ (by $\pi$) either side of $n_t/n_i = 1$.


Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added: Do you equate optically denser to higher relative permitivity or to lower relative impedance? Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. This is due to changes in dielectric constant or other physical properties. https://books.google.ca/books?id=k1brJjXmXOQC&pg=PA43&lpg=PA43&dq=light+reflection+impedance+phase+inversion&source=bl&ots=G3qHMfPksC&sig=hwt5bC3GuiJ6OU3uI7n0XSmFjR4&hl=en&sa=X&ei=RS6rT6uXM4Wg8QT23Kka&ved=0CFkQ6AEwAQ#v=onepage&q=light%20reflection%20impedance%20phase%20inversion&f=false

Added: This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.


I propose removing this image or finding a suitable one. This diagram is part of Young's experiment which illustrates the phenomena of diffraction and interference, but I am afraid they do not illustrate reflection. No reflected wave is represented there. As you might see the two waves come from different sources. The reflected wave should have the same angle as the incident wave (both respect to the normal to the surface) which is not the case of the two last diagrams. The first one could be interpreted as a reflection with phase change, but that is more confusing than clarifying.

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    $\begingroup$ Not true. The water wave at the wall of a pool does not get inverted after reflection. It would get inverted if the wall forced a zero of the wave at the boundary, but that's not the case. Waves get inverted in a string with a fixed end, for instance (because there, the wave is forced to be zero). $\endgroup$ – Telaclavo May 9 '12 at 23:36
  • $\begingroup$ Transmission Line Theory of light, optical and electromagnetic waves all support what I said about mismatched impedances with the standing wave created by a reflection. I gave the reader's digest description, Perhaps my water example was misleading. Check out Heaviside's transmission line theory based on Maxwell's equations. $\endgroup$ – Tony Stewart May 10 '12 at 2:40
  • $\begingroup$ I may have explained it poorly.. If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. THis is due to changes in dielectric constant or other physical properties. goo.gl/vTwQq $\endgroup$ – Tony Stewart May 10 '12 at 3:01
  • $\begingroup$ @Telaclavo close, but not quite. It's not a condition of forced zero, but a condition of forced continuity. The string doesn't have to be forced to be stationary at the discontinuity, but only that the string be rigidly fixed to the second medium (i.e. wall or thicker string). Similarly, the water wave will only reflect inverted if the transverse displacement of the wave was constrained be the same as the transverse displacement of the wall, which is not the case, of course. $\endgroup$ – Nathan Wiebe May 11 '12 at 4:13
  • $\begingroup$ @NathanWiebe Which sentence that I wrote is not correct? / "Forced continuity" exists almost everywhere (like in all points of a string which are not the ends). Every point of the string is rigidly fixed to the point of string next to it. That does not create any reflection. $\endgroup$ – Telaclavo May 11 '12 at 10:56

Without the phase change energy (and momentum) conservation would not be satisfied.

To see why this is true you can think of a simple Michelson interferometer; without one of the fields having a phase flip you could get constructive (or destructive) interference at both sides of the beam splitter which would result in twice (or none of) the energy which you sent into the interferometer coming back out. Now for the more mathematical explanation.

It is actually just a convention that the phase change happens on reflection from an optically denser medium. The actual requirement is more subtle and comes from energy conservation. To see this, consider a black box optical system which you know nothing about except that no energy is lost inside.

Black box optical system

The four fields must obey energy conservation which is expressed by $$ \begin{align} |E_1|^2+|E_2|^2&=|E_3|^2+|E_4|^2\\ &=|r_{31}E_1+t_{32}E_2|^2+|t_{41}E_1+r_{42}E_2|^2\\ &=(|r_{31}|^2+|t_{41}|^2)|E_1|^2+(|t_{32}|^2+|r_{42}|^2)|E_2|^2+2\Re[(r_{31}t_{32}^{*}+r_{42}^{*}t_{41})E_1E_2^*] \end{align} $$ The only way to satisfy this for all possible $E_i$ is by satisfying $$ |r_{31}||t_{32}|=|r_{42}||t_{41}|\qquad\&\qquad r_{31}t_{32}^*+r_{42}^{*}t_{41}=0 $$ If you write out the complex reflectivity/transmissivity coefficients in terms of their amplitude and phase, e.g. $r_{31}=|r_{31}|e^{i\phi_{31}}$, then these equations reduce to $$ |r_{31}||t_{32}|=|r_{42}||t_{41}|\qquad\&\qquad \phi_{31}-\phi_{32}+\phi_{42}-\phi_{41}=\pm\pi $$ This second equation is the one which must be satisfied by the phases. Our usual convention is to take $\phi_{31}=\pi$ and let all of the others be zero. Another convention which is attractive because it is symmetric is to let each transmitted field pick up $\pi/2$ of phase, $\phi_{41}=\phi_{32}=\pi/2$, with the others being zero.



It is easy to see why there is a sign change in the case where the electric field reflects from a conducting surface. The electric field would excite a current in the conducting surface, which in turn would force the electric field to be zero at the interface. Therefore, the reflected field must be such that it cancels the incident field at the surface. Hence, the negative sign.

enter image description here

In the case of a dielectric interface one can use a similar argument, but with some differences. In this case not all the power is reflected. Some of it is transmitted into the material on the other side of the interface. However, due to energy conservation, the power that is transmitted is always smaller that the incident power. Hence, the amplitude of the transmitted field is smaller. To satisfy the boundary conditions, the reflected field must have a negative sign so that it can be subtracted from the amplitude of the incident field to match that of the transmitted field.


The phase change happens because it is how waves behave.

I know that u are not satisfied with this answer but you can compare this with mechanical waves in a string which gives better intuition by use of newtons laws.