boundary conditionselectromagnetic-radiationelectromagnetismreflectionwaves
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.
The phase-shift occurs due to the complex nature of the reflection coefficient. The so-called critical angle is given by:
$sin(\theta_M) = \frac{n_2}{n_1}$
As long as $\theta<\theta_M$ we have only partial reflection and a real valued reflection coefficient $R$.
As soon as the critical angle is exceeded $(\theta>\theta_M)$, we have $\mid R \mid=1$ and total reflection of the light occurs. $R$ is now complex and a phase shift is imposed on the reflected light, we write:
$R=e^{2j\phi}$
wehre $\phi$ is the phase shift between the incident and reflected wave.
Now, we know what the value of $R$ is, both for TE and for TM modes, i.e. the Fresnel coefficients for the two different states of polarization of the electric and magnetic fields. To get the formula for the phase-shift you only need to equal this complex expression for $R$ to the Fresnel formulas and solve for $\phi$. The formula you wrote is the one corresponding to TE modes, so it's the TE phase-shift (with electric fields perpendicular to the plane of incidence spanned by the wave normal and the normal to the interface). For TM modes you would get exactly the same result but multiplied by a factor of $n_{1}^{2}/n_{2}^{2}$.
With coherent light from a laser one can assume that the path lengths of a Michelson interferometer are equal (for the purpose of your question) and due to the narrow bandwidth of the laser light one can also assume low dispersion (equal path lengths through the glass).
With everything assumed to be equal the only difference between the two paths that the light travels is:
There is a phase change for a reflection when a wave propagating in a lower-refractive index medium reflects from a higher-refractive index medium, but not in the opposite case.
Notice that the light travels through the glass twice, once from air to glass to mirror, and once from mirror to glass to air; one direction has increasing index and the other doesn't.
Reference:
"Phase shift between the transmitted and the reflected optical fields of a semireflecting lossless mirror is π/2" (1980), by Vittorio Degiorgio, in the American Journal of Physics 48, 81 (1980); https://doi.org/10.1119/1.12238
Assuming a lossless beamsplitter the energy conservation condition $\left | E_0 \right |^2 = \left | E_t \right |^2 + \left | E_r \right |^2$ gives $\phi^{'}_t - \phi^{''}_t = \phi^{'}_r - \phi^{'}_r + \pi$, thus:
$$\phi_R = \phi_T + \pi / 2$$
Best Answer
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.