Visible light - Being an Electromagnetic wave is reflected by glass (take mirror). Would all other waves in the electromagnetic spectrum be reflected in the same way by our simple mirror... For highly intense X-Rays & gamma rays, take that the mirror is more stronger such that it could resist for at least 30 seconds. Well If the mirror would reflect, then would the EM waves be the same after several (take 30) reflections..?

(At each reflection, the photons scatter in different directions... Would the intensity and energy of EM waves remain same?)


4 Answers 4


A metal (read mirror) behaves rather differently depending whether the frequency of the incoming light is above or below the plasmon frequency.

  • if the frequency is below the plasmon frequency, the electrons move in order to screen the electric field. This yields an approximate boundary condition $\mathbf E|_\text{boundary}=0$ and thus to reflection.

  • if the frequency is above the plasmon frequency, then the electrons cannot move (because of their inertia) and the wave essentially just goes through (just like in an ordinary dielectric).

As the plasmon frequency for typical metals is around $10^{15}\,$Hz, visible light is reflected and $X$-ray is transmitted.

  • $\begingroup$ What I was explaining is the result of the so-called Jellium model. In this model the charge of the ions is uniformly spread throughout the metal. In reality, the ions form a crystal which leads to diffraction but not total reflection. $\endgroup$
    – Fabian
    Aug 29, 2012 at 9:31
  • $\begingroup$ Dear @Fabian, I know this post is old but could you offer some clarifications? "if the frequency is below the plasmon frequency, the electrons move in order to screen the electric field." How does the screening occur (physically)? Next point: "if the frequency is above the plasmon frequency, then the electrons cannot move" But if the frequency of the incident wave is above the plasmon frequency, plasmons can be excited. What is the meaning of exciting plasmons, if electrons cannot move? I thought it's the oscillation of the free electron gas that is basically the plasmon modes. $\endgroup$ Aug 15, 2020 at 5:35

Reflection of X-rays or gamma-rays on an ordinary mirror will differ dramatically from reflection of visible light, as the wavelength of X/gamma is comparable or less than the typical distance between neighboring atoms of the mirror, so diffraction will be very important.

  • $\begingroup$ It is difficult to give a short answer for arbitrary wavelength, but at larger wavelength (than for visible light) reflection will be similar to reflection of visible light as long as skin depth is lesser than the thickness of the metal coating of the mirror, but I may miss some other caveats here. $\endgroup$
    – akhmeteli
    Aug 27, 2012 at 14:10

An electromagnetic wave reaching an interface is partially reflected, partially refracted. The fraction of it that gets refracted, i.e. the transmission coefficient, and the fraction of it that gets reflected, i.e. the reflection coefficient, depend on a number of things:

  • the refractive indices of both media,
  • the incidence angle, and
  • the polarization of the light.

Although the frequency of the electromagnetic wave does not come up explicitly here, it does implicitly through the refractive index, which varies with frequency, a phenomenon known as dispersion.

Once you have figured the refractive indices of your media for a given frequency, all you need is Fresnel's Equations to figure out what happens to your waves. At each reflection, only a fraction of the incident light will be reflected, and the fraction will be different for different polarizations. So the light after the first reflection will be less strong, and polarized differently, than before reflecting, and the differences will pile up after a few reflections.

Furthermore, if your reflections are not isolated, but happen between parallel surfaces, you can have interference between the various incident, refracted and reflected beams, leading to a further dependency of the outcome on wavelength. This is what makes soap bubbles iridescent, for example.

There really is no shorter answer to your question other than: find out what the refractive index for your frequency of choice is, work out the details, find out what happens...


As a radio engineer I will give you my answer. When a wave hits a reflector a situation called 'standing waves' occurs. The effect of which is to cancel or partly cancel the oncoming wave front because the phase of the reflected wave is often of an opposite. In radio transmission we try to avoid 'standing wave' as it means less of our power is transferred to the antenna. Usual measurements are normally given as SWR ( standing wave ratios) A high ratio of standing wave could damage our transmitting equipment.


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