From what I know of the original experiment, the beam-splitter was a half-silvered mirror, and one of the paths had a compensator in the form of a glass plate.

My questions are: (1) Since the glass compensator has a refractive index, wouldn't it reflect some of its own light away? Wouldn't this be disadvantageous, because some light is lost in one leg, causing an unequal amount of light in the two legs and thus a loss of fringe visibility?

(2) If glass itself can partially reflect light through its index of refraction, what's the exact necessity for using a half-silvered mirror as opposed to a glass plate?

I have possible ideas to some of these questions, but I would like to know what others have to say.


A bit of background first: The reason glass (in air) reflects light is not that it has a refractive index; it is that its refractive index is different from the refractive index of air. Immerse the glass in a fluid with the same index as the glass, and the glass will not reflect light. In air, a glass plate will reflect a few percent of the light that hits its front surface and another few percent of the light that hits its back surface.

The Answer: In a typical Michaelson interferometer, air is in the gaps between the glass components (i.e., between the mirrors, beamsplitters, and (if there are any) compensator); so some small fraction of the light is always reflected at each surface.

The purpose of a compensator is to make up for any path length differences in the two paths light travels in the interferometer. If the beamsplitter is a thinly silvered mirror, light reflecting off the silvered surface is deflected but not delayed. Light that passes through the beamsplitter has to go through the glass substrate. Because that substrate has a refractive index higher than air, light is delayed slightly in going through it. If the distance of travel through the substrate is, say, 1 cm and the refractive index of the substrate is 1.5, then the effective length of the portion of the light path through the substrate is 1.5 x 1 cm or 1.5 cm. That is, the length of the transmitted path is effectively increased by 0.5 cm. In order to ensure that both the reflected and transmitted light paths have the same length, a glass plate (a compensator plate) of suitable thickness can be put in the reflected path. Note that a cube beamsplitter does not require a compensator plate because both transmitted and reflected beams traverse the same thickness of glass.

Losing a little bit of light is usually not a problem. If the two beams must have precisely the same intensity, a variable attenuator in one beam can easily balance the intensities. And, it is common to use antireflection coatings on the glass surfaces to minimize losses and stray reflections.

Typically the reflection from an uncoated glass plate is a few percent: nowhere near the ~50% reflectivity desired in a typical interferometer. It is possible to use a carefully designed dielectric coating to obtain 50% reflectivity, though, at a fixed angle, so it is possible to make an interferometer without any silver (or aluminum, etc) coatings. If fringe contrast is not important, &/or if there is plenty of light power available, then it's entirely possible to make an interferometer using just a glass plate as a beamsplitter.

  • $\begingroup$ +1 very nice answer. I'm being picky: "If the beamsplitter is a thinly silvered mirror, light reflecting off the silvered surface is deflected but not delayed" significantly. $\endgroup$ – uhoh May 7 '18 at 4:37
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    $\begingroup$ Significantly is a little picky, yes, but correct. Pretty much everything the light encounters will affect its path length to some degree. A typical silver coating on a beamsplitter is a few tens of nanometers thick, so there could be a path length difference on the order of a few nanometers in the reflected beam vs the transmitted beam at the silver layer. $\endgroup$ – S. McGrew May 7 '18 at 5:03
  • $\begingroup$ Since it's the electron plasma's response, it's more complicated than that. But I get dizzy when I try to remember the difference between group and phase delays in reflection from metal surfaces. $\endgroup$ – uhoh May 7 '18 at 5:06

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