I have a new question about partial reflection. There appears to be two separate phenomena that affect the amount of reflection from glass. (1) As the glass thickness changes, partial reflection varies from 0 to 16%. (2) Fresnel's equations show that polarization and angle of incidence also effects the amount of reflection.

How do the two work together. In other words if the glass thickness should cause zero reflection while the angle of incidence should cause 100% reflection what happens?

  • $\begingroup$ You know I thought it was an OK question. There’s nothing worse than a down voter who has nothing at all to say. $\endgroup$ – Bill Alsept Jan 22 '18 at 5:50
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    $\begingroup$ I can think of many worse things in life than the loss of a few points of an entirely arbitrary point system on an internet site. $\endgroup$ – Kyle Kanos Jan 22 '18 at 11:08
  • $\begingroup$ You know what I mean. I didn’t care about the points. I was truly interested in the question. Thanks $\endgroup$ – Bill Alsept Jan 22 '18 at 15:43
  • $\begingroup$ It is pretty unclear exactly what you are changing when in the question. Now, in general, thin film optics equations tell you how it all interrelates. $\endgroup$ – Jon Custer Jan 23 '18 at 14:15
  • $\begingroup$ @JonCuster Thanks I posted this question back in July but have since realized the two types of reflection are separate phenomena and are derived differently. You can have zero partial reflection from two surfaces and at the same time have reflection do to the angle of incidence from one surface without effecting each other. $\endgroup$ – Bill Alsept Jan 23 '18 at 16:55

The Fresnel equations are derived from considering the three light rays at the interface and requiring that the electric and magnetic fields of the light be continuous at the interface:


The working is basically straightforward, though tedious, and you'll find it done on a thousand web sites so I won't reproduce it here.

However this assumes the dielectric is semi-infinite so we only need consider the three light rays when matching up the electric and magnetic fields. If we have a sheet of finite thickness we now have to include the light rays reflected off the bottom surface of the sheet:


This is a lot more complicated because:

  • we have to simultaneously match the electric and magnetic fields at both the top and bottom interfaces

  • we have a more light rays to consider at the top interface

  • the thickness of the sheet produces a phase shift that has to be included in the calculation

Presumably there is a closed form equation giving the reflected intensity at the top sheet, though I have never seen one. In practice when dealing with more than a simple interface things rapidly get so unwieldy that we usually solve the problem on a computer. In fact I did precisely this as part of my PhD though I was dealing with multiple sheets not just a single sheet. The method I used is described in Optical Properties of Thin Solid Films by O. S. Heavens, Butterworths Scientific Publications, London 1955. It is on Google Books, though sadly not a scan.

So the answer to your question is that there isn't a simple answer to your question. You can't treat the surface and the thickness of the sheet separately. You have to do the full (complicated) calculation to find out what the reflectivities of the two polarisations are.


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