The apparent color of water surfaces is fascinating, and much more complex than I realised (a combination of reflected light, refracted-and-transmitted light, absorption by vibrating H2O molecules, internal reflection by suspended particles, internal absorption by suspended particles, blue-shifting by vibrating-and-re-emitting H2O, ... etc).

But one thing I can't seem to find clear info on is the effects of polarization on color.

e.g. in this answer https://physics.stackexchange.com/a/423122/244448 :

"When light glances off a smooth surface, the surface reflects light with horizontal polarization more effectively than light with vertical polarization (see Brewster's Angle), so some skylight can end up darkened - or brightened - relative to other colors depending on the angle of polarization and the angle of incidence."

The key phrase: "darkened - or brightened - relative to other colors depending on the angle of polarization" is tantalising - experimentally, sometimes water surfaces appear darker than the sky, and sometimes they appear brighter, and there's no obvious pattern. But if the above quote is correct, and polarization both darkens and lightens colors, then polarization (which I cannot innately detect/see) could fully explain that.

NB: one of the problems I've encountered in exploring this is that - of course - most photographs of water are "fake" without recording this: in general, all photographers use polarizing lenses to some extent, from the universal sky-filters to the slightly more specialist rotating paired polarizing lenses on SLRs. This is not something that occured to me until I started digging in to polarization. This info doesn't (usually) appear in photo metadata, since it's outside the camera's electronic system.

(i.e. I'm finding that the only way I can investigate this phenomena is to physically travel to locations and bodies of water at times of day and weather conditions, which is severely limited by budget!)


2 Answers 2


The intensity of light after reflection is given by the Fresnel equations. The color is changed due to the different refraction indices of the water at different wavelengths. I'm expecting the most important deviations should occur near the Brewster angle (as each color has a different Brewster angle, so there should be very faint images where one of the colors is missing), but I don't think the differences would be noticeable beyond that (water's refraction index varies only ~1% in the visible spectrum).

  • $\begingroup$ Here's an example of a genuine (non-faked) polarized photo: en.wikipedia.org/wiki/File:Reflection_Polarizer2.jpg - see how the reflection colors are considerably different (and this is a relatively weak example - there are much more extreme examples, although it's hard to know which are real and which are fake). Can a 1% difference explain such changes? $\endgroup$
    – Adam
    Commented Nov 30, 2019 at 18:13
  • $\begingroup$ I was referring strictly to reflected light. In those pictures, you have 2 different optical images under the water surface: the bottom of that body of water and the reflection of the sky. From each you get s-polarized and p-polarized light. Generally both images will be visible under s-polarization, but for the p-polarization, the bottom will be more visible - especially near Brewster's angle, where there's (almost) no reflection. You could see then that the second picture is taken with a p-polarization, while the first is under s-polarization. There is no color change in those pictures. $\endgroup$
    – mostanes
    Commented Dec 1, 2019 at 13:32
  • $\begingroup$ The reflection clearly has a different color in the two images. $\endgroup$
    – Adam
    Commented Dec 3, 2019 at 1:25
  • $\begingroup$ Are you sure you're referring to the reflection? There's barely any reflected light in the second picture - how did you measure its color? Also, please use picture to disambiguate from optical images, especially since there are 2 different images, both of which are visible in the first picture, but only one visible clearly in the second picture (the sky's reflection is barely visible in the corners). $\endgroup$
    – mostanes
    Commented Dec 3, 2019 at 20:20
  • $\begingroup$ It's difficult to measure! That example isn't the easiest to work with, but it's one of the ones I have greatest faith is genuine. My process is twofold: using pixel-based color measurements, and (where possible), I've started doing photoshop-based reverse white-balancing, to account for photos being massively differently exposed once the colors change. Some of the polarized picture pairs I've seen I am unsure how much color change is due to camera auto-exposure settings causing a re-coloring. But some is definitely heavily shifted in blue and red. $\endgroup$
    – Adam
    Commented Dec 5, 2019 at 8:13

Imagine you take a library book down from its shelf to read it. Now you want to put it back. You can easily do so unless you rotate it by 90 degrees (for example). Now it is blocked. It can't go in. The principle is the same. Water acting like a filter can accept photons of light which would mean less are available for reflecting back to the observer who sees a darker image in the reflection than the original. The opposite would be true if larger numbers of photons bounce back (like the library book) due to their angle of approach.

  • $\begingroup$ This doesn't explain any of the color change that polarization causes, and that it is famously used for by photographers. $\endgroup$
    – Adam
    Commented Nov 30, 2019 at 18:07
  • $\begingroup$ Ok. Let's say you've got a very green reflection on a puddle from a neon night-club sign. Polarizer across a certain angle spectrum can block all the yellow light from entering your lens, scatter it away so your picture would look much bluer. $\endgroup$
    – Pagoda
    Commented Dec 2, 2019 at 12:04

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