In glass windows, you get light passing through and light reflecting off the front and back surfaces. Would you get more light reflecting off of triple or double pane windows? Because you have double or triple the amount of surfaces for the light to reflect off of.


3 Answers 3


Short answer - yes.

If you have a single-pane window, light that doesn't reflect off the front or rear surface will be transmitted. But a second pane might once again reflect some of this light, and that has a chance of making it all the way out to the front. The second pane doesn't "stop" any of the reflection off the first pane - so it can only add to the total reflection.

Here is a diagram that demonstrates this:

enter image description here

Reflections A and B will happen with just a single pane; C and D are added with a second pane, and you can keep going.

Calculating the theoretical total reflection is not too hard. Assuming normal incidence, the Fresnel reflection equations tell us the reflected intensity is

$$r = \left|\frac{n_1-n_2}{n_1+n_2}\right|^2$$

From the symmetry of this equation you can see that the reflection off the back surface is the same as the reflection off the front surface - but then that ray (marked B) needs to be transmitted - where of course $t = 1-r$.

This means that the intensity of A+B is given by $r+t^2r$ (A is just reflected, while B undergoes transmission twice and reflection once). For C, there are 4 transmissions and one reflection; for D, it is 6, etc.

Thus, for a stack of N window panes, the reflected intensity (if there is no absorption, and light is too incoherent for interference effects to be significant), the reflected intensity is

$$I = r \sum_{i=0}^{2N-1} t^{2i}$$

This is a geometrical series that sums to

$$I = \frac{r~(1-t^{4N})}{1-t^2}$$

It is clear from this equation that as $N$ gets bigger, the reflected intensity also gets bigger - although it's a law of diminishing returns. We can plug in some numbers...

For a typical case of glass with n=1.5, we get $r=0.04$ and $t=0.96$, so the fraction of reflected intensity for a single pane (A+B) would be $0.04\cdot(1+0.96^2) = 0.077$; a second pane would bring the fraction up to 0.142, and an infinite stack would give a reflected fraction of 0.510

Peripherally related is this question and associated answer.

  • $\begingroup$ It occurred to me that internal reflections inside each of the pieces of glass, or between pieces of glass, will further complicate things; but these change the "effective reflectivity" and don't change the underlying fact that the next pane can only ADD more to the reflected power. $\endgroup$
    – Floris
    Mar 7, 2017 at 18:57

You hit on half of the idea of optical thin film coatings. The other half is that light is a wave. Multiple reflections can interfere constructively or destructively. This can be used to make either an antireflection coating, or a mirror, or something more complex.

This shows the idea of an antireflection coating. Attribution below. The coating is 1/4 wavelength thick. Light that bounces off the back surface travels an extra 1/2 wavelength. It emerges out of phase with the front reflection. The two reflections cancel.

It doesn't really work on something as thick as a pane of glass. You can't make the front and back surfaces of glass exactly parallel and exactly the right separation.

Also, as WetSavannaAnimal pointed out, ordinary light does not have a perfectly stable wavelength. It is produced by a lot of individual short lived events in many different atoms, and each atom is independent. By the time a beam of light travels through a pane of glass twice, the light arriving at the top surface is from different events. There is no consistent relationship in the phases of light separated that far. So you don't get a cancellation that lasts longer than a fraction of a nanosecond.

A good quality laser has a very stable wavelength. With it, you can get interference patterns from the front and back surfaces of a pane. This is the original of the speckle that a laser produces when you shine it on a rough surface. You get interference from reflections from low and high points.

Illustration by DrBob - Own work (enwiki), CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=12305780

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    $\begingroup$ +1 for correctness, but I don't think this is the whole story here because the OP is most likely talking about light of coherence length shorter than either the thickness of the panes or their separation. $\endgroup$ Feb 26, 2017 at 0:05
  • $\begingroup$ @WetSavannaAnimalakaRodVance - Thanks. I updated my answer. $\endgroup$
    – mmesser314
    Feb 26, 2017 at 2:31

Double pane reflects more light, because the second pane adds reflections and light.

It is best seen by using a point source, as in the photo. Image A0 is due to a single reflection in the first pane, B0 is due to a single reflection in the second pane, C0 is due to two reflections in the second pane. The other images are due to multiple reflections inside a pane. So the A-images are light from the first pane, and the B- and C-images would disappear if the second pane was removed.

double pane window

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    $\begingroup$ This would be much improved by having the figure or description that tell us how the 'A', 'B', and 'C' groups are different and why the group spacing differs from the dot spacing inside the groups. $\endgroup$ Feb 25, 2017 at 20:00
  • $\begingroup$ Here's my guess. The surfaces of double pane are #1,2,3,4. The sequence of the reflecting surfaces is: A0:#1, A1:#2, A2:#2-1-2, A3:#1-2-1-2-1-2, ... B0:#3, B1:#3-1-2, B2:#3-1-2-1-2, B3:#3-1-2-1-2, ... C0:#3-1-3, C1:#3-1-3-1-2, C2:#3-1-3-1-2-1-2, ... A-dots are closer to each other than B-dots due to the angle of incidence and refraction. My guess may be wrong. Find a better solution by studying the reflections in your own double pane window. $\endgroup$
    – jkien
    Feb 25, 2017 at 23:38
  • $\begingroup$ I'm not asking because I am confused or uncertain about what's happening. I'm saying that without some explanation (preferably in the form of a figure as in mmesser314's answer) the value of this answer to people who don't already understand the phenomena is badly curtailed. $\endgroup$ Feb 26, 2017 at 4:44

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