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From this section on wikipedia, the phenomenon of an image appearing warped or slightly displaced when looking through water is due to its index of refraction (1.33) deviating from that of air (~1).

Also from wikipedia, the index of refraction for window glass is 1.55. Going by the above argument, the warping of an image should be even more pronounced looking through my window than through water.

I've considered that maybe it's because window glass is typically very thin, and therefore the light doesn't spend enough "travel time" scattering through a thin glass that would cause deviation from a straight line. Light refracts when entering, then refracts back by the reverse angle when it exits, and its overall displacement from a straight line due to time spent within the glass under a different index of refraction is negligible. On the other hand, if the window glass were for example 1 meter thick, I would expect to see image warping like I do with water.

Does this logic make sense? Or is another mechanism(s) at play?

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    $\begingroup$ You are looking at things in the water so refraction has happened at only one interface. If you had objects embedded in the glass you would see the effect. $\endgroup$
    – Lewis Miller
    Commented Oct 30, 2018 at 16:10
  • $\begingroup$ Consider something like a fish tank, where if you look from near perpendicular you can see right through the water and the glass. Things look normal through that if the wall is flat. $\endgroup$
    – JMac
    Commented Oct 30, 2018 at 17:01

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You are comparing an object submerged in water with an object in air behind a thin layer of glass.

For the object submerged in water, refraction takes place only once, so there is an obvious deflection, but for the glass, refraction takes place twice, from the air to glass and back to air, which does not result in a net deflection, only a small displacement. Therefore, your logic is correct.

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Your explanation does make sense. In fact, there is a formula: $x = t \sin i \left( 1 - \dfrac{\cos i}{\sqrt{\mu ^2 - \sin ^2 i}} \right)$

where $t = \text{Thickness of glass}$, $i = \text{Angle of incidence}$, $\mu = \text{Refractive index of glass}$.

As $t$ is very small in your case, the $x = \text{Lateral shift}$ will also be very very less.

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  • $\begingroup$ What is this formula called? $\endgroup$
    – Blaise
    Commented Oct 31, 2018 at 0:26
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    $\begingroup$ We called it "Formula for Lateral shift in glass slabs". Don't know if there is any specific name. $\endgroup$
    – Wrichik Basu
    Commented Nov 1, 2018 at 5:33

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