It is very difficult to hear underwater sounds from above the surface of a body of water. This is suggested to be due to the speed of sound being very different in air versus water, leading to most underwater sound being reflected back into the water, and not passing into the air.

Am I supposed to think of the air-water interface as some sort of plane mirror then? I have heard of Snells law for refraction, but am not sure how it applies to reflection in this case. If I am to think of the interface as a mirror, I might expect a very thin layer of water to have equal reflection/reflectivity compared to a thick one. Is this the case?

If not, how can I better think about what is going on? Also, is there interplay between the wavelength of the sound and the thickness of the water layer required for reflection?


2 Answers 2


Within reason, a thick sheet of water and a thin one will affect sound mostly the same.

With extremely large thicknesses, sound actually transmits better.

When the thickness approaches a wavelength of the sound in question, the equations get a bit more complicated. I'd be wary of venturing a guess in this regime.

  • $\begingroup$ I can't wrap my head around why there is reflection to begin with. I can use an analogy with light to see how it would refract, but not reflect. I have heard people say that the reflection is due to an impedance mismatch but what that really means I can't picture. Their logic being difference in speed of sound -> impedance mismatch -> reflection. $\endgroup$
    – user391339
    Apr 14, 2022 at 4:46
  • $\begingroup$ @user391339 Light also reflects from the surface of water. $\endgroup$
    – d_b
    Apr 14, 2022 at 15:53
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    $\begingroup$ @user391339 Consider a perfectly hard, immovable wall. A sound wave approaches it. The energy has to go somewhere. Since the energy is associated with a pressure wave, you end up building up some pressure on the surface. When a trough in the wave finally occurs, that pressure is released into the trough, resulting in a soundwave that is now traveling in the opposite direction. If there is an "impedance mismatch", its a situation somewhere between letting the soundwave propagate freely and the immovable wall case. $\endgroup$
    – Cort Ammon
    Apr 15, 2022 at 14:11
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    $\begingroup$ The same pattern can occur on the other side of the barrier too. If sound is traveling in a material like water, which can't compress all that much, and it hits a pocket of air, the water can't move far enough to really transmit all of the energy into the air, so it ends up not transmitting all of the energy into the air. What is left becomes a new sound wave traveling backrwards. $\endgroup$
    – Cort Ammon
    Apr 15, 2022 at 14:13
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    $\begingroup$ Also, I found this question about the impedance of thin materials. The answer does use frequency space notation (where the independent variables are $\theta$ and $\omega$ rather than $x$ and $v$). That might be a bit confusing, but it is the right units to explore that answer. $\endgroup$
    – Cort Ammon
    Apr 15, 2022 at 14:16

Yes but not directly because of the speed of sound. It's analogous to Snell's Law in optical reflectivity, or to VSWR in microwave components. The effective "index of refraction" for sound changes drastically at a water-air interface, thus leading to a lot of reflection and not much transmission.

This holds true for interfaces between depths of water with significantly different temperatures for the same reasons. Submarines take advantage of this to limit their sound "signature" from propagating to bad places like enemy sonar receivers.


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