Why is it that an underwater observer can see only a circular "window" and also can't see anything above the separating surface? Does the "window" depend on the depth?

  • $\begingroup$ This video shows what it looks like in a swimming pool: youtube.com/watch?v=FG6ryu0-C5w. Notice that the "window" resembles a fisheye lens: everything above water is visible, the field of view is 180° $\endgroup$ – jkien Mar 6 at 23:10

Beyond a certain angle total internal reflection occurs at the water-air interface. This is because Snell's law $\sin \theta_{air} = n \sin \theta_{water} $ has no solution for $\theta_{air} $ if $n \sin \theta_{water} >1$ or $\theta_{water} >\sin^{-1}(1/n) $.

If you are under water you will see a disk shaped area above you in which there is the image of the hemisphere above the surface.

At large depth this light fades because of absorption and scattering.

  • $\begingroup$ Yes. And at the critical angle, one can see the horizon (in principle, ignoring waves, low intensity). Beyond that angle, one see the reflection. $\endgroup$ – Pieter Mar 6 at 21:48
  • $\begingroup$ @my2cts Answer might be correct, but I came here looking for the exact same question...with respect to depth. So how does this answer talk about depth I was researching Snell's Window and the 200 meter light boundary to see if there was a connection. $\endgroup$ – Christopher Rucinski Jun 21 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.