4
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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 '19 at 23:10
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Yes. And at the critical angle, one can see the horizon (in principle, ignoring waves, low intensity). Beyond that angle, one see the reflection. $\endgroup$
    – user137289
    Mar 6 '19 at 21:48
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 21 '19 at 12:32
2
$\begingroup$

When a light ray passes from an optically denser to a rarer medium, it bends away from the normal at the point of incidence, in accordance with the Snell's law of refraction:

$$n_1\sin \theta_1=n_2 \sin \theta_2$$

where $n_1$ and $n_2$ are the refractive indices of the optically denser and rarer medium respectively, $\theta_1$ and $\theta_2$ are the angles of incidence and refraction respectively. You could see that if $n_1>n_2$ then in order for the product of the refractive index and the corresponding angle to be constant $\sin\theta_1<\sin\theta_2$ or $\theta_1<\theta_2$.

Now as you gradually increase $\theta_1$ then $\theta_2$ also increases. Since $\theta_2>\theta_1$, $\theta_2$ attains $90^\circ$ before $\theta_1$. The angle of incidence $\theta_1$ when the refracted ray grazes over the interface is known as the critical angle $\theta_c$. For even higher values of angle of incidence, refraction is no longer possible and total internal reflection takes place.

The following diagram clearly illustrates what happens when the angle of incidence is lesser than, equal to and greater than the critical angle.

enter image description here

Image source: Snell's window - Wikipedia

The same diagram is also true if we reverse the direction of light rays.

Imagine yourselves as the fish in the image below:

enter image description here

Image source: Earth Science Picture of the Day - Universities Space Research Association (USRA)

When you look up, you could see only an approximately circular region of the sky. Beyond this circular region you'd either see the reflection of the bottom of the swimming pool or any other water body, or if it's deep enough, it would be dark (as shown in this image). This circular region is known as the Snell's window. Called as "Snell's ..." in honour of the scientist Mr. Snell. Called as "... window" because you could see stuff outside the water only through this, in a similar way to the windows in our buildings.

Looking up from underwater, one sees the whole sky. But it doesn’t stretch $180^\circ$ from horizon to horizon like it does on the surface of water. Instead it's compressed into a circle about $97^\circ$ across, regardless of the observer’s depth. Why $97^\circ$? This is because, the critical angle of water-air interface is about $48.6^\circ$ and twice of which is $97.2^\circ$.

Irrespective of the depth, the angle $97^\circ$ remains the same. Just the area on the surface increases. However, as you move down intensity of light decreases due to scattering and absorption by the water molecules. Or in other words, the sky above becomes dimmer and dimmer as you go deeper.

$\endgroup$
0
$\begingroup$

When a light ray passes from an optically denser to a rarer medium, it bends away from the normal at the point of incidence, in accordance with the Snell's law of refraction:

$$ n_1\sin θ_1 =n_2\sin θ_2$$

where $n_1$ and $n_2$ are the refractive indices of the optically denser and rarer medium respectively, $θ_1$ and $θ_2$ are the angles of incidence and refraction respectively. You could see that if $n_1>n_2$ then in order for the product of the refractive index and the corresponding angle to be constant $\sin θ_1<\sin θ_2$ or $θ_1<θ_2$.

Now as you gradually increase $θ_1$ then $θ_2$ also increases. Since $θ_2>θ_1$, $θ_2$ attains $90^\circ$ before $θ_1$. The angle of incidence $θ_1$ when the refracted ray grazes over the interface is known as the critical angle $θ_c$. For even higher values of angle of incidence, refraction is no longer possible and total internal reflection takes pl

$\endgroup$
1

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.