Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a digital watch, rated to go underwater to $100 \rm m$. When it is underwater it can be read normally, up until you reach a certain angle, then suddenly, it becomes almost like a mirror, reflecting all the light that hits it, almost perfectly (you cannot read the digits, the entire surface becomes reflective.)

I've seen this happen with other waterproof watches too, so I don't think it's unique to mine or the specific model. I'm wondering what causes this? My physics teacher was stumped when I told him about this (we're doing lenses and imaging in physics right now.) I think it has something to do with internal refraction. I haven't been able to measure the angle it becomes reflective accurately, I estimate about 30-40 degrees. Near this critical point, it can be half and half reflective, where only half becomes a mirror, but it's always either reflective or not - never in between being reflective and non-reflective.

share|cite|improve this question
up vote 13 down vote accepted

What you are observing is total internal reflection. Snell's law tells you that, for a ray transmitting through a surface $n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}$, where $\theta$ represents the angle of reflection from the surface, $n$ represents the index of refraction of the substance in question, and the labels 1 and 2 represent the source medium and the destination medium.

If $n_{1}>n_{2}$, as would be the case for light leaving water ($n\approx 1.33$)and entering the air ($n\approx 1$)inside of your watch, simple algebra will tell you that there is a range of $\theta_{1}$ at which you will find that Snell's law predicts $\sin \theta_{2}>1$. For this range of angles, since you can't solve for $\theta_{2}$, light cannot be transmitted, and must be reflected. So the watch looks like a mirror. In fact, if you flip over, and look at the surface of the water, you will find a portion of the surface of the water looks like a mirror, too!

share|cite|improve this answer
Good guess, but you've got your $n$'s mixed up. There is no water-air surface here, because there's glass or plastic (the watchplate) in between the water and the air! The index of plastics and glasses is usually higher than water, so unless this watchplate happens to have $n\approx 1.2$, then there must be something else going on. – ptomato Dec 1 '10 at 15:56
@ptomato, @jerry : Actually , you have 3 indexes $n_w, n_a, n_g$, for water, glass and air. Applying the refraction law at each inreface, we have $n_w\sin\theta_w=n_g\sin\theta_g=n_a\sin\theta_a$ which imply the relation given by Jerry. In other words, you can neglect the glass if there is no total reflection at the first interface. – Frédéric Grosshans Dec 1 '10 at 17:20
@Frédéric, @Jerry: if the total reflection happens at the glass-air surface, then you don't even need to be underwater to see the effect. I wanted to try some experiments, but I didn't have a watch handy. I'll see if I can find one this evening. – ptomato Dec 1 '10 at 18:18
I tried it myself. Underwater the faceplate turned opaque at an angle of about 30 degrees like in the original post. I tried to observe the effect in air, but couldn't - I saw Fresnel reflection, but up to an angle of about 80 degrees I could still see the digits under the faceplate at least faintly. I couldn't observe all the way to 90 because the watch housing sticks up around the faceplate a little. – ptomato Dec 1 '10 at 22:40
@jerry, @ptomato : for the watch in the air, you have no chance to see total internal reflection (that is if you live in the air, not in the glass !). If there would be tolal internal reflexion it would be at the glass-air interface, but before coming into the glass, the light-ray would come from the air, and that itself guarantees that their angle inside the glass is below the critical angle. The same effect in water makes sure that the observed angle is the water-air angle, even if the reflection itself hapeens at the glass-air interface – Frédéric Grosshans Dec 2 '10 at 9:22

I feel that the explanation for this needs a picture:

enter image description here

On the left is the situation you have outside of the water. When you look at the watch, the light you see was refracted through the glass, so that the internal angle is less than the critical angle (where total internal reflection occurs). On the right you see the situation where you are observing from inside the water. Key difference is that there is a much smaller refractive index mismatch at the first interface (water/glass) so the light travels almost straight; consequently, it is capable of reaching the second interface at an angle greater than the critical angle, and undergoing total internal reflection.

If we assume the refractive index of the three media (top to bottom) to be $n_1$, $n_2$ and $n_3$, then the condition for total internal reflection can be calculated by working backwards:

$$\sin\theta_{2} =\frac{n_3}{n_2}$$

for the critical angle at the interface between 2 and 3 (where $\sin\theta_3=1$)


Substituting, we find the angle $\theta_1$ at which the watch will look like a mirror is given by

$$\sin\theta_1 = \frac{n_3}{n_1}$$

In other words - when $n_1$ (the refractive index of the medium with the observer) is greater than $n_3$ (the refractive index of the medium inside the watch) then it is possible that you cannot see the dial.

This problem has been solved by at least one manufacturer of extreme diving watches, by filling the mechanism with a clear fluid with the same refractive index as the (sapphire) "glass":

enter image description here

source: SINN catalog. The watch in the middle of the image has this technology - the others don't. This has the added advantage of making the watch extremely pressure resistant, since the liquid inside is highly incompressible. The watch is rated for 200 bar (2000 m depth). That's not a depth any human should try to dive to.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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