# Why does my watch act like a mirror under water?

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.

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## 1 Answer

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!

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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
Alternately, if the index of refraction of the glass/plastic is similar to that of water, then the dominant reflection event would happen at the glass/air interface, and the argument above would map over. The qualitative behaviour described above leads me to believe that this has to be total internal reflection at some surface. The only other plausible mechanism is thin-film diffraction, but that effect is wavelength dependent, and you would see several maxima as you rotated the watch, not just one critical angle. –  Jerry Schirmer Dec 1 '10 at 16:33
@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
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