# Reflections in a glass of water

I've noticed that if you take a full glass of water and look from above, through the water, you can't see through the glass sides - instead, you see a reflection. I tried with a laser pointer and the case was the same, the beam was just reflected from the water-glass interface. Even when looking from a variety of angles the case is the same. With no water in the glass, you can see through it fine.

Why is this?

I thought maybe TIR, but then $n_{\text{water}} \approx 1.3$ and $n_{\text{glass}} \approx 1.5$. So the light is going from a lower to high refractive index material, which shouldn't give TIR.

A little research (Wikipedia) suggests that you can get Frustrated TIR from tightly grasping the glass, so photons can tunnel through from inside the glass of water, get reflected by skin and then tunnel back to get to your eye (again, I think this is right).

However, I wonder what physical mechanism prevents light from escaping a glass of water?

• I would guess it's total internal reflection at the glass-air boundary Jun 12, 2013 at 8:28

The light rays get "straighter", closer to the normal direction of the boundary plane, when they travel from lower $n$ (water, $1.3$) to higher $n$ (glass, $1.5$).
However, the photons sent by the objects you see in the reflection are also reaching the glass-air boundary, as John Rennie pointed, out, and if they continued, they would travel from a higher $n$ (glass, $1.5$) to a lower $n$ (air, $1$). For angles (between the normal direction and the light ray's direction) greater than the critical angle $$\theta_c=\arcsin(1.00/1.50) = 41.8^\circ$$ the photons can't get through and they will reflect from the glass-air boundary back to the glass and some of them will make it to the water (there may also be a total internal reflection when going from glass to water! the critical angle is $\arcsin(1.3/1.5)=60.1^\circ$). These photons that get through the water back to your eye (the water-air critical angle is $\arcsin(1/1.3)=50.3^\circ$) will communicate the reflected image of the object that emitted the light we were tracing.
• A metallic mirror is a conductor so it sets the electric field to zero - because of Ohm's law, $j=\sigma E$, a nonzero $E$ would immediately create currents that would compensate the distribution of charges. With the $\vec E = 0$ boundary conditions at the mirror (and with some related boundary conditions for the magnetic fields), the only allowed waves are those that vanish inside the metal but get reflected on the side of the air. Jun 13, 2013 at 10:57