# Under water light reflection of air in a cup vs water in a cup

The last time I was on vacation I was drinking on the swimming pool, and after I was finished with the drinks I started playing with the 2 plastic cups I had, and noticed that the air produces a reflection effect with the cups.

In this picture, both cups are full of water and both cups are completely transparent: In this picture, the left cup is filled with water, and the right cup is filled with air. And as you can see the right cup is reflecting the light. Also, I noticed that this effect was visible only when I was above the surface of the water. When I was under water both cups were completely transparent.

I have to add that most of the light is coming from an underwater light.

I would like to understand why the air produces a difference on how the light is reflected.

That's a beautiful question and shows a deep interest and I encourage you to keep on that way of thinking

I only have a long answer for you, so bear with me.

The long answer depends on an understanding for :

1. Optical density and light speeds.
2. Angles of light incidence between media.

1. A fundamental concept in optics called: optical density

optical density is a property of transparent material (different from normal density) and it's a measure of how fast light goes in said material.

usually given symbol $$n$$.

why would light speed vary? light behavior on the microscopic scale is quantized, i.e. light has bits and parts of energy that run around like marbles. This energy in return is controlled by : (this will look familiar I hope)

$$c = \nu \lambda$$ , where c is the constant = 3x$$10^{8}$$ in vacuum, $$\lambda$$ is the wavenlenght and $$\nu$$ is the frequancy. (1)

Now what would these bits of energy do ? They interact with other bits, more precisely here: electrons.

The photons going thru a transparent medium are like that one popular guy in a party who walks in says hi to this grin to that gets high fives and high fives, and by the time he's at the backyard, he has exchanged a good amount of handshakes and body touching with those in his paths AND has taken longer to go there because he keeps on stopping to say hi unlike an invisible introvert who walks between the crowd to the backyard.

Similarly, the electrons going in their paths are met by a transparent material, so they exchange parts of their energy and get them back as they go, this makes them slower.

How would that affect the equation (1)? It won't. We repute it as

$$v = \nu \lambda$$ , $$c > v$$

Stop here and move to our 2nd :

1. The angels of incidence. Light going from a transparent medium to another adjusts its path according to the equation (whose proof will take longer but I've attached some sources).

This is what we call Snell's equation for light refraction.

It has a special case when light moves from medium X to medium Y where $$n$$ X > $$n$$ Y.

In this case explained briefly by an image I found : We're interested in the case of the critical angle, the light falls with an angle greater than it, it no longer refracts or changes its path, it merely reflects in the same medium and doesn't path to the other. (Total internal reflection.)

I first make some assumptions that I will explain one by one :

$$n$$ of air is givin 1

$$n$$ of fresh water is given around 1.3 (with some variation here since is this pool water )

$$n$$ of cup material is more than 1 and slighty, very slightly more than that of water

With one disclaimer here : I'm not sure of the angle at which light fell on in each case so I will guess it roughly as I go from the given.

Light in this case [A] went : water (source) > cup material > water in cube.

We've assumed that $$n$$ cup material is larger than the water so the angle of incidence wouldn't matter here, you haven't experienced any TIR in that case therfore light simply followed snell law. I suppose the refraction is too little to be noticed.

case [B] : 1st cup like case A

2nd cup is empty, path is : water (source ) > cup material > air > .....should be more Cup Material but it's: more air!

This here is a case of TIR. The light went from the material to air, walked a little in air and faced by cup boundary .. at an angle that's more than the critical angle so it simply went back .. reflected as you said.

Beautiful, isn't it?

• Long answers are not always better than short ones. – sammy gerbil Apr 22 '17 at 2:14
• The TIR happens at the interface plastic-air and not at the air-plastic interface. So the light does not "walk a little in air". Unless you mean the evanescent wave at the same plastic -air interface. – nasu Oct 2 '18 at 3:46