# Reflection of light on semi-infinite solid

The following statement is often made concerning reflection on glass:

"When light is normally incident on a glass surface, about 4% gets reflected and the rest is transmitted. The reflected wave is 180° out of phase with the incident wave."

In my opinion, this statement is only true if the slab is finite, since then another reflected wave (this one in phase with the incident wave) from the other surface of the slab interferes and explains why the intensity of the measured reflected light is between 0% and 16%, depending on the thickness. (There is also the fact that the measured reflected wave is generally not 180° out of phase. It is 180° for maximum reflection only.)

Thus, if the statement is made with reference to semi-infinite glass (which has only one surface), which is often the case, it must be wrong.

Am I right?

• The reflected light will vary with the thickness. Look at the details of interference on thin layers. – CuriousOne Jan 14 '16 at 19:17
• You can crank through the equations for a semi-infinite slab. What is the difficulty, precisely? – Jon Custer Jan 14 '16 at 20:11
• There are two parts to the statement: "4% reflected" and "180 degree phase shift". Which of these statements do you think is wrong? It seems to me to be true for an semi-infinite slab, and may not be true when secondary reflections (from the second surface) are introduced. Why do you think it's the other way around? – Floris Jan 14 '16 at 20:57
• In his book QED, Feynman shows that you get the same answer for the amount of light reflected by a slab of glass whether you suppose that only the front and back surfaces reflect light (with 180° and 0° phase shifts) or you divide the slab in thin layers and add reflected light by each (with 90° phase shifts). But if you divide a semi-infinite slab in thin layers and start to add, the sum does not "converge" towards a 180° phase shift, as if there was only the front surface. – courno Jan 14 '16 at 23:50

That statement comes from Fresnel equations, which say that when a wave pass from $n_1=1$ to $n_2=1.5$, the reflectivity is $R=(\frac{n_1-n_2}{n_1+n_2})^2=0.04$, without considering a second surface. You can surely take the second reflection in consideration and use interference, but the 4% from Fresnel is about a single surface (to eliminate the second one, often you find it diffusive, or wedged, or absorbent etc.). See also this question.