# How to calculate the fraction of light rays reflected?

The smooth surface is air-liquid interface and the rough surface is let's say a t-shirt.

The image above shows the probability of each reflection, and the $$p$$ value is the probability of the light rays totally reflected by the air-liquid interface. So this means $$p$$ represents the fraction of light rays that goes in an angle greater than the critical angle and we are after the intensity (or amount I don't really know how I should call it) of light in the yellow shaded area:

The article I am reading finds $$p$$ in the following way but I do not understand it. Can anyone please explain it in a way that can be clearly understood?

• Can you post a link of the article? Apr 8, 2020 at 22:04
• @boyfarrell Finally someone is here! :) Here you go: sci-hub.si/10.1364/AO.27.001278 Apr 8, 2020 at 22:06
• They are solid angle integrals. Will write and answer tomorrow Apr 8, 2020 at 22:09
• @boyfarrell Thank you! I am a high school student and don't have much information about them so it would be very helpful if you explain each and every step clearly. Good night! Apr 8, 2020 at 22:11
• Well, I will see how much time I have and write accordingly. Apr 8, 2020 at 22:27

Lambertian phase function is,

$$\Phi(\theta) = \cos(\theta)$$

This means at normal $$\theta=0$$ maximum intensity is reflected form the cloth surface, which falls of to zero when $$\theta=\pi/2$$.

Reflection from the surface is diffuse, meaning that a beam of light entering at a constant angle, will be reflected into all angles and will be distributed according to the phase function.

This requires integrating over all angles to calculate to total intensity reflected.

The integral in the numerator looks like this,

The integrals adds up all possible directions of reflected light outside the cone.

$$\int \Phi(\theta) d\Omega$$

Element of solid angle is defined as

$$d\Omega = \sin(\theta)d\theta d\phi$$

Substituting in and including the correct limits,

$$\int_0^{2\pi} \int_{\theta_c}^{\pi/2} \cos(\theta)\sin(\theta) d\theta d\phi$$

We can integrate directly in $$\phi$$,

$$2\pi \int_{\theta_c}^{\pi/2} \cos(\theta)\sin(\theta) d\theta$$

This is what is written above.

The second integrals is just adding up all possible reflected directions in the hemisphere; it’s used to normalise the reflected value to give a probability.

• I think I understand why you put these limits but could you please explain why did you choose these ones? Especially for the first integral. Is it because it's a hemisphere with a radius of $\frac{thickness\space of\space the\space liquid\space layer}{2}\space \space \space$? Apr 13, 2020 at 14:50
• These are not length integrals they are direction (solid angle) integrals. Imagine the integral is sweeping out a “volume”, if you use the limits above it sweeps out the yellow section in the diagram. This is what I was trying to show with that. Apr 13, 2020 at 14:54
• Thank you! I don't know if I was clear so I am asking again: By making the second integral's boundaries $\frac{\pi}{2}$ to $\theta_c$ we kind of calculated the area in two dimensions. And by making the first integral's boundaries $2\pi$ to $0$, we integrated the value (I don't really know what it is :/) over the second angle of the solid angle. I am sorry this is not my first language but what I want to say at short is that solid angle has two angles and that's why we integrate it twice. Is that it? Apr 13, 2020 at 15:03
• Yes! I that’s correct. A solid angle defines a “volume” of directions from a starting point. The first integral in actually over $\theta$, it sweeps out a “standard” angle (it looks like a “plane” of directions from $\theta_c$ to $\pi/2$). The second integral, over $\phi$, rotates the “plane” to define a “volume” of angles: a solid angle. Yes a solid angle is defined by a two angular ranges: one over $\theta$ the other over $\phi$. Apr 13, 2020 at 15:15