Can anyone please explain how it subtends such an angle? What I understand from this is something like this: https://prnt.sc/ruroj6. However, I can't understand how it is expressed mathematically. It was the grandson of the Ångström that the unit is named after. Huh -- didn't know Angstrom was so recent, seems like the unit would have been named long ago. The question of why a soaked fabric is more transparent is slightly different. The fabric is made of many fibers with air in between and this structure causes light to bounce around many times inside, making it hard for light to get through. When we soak the fabric, we are replacing the air with water, which has a closer index of refraction to the fibers, making the inside reflections less important and more light goes through. The swedish physicist Anders Angstrom was the first person to study this effect. He was intrigued by the darkening of soil when in contact with water and published the first proposed explanation of the effect in 1925.  What we are doing here is calculating the ratio of radiance (power per unit of solid angle per unit of area) that is totally internally reflected (which corresponds in the figure to the blue area)  $$ p = \frac{R_{blue}}{R_{total}} = \frac{\int^{\pi/2}_{\theta_c}\int^{2\pi}_{0}\frac{\phi_{source}}{4\pi r^2 }\cos \theta d\Omega}{\int^{\pi/2}_{0}\int^{2\pi}_{0}\frac{\phi_{source}}{4\pi r^2 }\cos \theta d\Omega} = \frac{2\pi \int^{\pi/2}_{\theta_c}\cos \theta \sin \theta d\theta}{ 2\pi \int^{\pi/2}_{0}\cos \theta \sin \theta d\theta} $$ where $d\Omega = \sin \theta d\theta d\psi$ is the differential solid angle. The integral is easy to solve since $\frac{d}{d\theta} \cos^2 \theta = -2 \cos \theta \sin \theta$ and so $p = \cos^2 \theta_c$. Saying that in a Lambertian Surface the intensity reflected at angle $\theta$ is proportional to $\cos \theta$ (first derived by Johann Heinrich Lambert in 1760) is equivalent to saying that the surface has the same radiance (power transmitted per unit solid angle per unit projected area) when viewed from any angle.  To prove that, let's analyze the following diagram representing a Lambertian Surface. The radiance at A is $R_A = \frac{\phi_{source}}{4\pi x_0^2} $ where $\phi_{source}$ is the luminous flux of the source. At B, the radiance is $R_B= \frac{\phi_{source}}{4\pi x^2}\cos \theta$ since the plane of incidence is tilted at an angle $\theta$ from the normal direction. We just proved that for a degree $\theta$, $R\propto \cos \theta$! Note that if you isolate the term $a(1-R_l)$ we get $$ \frac{A}{a(1-R_l)} = 1+(1-a)p+(1-a)^2p^2+(1-a)^3p^3+ \ldots $$ which is the same as the geometric series $\sum^{\infty}_{n=0} x^n = \frac{1}{1-x}, |x|<1$ with $x = p(1-a)$. Thus $$ A = \frac{(1-R_l)a}{1-p(1-a)} $$ When light hits a surface we can decompose the propagating electric and magnetic fields in 2 components: the component perpendicular to the surface and the component parallel to the surface. The reflectivities for these 2 components are not the same and in particular, there's a small but non neglectable reflectivity for $\theta < \theta_c$ - which means that there's a percentage of light that is reflected in the liquid-air transition even at angles below the $\theta_c$. In the following to graphs we can see in blue the reflectivity for the liquid-air transition for the 2 polarizations: perpendicular and parallel. As we can see for both polarizations the reflectivity for $\theta<\theta_c$ is small but not zero, and so we also have to take that into account.  ![]()  The first graph corresponds to the polarization where the electric field vector is perpendicular to the plane of incidence and the other with its electric field vector parallel to the plane of incidence. In the following image we can see the difference between *Diffuse reflection*, where light is reflected from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of *specular reflection*. For ideal diffuse reflecting surfaces the intensity of emitted light (luminance) is the same regardless of the direction we look in the upper half plane.  All the light rays diffused in the blue regions are going to be totally reflected as they hit the liquid-air transition making it impossible to escape to the air.  Note: If you don't know what total internal reflection is you can quickly learn [here](https://courses.lumenlearning.com/physics/chapter/25-4-total-internal-reflection/).