Does the Fundamental Theorem of Radiometry mean light can be perceived stronger underwater than above?

Question

I'm not a physicist, but trying to understand the propagation of light for rendering purposes. So far, a lot has made sense, but I've stumbled upon something called the "Fundamental Theorem of Radiometry": $$ \dfrac{L_1}{L_2} = \dfrac{n^2_1}{n^2_2} $$ Where $L_1, L_2$ are the radiance and $n_1,n_2$ are the refractive index in element 1 (e.g. air) and 2 (e.g. water) respectively along a ray. A better definition is found here in section 2.5.4.2: https://dirsig.cis.rit.edu/docs/WaterManual.pdf

Since air has an IoR of $\approx 1$ and water of $\approx 1.34$, this would mean that $$L_2 \approx L_1 * \dfrac{1.34^2}{1^2} $$, so the radiance becomes larger in water. This is also explained in Figure 7 of this source: https://www.oceanopticsbook.info/view/light-and-radiometry/visualizing-radiances#x1-87

My question as layman would now be: Does this also mean a sensor (e.g. camera) would perceive incoming light in a brighter fashion right below the water surface than right above the surface?

This somehow contradicts with my understanding of the preservation of energy, but I realized it has something to do with the solid angle of the beam decreasing in water, but i don't know how this even matters for a single, linear traveling beam, or$-$from the perspective of a sensor$-$a measurement over a larger angle for all incoming light.

Answer 1

I try to answer, but my feeling is you mainly answered yourself already correctly above.

Radiance is $W/(m^2 sr)$ and the solid angle (in unit sr) decreases when $L$ enters the water. So $L$ increases as the denominator gets smaller, the optical power (nominator) and thus the energy is of course preserved. (Actually, parts of the light will be reflected at the interface, so you will loose some photons.) The decrease of the solid angle is explicitly mentioned in the ocean optics webbook you have linked above.

When you have a hypothetical single ray, it follows just Snell's law. However, each light source, even a laser, has some solid angle and therefore contains a bunch of diverging rays. If you apply Snell's law to all rays coming from the upper half-dome (e.g. uniform skylight is the light source), you will find that they are limited to Snell's window below the water.

Good question for the camera. Each pixel has some acceptance angle determined by its size and the focal length of the lens, so it measures the photons coming from a certain solid angle or field of view (around the direction the pixel sees). As single rays gets closer, more photons are in a given solid angle, which a pixel sees (pixel gets brighter). However, the solid angle a pixel sees (its acceptance angle) changes as well because you no longer have the air/glass interface at your lens, but a water/glass interface. This increases the effective focal length and as a result the complete field of view of the camera as well as the field of view of a single pixel gets smaller (pixel gets again darker). No idea if these effects exactly cancel out each other. However, in addition you have absorption under water and some photons are reflected at the surface according to Fresnel coefficient.