Seewolf

No, Rayleigh scattering has nothing to do with the blue color of water since it is not efficient. Scattering at objects that are much smaller than the wavelength of light, such as molecules, does depend on the wavelength (Rayleigh scattering). For blue light, it is a little bit less inefficient than for red light. This is what causes the blue color of the sky, but not the blue color of water. If you look at the sun at sunrise or sunset it is red because the blue light passing through a huge column of the atmosphere is scattered away. During the day the sun is yellow, more or less the natural color defined by the surface temperature of the sun. The column density of molecules for light passing perpendicularly through the atmosphere is too low to produce a noticeable effect. If the blue color of water was caused by Rayleigh scattering, a distant dive lamp would look red, since the red light could pass the water and the blue light would be scattered. Just like the sun at sunset. The blue color of water is caused by the absorption of light, not by scattering, just like a dye. It is the fourth harmonic of the OH stretching vibration of water that causes the absorption. Interestingly, the heavy isotopomer of water, D2O, is colorless, since the overtone of the OD stretching vibration is shifted out of the visible region. If it was scattering, it should be the same as normal water H2O.

OK, let’s analyze backscatter from a scientific point of view. First of all, scattering has absolutely nothing to do with reflection. While reflection is directional (that’s how a mirror works), scattering is (almost) not. A particle, that is illuminated by a light source becomes a light source by itself and is irradiating more or less uniformly in all directions. Therefore, it is irrelevant from which angle a scattering particle is illuminated, all that matters is the intensity of the irradiating light at the place where the scattering particle is located. You can easily prove this by yourself. Use a narrow beam dive light to irradiate a glass of water. There is no scattering, as long as there are no particles that scatter the light in the water (not completely true, I only consider Tyndall scattering). Now add a couple of drops of milk to the water, and you will see the light beam (Tyndall effect). You can move around the dive light and the intensity of the scattering will stay the same, irrespective of the angle of irradiation. How can we reduce scattering? (i) Don’t irradiate the scattering particle and (ii) pull back your strobe. (i) is trivial, but why does (ii) work? Let’s assume you have an ideal strobe, then the inverse-square law holds for the light intensity as a function of distance. If you double the distance between strobe and object to be illuminated, the light intensity decreases to ¼. This is true for both scattering particles and the object you want to photograph. Now let’s do a “Gedankenexperiment”. You have your strobe, a scattering particle, and your object in line. The distance between strobe and scattering particle is 10 cm and the distance between strobe and object is 100 cm. In this case, the light intensity (number of photons per unit area) at the scattering particle is 100 times larger than at the object (because of the inverse square law and 10 times larger distance). If you now pull back the strobe by 80 cm, the distance between the strobe and scattering particle is 90 cm, the distance between scattering particle and object still 90 cm, and the distance between strobe and object 180 cm. Now the light intensity at the scattering particle is only 4 times that at the object and not 100 times! This is also true if you increase the power of the strobe to compensate for the larger distance to the object. Compared to the object, the scattering particle looks now much less bright. The final “Gedankenexperiment” is to pull back the strobe to an infinite distance. We do have such a strobe, it is called the sun. In this case the light intensity is the same at the scattering particle and at the object. We still see the scattering, but it is much less intense than by using a strobe considering the inverse square law. Of course, this is very much simplified. I am only talking about Tyndall scattering of particles larger than the wavelength of light, not about Rayleigh scattering at molecules.

I am not sure if this question has been answered. The upgrade Z330 Type 1 to Type 2 can be done by an authorized Inon service center. I paid 119 € for the upgrade in 2021 (uwcamerastore.com).

I made a quick check and can confirm this. The Teleplus HDpro + 8-15mm does not work with the Z8 (F error) while the same combination works flawlessly with the D500. Too bad, since I decided to buy a housing for the Z8 and sell my Aquatica AD500. I still have an older N-AF Teleplus Pro 300 but this not even worked with the D500 + 8-15mm (Tokina 10-17mm was OK). The TC might be defect, so I replaced it with the HDpro.

Just migrated from the old forum, and I am happy to see such an active community. Shooting with a Nikon D500 and Aquatica housing. Just decided to move to the Z8. Happy diving and best wishes to all!