Here is what appears to be a very informative document entitled "Underwater Radio Communication" by Lloyd Butler VK5BR. Here are some interesting extracts.
- Attenuation (α) in dB/metre = 0. 0173 √(fσ)
- where f = frequency in hertz and
- σ = conductivity in mhos/metre (siemens per metre)
Here's a useful graph linked with the above formula. It basically plots attenuation against a base of frequency for "fresh water and salt water. As an example, at 100kHz, Adelaide fresh water attenuates at about 1.5 dB/m whereas sea water at 100kHz is about 10 dB/m.
The document then goes on to discuss loss due to water/air interface (surprisingly high) but improves dramatically with frequency. Another interesting point formulates the wavelength of a transmission in a conductive medium: -
- Wavelength (λ) in metres = 1000 √{10/(fσ)}
For instance, in sea water, wavelength at 10 kHz is only 15.8 metres compared to 30 km
in space. The upshot of this is that it makes antenna design at lower frequencies much easier because of considerable size reductions.
The document then goes on to discuss several scenarios of transmission and reception.
Regards the situation in the question, I'm thinking that 100 MHz might be pretty good for 2m depth because, although the attenuation is high at about 80 dB for the two metre depth, the interface loss at the surface is quite low at about 10 dB. This is a total loss of about 90 dB and would seem "do-able" for a transmission power of 1 watt. This is the fresh-water scenario. For the seawater scenario, the attenuation is significantly worse at 346 dB!!
If considering seawater then you must go significantly lower in frequency - maybe something like 1MHz.
