Though a little portion of EM radiation reflects back after absorption, it is not strong enough to make it all the way back to atmosphere, also the refractive index of still water might play some role in changes of return signal.

If the water is turbid, you will get multiple reflections!

Water bodies cannot be penetrated by microwaves, a medium is needed (soil, vegetation, the dryer the better).

• In case of calm and open water bodies, the incoming signal gets mirrored at the surface. Due to the side looking geometry of the sensor, a large proportion of the waves are just heading on in look direction instead of returning to the sensor. This leads to a very small proportion of backscatter for these surfaces (low values, black pixels).
• As Sandeep explained, if your water surface is rough, you may have parts with higher backscatter due to the incoming angle.
• If the water is vegetated by reed, for example, the opposite is the case: Instead of surface reflection you have double bounce effects and a comparably high proportion of backscatter (high values, bright pixels).

I recommend the paper of Stian Solbø and Inger Solheim:

Article Towards Operational Flood Mapping with Satellite SAR

Thanks for the good explanations provided.

Since light is an EM wave, consider what happens with light. If smooth, it is highly reflective and light specularly scatters in forward direction. If rougher, it partially penetrates but bends rays towards vertical de to much higher index of refraction. If highly conductive (i.e., sea water), penetration is minimal and wave propagates along surface.

Essentially, yes, as you say, water has a high reflectivity. If the surface is calm then there is a strong specular reflection (angle of incidence = angle of reflection) and so when observed away from normal incidence, radar reflections are largely concentrated in a narrow region which is away from the angle of illumination, hence the backscatter towards the radar is low and the water surface appears dark in a SAR image. As the surface becomes rougher (wind causing ripples and waves, and eventually breaking waves), the water surface has many facets in many directions and so the reflection becomes more diffuse. The backscatter back towards the radar tends to increase with increasing sea-state (roughness). Backscatter is also dependent on the look angle with respect to wave (and hence wind) orientation and polarisation, although polarisation becomes less of an issue with increasing sea-state.

OK, now for the aborption issue. A pure dielctric is a material having zero conductivity. An incident wave will either be reflected from its surface or transmitted into the dielectric but none would be absorbed. Hence the transmitted portion would pass staright through the dielectric. Most practical dielectrics, however, are not pure; they have some conductivity which causes absorption of the signal transmitted into the dielectric. The further the signal penetrates the more it is absorbed and the signal experiences a loss in field strength which has an exponential decay with distance. We may therefore call this a lossy dielectric. The greater the conductivity of the dielectric, the greater the loss and so the more rapid is the exponential decay of the signal attempting to pass through it. Water, and sea-water in particular, are very lossy dielctrics in the RF/microwave band. The exponential decay is very rapid. Although, in theory, an exponential decay never reaches zero, in practice the decay is so rapid that all the signal transmitted into water is absorbed over a short distance. At low RF, sea-water absorption is almost 100% after a few metres but at GHz frequencies a few cm or even mm will provide "total" absorption. So it is not strictly true to say that RF cannot penetrate water, but for most practical applications the penetration is so low that we simply say that microwave signals cannot go through water. It is a similar story for materials containing water such as many natural materials (plants, wood, soil, skin and other tissues). The rate of absorption (or exponential decay) depends on the water content and is typically high. The energy absorbed by the water is disipated as heat and is the source of RF burn hazards and the principle behind microwave cooking.

In the case of water droplets in the atmosphere, there is reflection from the water droplets forming the cloud and there is a loss in the signal that penetrates the cloud due to absorption in the water itself, but there is also scattering of the signal from the droplets. This scattering deflects the signal away from its intended path and so constitutes a two-way loss (but not by absorption) - less of the transmitted signal reaches the target and less of the reflection reaches the radar receiver. The scattering through 180 degrees constitues the reflected signal that may be detected by a radar. The degree of scattering depends on the ratio of the circumference of the water droplets with respect to wavelength and reaches a peak when this ratio =1 (circumference = wavelength) as happens with rain at mm-wave frequencies. Both absorption and scatter also depend on the density of the water droplets in the atmosphere. Cloud has droplets of

Thank you Clive for your detailed explanation

Vincent Ferrer Thank you for the question

In the case of open water bodies, the incoming signal gets mirrored on the surface. Due to the side-looking geometry of the sensor, a large proportion of the waves are just heading on in look direction instead of returning to the sensor. This leads to a very small proportion of backscatter for these surfaces (black pixels have low values).

As Sandeep explained, if your water surface is rough, you may have parts with higher backscatter due to the incoming angle.

If the water is vegetated by reed, for example, the opposite is the case: Instead of surface reflection, you have double bounce effects and a comparably high proportion of backscatter (bright pixels have high values).