Light penetrating the water surface is absorbed and scattered by the water molecules, plus suspended and dissolved particles. Even very clear water attenuates light at a significant rate. Even in very clear water less than one tenth of the light that penetrates the water surface still remains at a depth of 30m (Dustan, 1982), and in less clear water, 90% attenuation may occur at shallower depths, such as 15m on an inshore reef (Fabricius and Alderslade, 2001).the location (both latitude and depth) of an organism will determine the maximum intensity the organism can receive; however, at any particular time of day or from day to day, month to month, both the intensity and spectrum of light an organism receives varies greatly.

Ref: Under water lighting condtions by Andrew Trevor-Jones

www.reefkeeping.com/issues/2002-09/atj/feature/

Hello,

Maddali has given you a complete and clear answer, I consider. In coastal waters with a high load of suspended matter, attenuation may reach about 90% at 1 m depth only. Nonetheless, it also depends on the year period because of the azimuth angle with which the rays impinge on the water surface, which in turn makes every wavelength undergo a different reflexion.

Hope this may serve. Regards

As you can see there is a great variation in the attenuation rate, and I have lerned that it depends for the water condition (obviuosly). By the way, there is in the NOAA a photographic library with images from several parts of the ocean specifically dedicated for sea color observation from the space.  See for example, http://seawifs.gsfc.nasa.gov, http://oceanexplorer.noaa.gov/explorations/04deepscope/background/deeplight/deeplight.html

Saludos

Hello all,

Thanks a lot for your valuable inputs and this definitely would help a lot. However, since I am working in the Bay of Bengal with high Suspended load and DOC, this is expected to have a very high attenuation of light in the surface 10-15m itself.  Dr. Maddali has given a very good link where they describe the relation in details including the Fig. 12. Nevertheless, hanks a lot again to all of you for sharing your views,

Cheers,

Haimanti

Hi Haimanti,

In the Bay of Bengal you will probably lose most light by attenuation and scattering in the first meters, but I guess your system has a large temporal and spatial variability. You can get some indirect insight from the biological communities on the seabed and in the water column. They need light to grow so the can be indirect indicators of light dynamics water columnas they are actually as temporal and spatial integrators of the light conditions. Maybe a high-resolution chlorophyll satellite picture or the extent of sea-grasses on the sea-bed of the bay of Bengal will be highly informative for you

Best

Hi Miquel,

Thanks for your answer. Yes, you are absolutely right. I am working on a large set of HPLC pigment data from a monsoon coastal cruise in the Bay of Bengal. My aim is to understand the variation in light harvesting carotenoids to photoprotective carotenoids ratios in different depths and nutrient variability. Additionally, we  conducted some experiments under different light conditions with natural phytoplankton communities and this is highly interesting that these guys are having high potential to enhance their light harvesting pigment concentrations when light availability is low. It has already been published that in this bay productivity is low due to light. however, nutrient could also limit the same. So I am trying to understand from both field and experimental data, up to which level they can maximize their light capturing capacity.

By the way, we do not have many sea grass bed here due to high suspended load and less clarity of the water column. they are only restricted to the southern side close to the coral reefs,

Thanks and regards,

Haimanti

Roughly, Attenuation coefficient, k= 1.7/Zsd it will come in m-1,

SD= Secchi disk depth

The depth of the euphotic zone (depth of light) was calculated as Zp = 4.61/k

Is this correct?

Richard H. French, James J. Cooper, and Steven Vi&, 1982, SECCHI DISC RELATIONSHIPS

Hi Dip,

Thanks for your input, yes,  you are correct and this was included in my thesis long time back, this is valid when you have a secchi disk measurement and can calculate the attenuation from that. However, I do not have a secchi disk measurements here for this data set and hence cannot calculate this. Nevertheless, thanks for your suggestion.

Cheers

Haimanti

It is a pity that you lack SD from this sampling because as you are aware, this is exactly what you need.

There still a slight possible solution: did you take pictures during the cruise of the instruments going down on the water?

I you have some way to know the depth of the instruments (or any object) on the water at the time the picture was taken you can use them as a surrogates of Secchi Disks (for instance if you where using a rosetta of Noskin bottles with a CTD sending real-time data to the boat). From the loss of brightness on the picture of the object and the depth you can calculate actually a light attenuation coefficient. I know that this is equivalent to killing ants with a gun, but when you lack some critical data sometimes desperate measures are worth to be taken!

Best and good luck!

Hi Miquel,

Thanks for your idea. however, we had a PAR sensor and unfortunately it did not work and that's why we miss this data. Nevertheless, I am not sure if this would be possible to configure the way you told. We have all other parameters and it is not too bad without the PAR.

Cheers,

Haimanti

Dear Haimanti,

among the response your received, none mentions some welle-known laws of solar light penetration. The mostly used in numerical models is the Beer Lamber law which can take the following form (Paulson and Simpson 1977 for instance), : 

I = I0[Rexp(z/D1)+(1-R)exp(z/D2)]

where I0 is the incident less reflected irrandiance at the surface, R a partitionning constant between IR and visible radiation, z the depth you consider and D1 and D2 extention length for the IR and visible lights respectively. As you use water with high suspended matter load, you can use the following parameters (type III waters from the Jerlov 1969 classification):

R=0.78

D1=1.4 m

D2=7.9 m

Then you can easily compute the fraction of light at different depths.

Hope this helps you!

Regards,

G.C.

i,

Hello Dr. Caniaux,

Thank you very much for your highly valuable input. The fact is I was looking for such an empirical formula to compute the light penetration with depth. This is exactly what you provided. Moreover, I also found out the reference by Simpson and Dickey, 1981 where they have also explained the same constants for type III waters by Jerlov as you have given  here. This indeed a great help. However, I was wondering do you have any such empirical equation where I can feed suspended sediment and DOC data and can also compute this constants given here? Please let me know.

Thanks a lot again,

Regards,

Haimanti

Dear Haimanti,

As far as I know, the evaluation of the coefficients R, D1 and D2 where obtained by Jerlov with Secchi disks (an empirical method but that works quite well). You can also find in the litterature only one exponential (by neglecting the IR radiation) or else parameterization function of chlorophyle (see for instance: Ohlmann 2003, J. Climate, 16, 1337; or Sweeney et al., 2005, JPO, 35, 1103). But for me the best performance can be obtained with the old results of Jerlov, Paulson and Simpson, or Simson and Dickey. Of course if you have PAR data this is the best to evaluate new coefficients for very specific cases.

All the best

G.C.

Hello Dr. Caniaux,

Thanks a lot for all these suggestions and I will try to do this later with a set of known PAR values. However, now when I tried to calculate the values of with the given formula at the depth of 10m, with a surface incident light of 1100 umol/m2/sec, the calculated value of I is coming almost 10^6 scale. I am not sure if I am making a mistake or there is any unit for "Io" for the given formula? Kindly suggest?

I = 1100 [0.78exp(10/1.4)+(1-0.78)exp(10/7.9)] =10.85 x10^5

Thanking you again,

Best regards,

Haimanti

Dear Haimanti,

In the formula, I0 is the incoming quantity, so your figure is right. But attention, in the exponential this is the depth that is required... with a negative sign!!!

so I=1100[0.78exp(-10/1.4)+(1-0.78)exp(-10/7.9)]=68.92

Regards,

G.C.

Dear G. Caniaux,

Thanks a lot for your input. Now I got it........

Best regards,

Haimanti