How Do You Calculate Light Intensity After Passing Through Water?

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SUMMARY

The calculation of light intensity after passing through water involves using the equation I=Ioe-A, where I represents the intensity after passing through the medium, Io is the initial intensity, and A is the absorbance. In this case, the water attenuation coefficient is 0.23222, and the depth is 3.3 cm. The user incorrectly calculated the intensity as I=10000*9.9, resulting in an unrealistic increase in intensity. The correct approach should ensure that the result for 10^(-A) remains less than 1 when A is greater than 0.

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  • Understanding of light intensity and lux measurement
  • Familiarity with the concept of absorbance in optics
  • Knowledge of the water attenuation coefficient
  • Basic algebra for manipulating equations
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I need to calculate the intensity of a 10000 lux light source traveling through 3.3 cm of water.

I have attempted to solve this using the equation I=Ioe-A where i is the intensity, Io is the initial intensity, A is the absorbance calculated as the water attenuation coefficient times the depth, using the water attenuation coefficient of 0.23222 (this could be wrong?)

This equation can be simplified into I=Io*10^-A
however when i plug my numbers into this i seem to get I=10000*9.9 which means that the intensity has increased tenfold in 3cm of water.
Somthing tells me I am wrong...

Any help would be greatly appreciated

P.S. thanks to ehild for last response
 
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How could you get a number greater than 1 as the result for 10^(-A) (A>0)?

ehild
 

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