Question

A lake has a light attenuation constant of -0.0798/m (we want to figure out the depth at which 60% of the light is attenuated (where 40% of the initial intensity still remains).
a) What is this depth in meters?
b) How does this depth compare to what you would expect to find in the Chesapeake Bay?
c) Would the attenuation coefficient absolute value (remember absolute value ignores positive or negative) go up or down if it were measured in the Chesapeake Bay?

Answer 1

a]

The attenuation would follow Beer's law which is:

here, I/I_o = 0.4

so,

taking ln on both sides give

ln(0.4) = - 0.0798d

=> d = - (1/0.0798) ln (0.4) = 11.48 m.

this is the depth when the intensity drops to 40%.

b] the depth of the Chesapeake Bay is about 53 m.

so, the depth of the Chesapeake Bay is about 4.6 times the depth calculated above.

c] Since the depth of Chesapeake Bay is more than the calculated value above, the light must attenuate less to reach a farther distance there.

This will happen when the absolute value of the attenuation coefficient decreases. So, the absolute value of attenuation coefficient will be lesser in Chesapeake Bay.