1. Assignment #5
Absorption
Due 3/3/11

2. Problem 2
The transmission of light can be calculated from Beer’s law

3. 1a. Transmission for two depths
350 nm
0.12
500 nm
0.04
650 nm
0.33
T for 1 m
0.89
0.96
0.72
T for 10 m
0.30
0.67
1b. At 1 m, there is a little less red and UV, but most of the wavelengths are still present. However, after 10 m, the light covers a narrower spectral range as nearly all the red light and most of the UV light have been absorbed. Only the blue-green light is transmitted and so remains.

4. How do we determine attenuation coefficient
Measure light of particular wavelength and how it is transmitted to different depths
Depth (m)
I (350 nm) 1 10
0.58 20
0.35 30
0.25

5. Beer’s law T = e-l If we take the log of both sides:
ln T = -l
We can multiply each side by -1
-ln T = -ln(I/I0) = l
So a plot of -ln T versus depth will give us a line with slope 

6. = 0.05 based on slope of line

7. Alternative We could define Beer’s law by T = 10-l
In that case, we would need to fit the original data by plotting
-log T versus depth, l

8. In that case, the slope of log T versus depth gives =0.022

9. For consistency I will always define Beer’s law as T = e-l
So don’t use 10-l

10. FishBase: Fish at depth viewer

11. Problem 3. A. This is a parrot fish
B. In the original photo it has lots of bright reds and greens.
C. The reds all go away with depth. This occurs pretty quickly so they are gone by 10 m. This leaves greens and then only blues.
D. This agrees with how light is transmitted through water - the blue-green wavelengths are transmitted best and only they remain at great depths.

12. Problem 4 A. Lobsters B. Frogs
T = exp(-l)= exp( um-1 * 240 um)
= exp (-0.696) = e = 0.50
A=1-T = 0.50
B. Frogs
T = exp(-l)= exp( um-1 * 86 um)
= exp (-1.29) = e-1.29 = 0.28
A=1-T = 0.72

13. Problem 3. Length (um) Attenuation coefficient (um-1) A Lobster 240
0.0029
0.50
Frog 86
0.015
0.72
What do you notice about invert and vert photoreceptors?

14. Problem 3c. The fraction of light absorbed = 1-I/I0
This is the same as the probability that one photon is absorbed
If A=0.72, how likely is it that a photon is absorbed? very likely! I I0

15. What about humans? Cone Attenuation coefficient Length Absorption
S (blue)
0.037 um-1
40 um
0.77
M (green)
0.032 40
0.72
L (red)
0.027
0.66
I was not able to find data on the length other than they are typically um. I therefore chose to use the same length for all.
Data from Bowmaker and Dartnall 1980

16. Human visual pigments
Instead of blue, green and red cones, we sometimes call these short wavelength sensitive (SWS), medium wavelength sensitive (MWS) and long wavelength sensitive (LWS). Why is this a good idea?

17. Problem 4a. Green Green is absorbed by the MWS cone and also by the LWS cone with minor absorbance by SWS cone

18. Problem 4b. Yellow Yellow is absorbed by the MWS cone but even more so by the LWS cone. Just a tiny absorbance by SWS cone.

19. Problem 4c. Red Red is absorbed by the LWS cone and also just a bit by the MWS cone. No absorption by SWS cone.

20. 4d.
Different cones absorb different amounts of light based on their visual pigments.
It is this differential absorption which enables a comparison of the cone outputs to determine the color.