absorption coefficient µa [cm-1]
scattering coefficient µs [cm-1]
anisotropy of scatter g [dimensionless]
reduced scattering coefficient µs' = µs(1-g) [cm-1]
D = 1/3/(µa+µs')
δ = sqrt(D/µa)
T(r) = exp(-r/δ)/(4πDr) , transport [cm-2]
P [W] = power of point source of light.
t [s] = exposure time.
Q [J/cm3] = PtµaT, energy deposition [J/cm3]
ΔTemp = Q/(4.18 J/(g°C), temperature rise for aqueous medium
Problem 1: You deliver 100-mW 0.1-s pulse of power through an optical fiber that is inserted into a gel made with added ink and milk, so it is light scattering. The optical properties of the gel are µa 0.1 [cm-1], µs' = 10 [cm-1]. The fiber throws light forward such that the source for light diffusion is a distance 1/(µa + µs') in front of the fiber, and this point source is assigned the position r = 0.
An absorbing sphere, 100-µm diameter, is centered at a position 5 mm from the point source, at r = 0.5 cm. Its absorption is µa.sphere = 10 cm-1.
Question: What is the temperature rise in the surface of the absorbing object at the end of the pulse?
Ignore the temperature distribution within the sphere.
Ignore thermal diffusion.
hint: the temperature rise is very small.
Answer: A MATLAB program yields the answer dTemp in degrees.