absorption coefficient µ_{a} [cm^{-1}]

scattering coefficient µ_{s} [cm^{-1}]

anisotropy of scatter g [dimensionless]

reduced scattering coefficient µ_{s}' = µ_{s}(1-g) [cm^{-1}]

D = 1/3/(µ_{a}+µ_{s}')

δ = sqrt(D/µ_{a})

T(r) = exp(-r/δ)/(4πDr) , transport [cm^{-2}]

P [W] = power of point source of light.

t [s] = exposure time.

Q [J/cm^{3}] = Ptµ_{a}T, energy deposition [J/cm^{3}]

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 µ_{a} 0.1 [cm^{-1}], µ_{s}' = 10 [cm^{-1}]. The fiber throws light forward such that the source for light diffusion is a distance 1/(µ_{a} + µ_{s}') 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 µ_{a.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.