PROBLEMS
1. In a purely absorbing (non scattering) medium with absorption coefficient µa,
a. what percentage of light is left after a light beam propagates
a length of L?
b. derive the average length of survival of a photon.
2. In a purely scattering (non absorbing) medium with scattering coefficient µs,
a. what percentage of light has not been scattered after the original light
beam propagates a length of L?
b. derive the average length of survival of a photon.
3. What happens to the angles of incidence, reflection, and refraction at the critical
angle?
a. Is there a critical angle for the interface between water and air, going
from water to air?
b. Is there a critical angle for the interface between air and water, going
from air to water?
4. In a scattering medium with absorption coefficient µa and scattering coefficient
µs, what percentage of light has survived scattering and absorption after the
original light beam propagates a length of L? Of the percentage that has been
absorbed and scattered, what is the percentage that has been absorbed?
5. A thin sheet of clear tissue, such as a section of the cornea of the eye, (n = 1.33)
is inserted normally into one beam of a Michelson interferometer. Using 589
nanometers of light, the fringe pattern is found to shift by 50 fringes. Determine
the thickness of the tissue section.
6. Calculate the decay of ballistic light after penetrating a biological tissue 30 mfp
thick. If the scattering coefficient of the tissue is 100cm-1, calculate the
corresponding thickness in cm.
7. Given a red helium neon laser with a wavelength of 633 nm, determine the
velocity of the light traveling through clear tissue such as the cornea that has an
index of refraction of 1.33. How would this velocity change in glass that has an
index of refraction of 1.5? Explain the significance of this result.
8. A 5-mW collimated laser beam passes through a 4-cm non-absorbing, scattering
medium. The collimated transmission was measured through a small aperture to
be 0.035 mW. Calculate the scattering coefficient.
9. An Nd:YAG laser operating at 1064 nm with a Gaussian beam is to be used for
performing a portwine stain treatment. Assume a heuristic model for light
distribution in a homogeneous material such that I(r, z) = I0exp(-µtz)exp[2(r/ωb)2], where µt = µa + µs’. For the following multilayer case, at r = 0 give the
expressions and graph intensity I and the rate of heat generation Q in the tissue
by the laser, I(r = 0, z) and Q(r = 0, z). Assume I0 = 100. The absorption and
scattering coefficients of each skin layer and blood vessel for this wavelength are
given in Table below.
10. In the near-infrared region of the optical spectrum between 600–1100
nanometers there are well-known absorbance peaks for oxygenated and
deoxygenated hemoglobin. A big assumption can be made for the moment that
in this region the dominant optical signal is due to absorption (which is not
generally the case). Given that you propagate through a 2-cm sample of tissue,
you need to calculate three parameters: the concentration of oxyhemoglobin and
deoxyhemoglobin, and a background blood absorbance. At three wavelengths
(758 nm, 798 nm, 898 nm), you can measure the extinction coefficients for
oxygenated (1,533, 0.702, and 0.721mM-1 cm-1) and deoxygenated hemoglobin
(0.444, 0.702, and 1.123mM-1 cm-1) respectively. The total absorption coefficients
at these three wavelengths in the tissue can be measured using a time-resolved
system as (0.293 cm-1, 0.1704 cm-1, 0.1903 cm-1), respectively. Calculate the
oxygenated and deoxygenated hemoglobin levels given these parameters. Is this
reasonable from a physiologic point of view? Explain.
11. A fiberoptic probe is used to perform a diagnostic and therapeutic procedure on
a cancerous lesion in the esophagus. The probe first determines the boundary of
the suspected cancerous lesion by measuring the reflectance of the light at the
interface, where it is known that the refractive index of the cancer tissue is 1.2.
The normal esophageal tissue refractive index is 1.4. The fiber is polished at an
angle of 300 (angle of incidence). Helium-neon (HeNe) laser light is propagated
down a fiberoptic probe. The power density at the fiber tip is 1 Wm-2 at the
interface with the tissue. The optical properties of the cancerous tissue at this
wavelength are given as µa= 3 cm-1and µs=150 cm-1. The anisotropy factor g is
0.9. Assume a semi-infinite geometry. Calculate the reflectance for the two
tissues, normal and cancerous, at the interface between the sapphire fiber optic (n
= 1.7) and the tissue.
12. Given a red helium neon laser with a wavelength of 633 nm, determine the
velocity of the light traveling through clear tissue such as the cornea that has an
index of refraction of 1.33. How would this velocity change in glass that has an
index of refraction of 1.5? Explain the significance of this result.
13. Given a green light beam that hits the cornea of the eye, making an angle of 258
with the normal, as depicted in the following figure, (a) determine the output
angle from the front surface of the cornea into the cornea given the index of
refraction of air is 1.000, and that of the cornea is 1.376; and (b) what can be said
about how the cornea bends the light?
14. For a Gaussian laser beam irradiating at a wavelength of 2.1 mm the absorption
coefficient is estimated to be 25 cm-1 and scattering is negligible. The radial
profile of light intensity and the rate of heat generation at the tissue surface, z =
0, and its axial profile along the center axis of the beam, r = 0, can be found
using the previous equations. Graph I(r, 0) for r ranging from -2ω0 to 2ω0 and
I(0, z) for z = 0 to z = 5/µa, as well as a graph of QL(r, 0) and QL(0, z) are
shown on the next page. For simplicity, I0 = 1W=cm2.
Energy absorbed :
Total intensity :
QL = µa I
I(z) = I0 exp(−µa z)
Beam profile :
f (r) = exp(−
2r 2
)
ω 02
Plot of the radial and axial profile of light intensity, I, and volumetric rate of
absorption, QL in a purely absorbing tissue.
QL (r,z) = µa I(z) / f (r)
15. It is desired to measure the concentration of an absorber in a scattering medium
with known scattering coefficient. If the reduced scattering coefficient µs’ is
known and the relative intensity at a distance r0 from an isotropic source can be
measured, an algebraic equation may be solved for the absorption coefficient
based on the diffusion approximation given that r0 is large enough for diffusion
approximation to be valid.
16. A 5-mW collimated laser beam passes through a 4-cm non absorbing, scattering
medium. The collimated transmission was measured through a small aperture to
be 0.035 mW. Calculate the scattering coefficient.
17. A thin sheet of clear tissue, such as a section of the cornea of the eye, (n = 1.33)
is inserted normally into one beam of a Michelson interferometer. Using 589
nanometers of light, the fringe pattern is found to shift by 50 fringes. Determine
the thickness of the tissue section.
18. Calculate the decay of ballistic light after penetrating a biological tissue 30 mfp
thick. If the scattering coefficient of the tissue is 100 cm-1, calculate the
corresponding thickness in cm
19. Using diffusion theory the light penetration in biological tissue can be
approximated to the first order of accuracy. A 50 mW HeNe laser is launched
into the tissue through a fiberoptic light guide with a diffusing tip at the end,
inserted into the tissue at 10 cm depth. Assume that the absorption coefficient is
0.1 cm-1, the scattering coefficient of the tissue is 100 cm-1, and the scattering
anisotropy factor 0.9 at a wavelength of 632.8 nm. Calculate the light fluence rate
5 cm from the source.
20. A biological tissue is given with an absorption coefficient µa = 0.1 cm-1, since the
scattering coefficient and scattering anisotropy factor are not known, these two
are combined as reduced scattering coefficient. Using the diffusion
approximation, draw in a single graph the logarithm of relative radiance versus
radius for four different values of reduced scattering coefficient µs’ = 10, 50, and
100 cm-1, for r ranging from 0.1 to 2 cm.
21. In a tissue where absorption is less dominant and scattering is large the impact of
a collimated monochromatic light source give rise to a diffusion process. Instead
of perpendicular impact the light source is tilted 45o known as the oblige angle
technique, figure 1, left panel. The resulting reflectance curve along the impact
plane is shown in figure 1, right panel. C is the centre line of intensity curve.
Explain theoretically how absorption, scattering coefficient and diffusion can be
obtained in this setup. From the figures calculate the values. n=1.33
Figure 1: Left panel: Theoretical description of the oblique angle (αi) technique. Right panel: resulting
reflectance curve in the impact plane.
PROBLEMS
1. In a purely absorbing (non scattering) medium with absorption coefficient µa,
a. The probability for survival is: T ( x ) = I ( x ) , at distance L, e−µ L ×100%
a
I0
2. In a purely scattering (non absorbing) medium with scattering coefficient µs,
a. The probability for survival is: T ( x ) = I ( x ) , at distance L, = e−µ L ×100%
s
I0
3. What happens to the angles of incidence, reflection, and refraction at the critical
angle?
a. Yes. Incidence and reflection angle do not change. Snells law.
b. No. Incidence and reflection angle do not change. Snells law.
4. The probability for survival is: T ( x ) = I ( x ) , at distance L, e−(µ +µ )L ×100%
a
s
I0
5. 2dcos(θ) = mλ, and thus d = (50*.589)/2 = 14.72 µm, which is the calculated
optical path length. However, the physical length of the tissue must take into
account the index of refraction of the sample and the air. Thus, the equation for
physical path length is L = d/(ns-nair) = 14.72/(1.33-1.0) = 44.6 micrometers.
6. Based on Beer’s law, the decay is exp(-30) = 9.4x10-14. The thickness is 30/100 =
0.3 cm.
7. Rearranging the preceding equations,
ctissue = c0/ntissue =2.998E8/1.33 = 2.25E8 m/s for clear tissue while
cglass = c0/nglass = 2.998E8/1.5 = 2.00E8 m/s for glass
The significance of this result is that light travels faster through a material with a
lower index of refraction such as clear tissue compared to glass. This has many
implications when it comes to determining the angle of reflection and refraction
of light through differing materials and clear tissues (e.g., cornea,aqueous humor,
and lens of the eye).
8. Based on Lambert Beer’s law, µs = ln(5/0.035)/4 = 1.2 cm-1
9. Epidermis :
'
I epi (z) = I 0 e−µ z = I 0 e−(µa + µs )z = I 0 e−10,1*10
−4
z
Dermis :
'
I der (z) = I 0d e−µ (z−65) = I 0d e−(µa + µs )(z−65) = I 0 e−10,1*10
Blood :
−4
65 −10,5(z−65)*10−4
e
'
−4
I blood (z) = I 0d e−µ (z−365) = I 0d e−(µa + µs )(z−365) = I 0 e−10,5(365)*10 e−40(z−365)*10
Heating :
Qepi = µ a−epi I epi
Qder = µ a−der I der
−4
Qblood = µ a−blood I blood
10. Absorbance A = µ z = z∑ ε i Ci
i
and define Cox = Concentration of Oxyhemoglobin, Cdox = Concentration of
Deoxyhemoglobin, and Cother as the concentration of the background. These are
our three unknowns. We can create three equations using the equation above:
one equation for each wavelength. However, we do not know the ε of the
background for each wavelength. In principle, for each wavelength, εother will be
different. In that case, we would have an additional 3 unknowns and the system
of equations would be underdetermined. In order to solve the equations we need
to ASSUME that εother is the same for all wavelengths. With this assumption, we
can
combine
Cother
εother
=
µother.
The background blood attenuation coefficent. To calculate the background blood
absorbance, we calculate Aother = µother L, where L is the optical path length.
758nm : 0.2930 = 1.533Cox + 0.444Cdox + µother
798nm : 0.1704 = 0.702Cox + 0.702Cdox + µother
898nm : 0.1903 = 0.721Cox + 1.123Cdox + µother
C = ε −1µ
Cox = 0.04 mM
0.04
C = 0.16
0.03
Cdex = 0.16 mM
µother = 0.03 cm −1
Sox =
Cox
0.04
=
= 20%
Cox + Cdox
0.2
11. The reflectance
⎛ n cosθ1 − n2 cosθ 2 ⎞
RS = ⎜ 1
⎝ n1 cosθ1 + n2 cosθ 2 ⎟⎠
Snells law
2
⎛ n cosθ 2 − n2 cosθ1 ⎞
RP = ⎜ 1
⎝ n1 cosθ 2 + n2 cosθ1 ⎟⎠
⎛n ⎞
sin θ1 n2
= ⇒ θ 2 = arcsin ⎜ 1 ⎟ sin θ1
sin θ 2 n1
⎝ n2 ⎠
Rnormal = 0,0196
Rcancerous = 0,0729
2
12. Rearranging the preceding equations,
ctissue = c0/ntissue =2.998E8/1.33 = 2.25E8 m/s for clear tissue while
cglass = c0/nglass = 2.998E8/1.5 = 2.00E8 m/s for glass
The significance of this result is that light travels faster through a material with a
lower index of refraction such as clear tissue compared to glass. This has many
implications when it comes to determining the angle of reflection and refraction
of light through differing materials and clear tissues (e.g., cornea,aqueous humor,
and lens of the eye).
13. (a) Snell’s law, n1sin (q1) =n2sin (q2), can be rearranged so as to calculate q2 in
the ens. θ2 = sin-1 (1.000 sin (25)/1.376) = 17.89 degrees
(b) It has been shown and can be said that the cornea tends to bend the light
toward the normal as it passes through. This makes sense because the eye is
made to bend the light so that it can pass through the center iris and lens toward
the retina to be imaged by the brain.
14.
Energy absorbed :
QL = µa I
Total intensity :
I(z) = I0 exp(−µa z)
Beam profile :
2r 2
f (r) = exp(− 2 )
ω0
Plot of the radial and axial profile of light intensity, I, and volumetric rate of
absorption QL in a purely absorbing tissue.
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15. Given scattering coefficient µs’, r0 and F(r0)/F0, an algebraic equation for ma
based on the diffusion approximation is required.
φ (r) = φ 0 exp(−r /δ ) /(4 πDr)
log[φ (r) / φ 0 ] = −r /δ − [log(4) + log(π ) + log(D) + log(r)]
D = 1/3µt '
δ = D / µa
The lefthand side of the above equation is a known value, e.g. k1
k1 = −r /δ − [log(4) + log(π ) + log(D) + log(r)]
k1 + log(4) + log(π ) + log(r) = −r /δ − log(D)
The lefthand side of the above is again a known value, e.g. k 2
k 2 = −r / D / µa − log(D)
k 2 = −r µa /D − log(D)
k 2 = −r µa /1/3µt ' − log(1/3µt ' )
k 2 = −r 3µa µt ' − log(1/3µt ' )
k 2 = −r 3µa (µa + µs' ) − log(1/3(µa + µs' ))
Solving this equation will give the absorption and scattering properties.
16. Based on Lambert Beer’s law, ms = ln(5/0.035)/4 = 1.2 cm-1
17. 2dcos(θ) = mL, and thus d = (50*.589)/2 = 14.72 mm, which is the calculated
optical path length. However, the physical length of the tissue must take into
account the index of refraction of the sample and the air. Thus, the equation for
physical path length is L = d/(ns - nair) = 14.72/(1.33 – 1.0) = 44.6 micrometers.
18. Based on Beer’s law, the decay is exp (- 30) = 9.4x10-14. The thickness is 30/100
= 0.3 cm.
19. Solution
φ0 e−r /δ
φ (r ) =
⋅
4π D r
D
δ=
= effective penetration depth
µa
D=
1
3 (µ a + (1 − g ) µ s )
µ a = 0.1 cm −1
µ s' = (1 − 0.9 )100 = 10 cm −1
D=
δ=
1
1
=
3 ( 0.1+10 ) 30.3
1
1
=
0.1⋅ 30.3
3.03
φ (5) =
50 ⋅ 30.3 e−5 3.03
⋅
= 0, 004 mW/cm 2
4π
5
20. Solution
21. The distance of the buried source to the point of incidence is most accurately
determined by ds
ds = 3D =
sin (α t )
1
and
Δx
=
0.35 µa + µ s'
0.35 µa + µ s'
µa
D
1
µeff = − = slope in the log scale of right fgure
δ
Δx
D=
3sin (α t )
µeff =
2
µeff
Δx
µa =
3sin (α t )
µ s' =
sin (α t )
− 0.35 µa
Δx