Biophotonics_Concepts_to_Applications

monotonically and reaches a value of unity at 90°. This condition of unity reflec-
tance means that the wave is traveling parallel to the material interface so that there
is no reflected wave anymore.

Example 6.2Consider an opticalfiber that illuminates a tissue area as shown

in Fig.6.4. Let thefiber axis make a 60° angle with respect to the normal to the

tissue surface. Suppose thefiber NA is such that the light emerging from the

fiber illuminates an oval area with the near and far edges of the beam making

angles of 50° and 70°, respectively. What are the perpendicular reflectances for

points at each of these angles? Let nair= 1.000 and ntissue= 1.400.

Solution: Consider the perpendicular reflectance coefficient expression given

in Prob. 2.10. In this problem n 21 =n 2 /n 1 =ntissue/nair= 1.400.

(a) Forθ 1 = 50° and with n 212 = 1.960 the perpendicular reflectance is

R?¼ðÞr?^2 ¼

cosh 1  n^221 sin^2 h 1

 1 = 2

cosh 1 þ n^221 sin^2 h 1

 1 = 2

"# 2

¼

cos 50 1 : 960 sin^250 

 1 = 2

cos 50þ 1 : 960 sin^250 

 1 = 2

"# 2

¼ 0 : 085 ¼ 8 : 5 %

(b) Similarly, forθ 1 = 70° the perpendicular reflectance is 25.4 %.

(^0102030405060708090)
Incident angle from normal (degrees)
Reflectance
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Parallel polarized
Circularly polarized or unpolarized
Perpendicularly polarized
Brewster angle (53°)
Fig. 6.3 Perpendicular, average (circular or unpolarized), and parallel reflectances for an air-water
interface
152 6 Light-Tissue Interactions