Biophotonics_Concepts_to_Applications

R?¼

E0r

E0i

 2

x

¼Rx¼r^2 x ð 2 : 31 Þ

Rjj¼

E0r

E0i

 2

y

¼Ry¼r^2 y ð 2 : 32 Þ

T?¼

n 2 cosh 2

n 1 cosh 1

E0t

E0i

 2

x

¼Tx¼

n 2 cosh 2

n 1 cosh 1

t^2 x ð 2 : 33 Þ

Tjj¼

n 2 cosh 2

n 1 cosh 1

E0t

E0i

 2

y

¼Ty¼

n 2 cosh 2

n 1 cosh 1

t^2 y ð 2 : 34 Þ

The expression for T is a bit more complex compared to R because the shape of
the incident light beam changes upon entering the second material and the speeds at
which energy is transported into and out of the interface are different.
If light is incident perpendicularly on the material interface, then substituting
Eq. (2.29) into Eqs. (2.31) and (2.32) yields the following expression for the
reflectance R

R¼R?ðh 1 ¼ 0 Þ¼Rjjðh 1 ¼ 0 Þ¼

n 1 n 2

n 1 þn 2

 2

ð 2 : 35 Þ

and substituting Eq. (2.30) into Eqs. (2.33) and (2.34) yields the following
expression for the transmittance T

T¼T?ðh 2 ¼ 0 Þ¼Tjjðh 2 ¼ 0 Þ¼

4n 1 n 2

ðÞn 1 þn 22

ð 2 : 36 Þ

Example 2.9Consider the case described in Example2.8in which light

traveling in air (nair= 1.00) is incident perpendicularly on a smooth tissue

sample that has a refractive index ntissue= 1.35. What are the reflectance and

transmittance values?

Solution: From Eq. (2.35) and Example2.8the reflectance is

R¼½ŠðÞ 1 : 35  1 : 00 =ðÞ 1 : 35 þ 1 : 00 2 ¼ðÞ 0 : 1492 ¼ 0 :022 or 2: 2 %:

From Eq. (2.36) the transmittance is

T¼ 41 ðÞ: 00 ðÞ 1 : 35 =ðÞ 1 : 00 þ 1 : 352 ¼ 0 :978 or 97: 8 %

Note that R + T = 1.00.

42 2 Basic Principles of Light