rjj¼ry¼
E0r
E0i
y
¼
n 2 cosh 1 n 1 cosh 2
n 1 cosh 2 þn 2 cosh 1
ð 2 : 26 Þ
t?¼tx¼
E0t
E0i
x
¼
2n 1 cosh 1
n 1 cosh 1 þn 2 cosh 2
ð 2 : 27 Þ
tjj¼ty¼
E0t
E0i
y
¼
2n 1 cosh 1
n 1 cosh 2 þn 2 cosh 1
ð 2 : 28 Þ
If light is incident perpendicularly on the material interface, then the angles are
h 1 ¼h 2 ¼0. From Eqs. (2.25) and (2.26) it follows that the reflection coefficients
are
rxðh 1 ¼ 0 Þ¼ryðh 2 ¼ 0 Þ¼
n 1 n 2
n 1 þn 2
ð 2 : 29 Þ
Similarly, forθ 1 =θ 2 = 0, the transmission coefficients are
txðh 1 ¼ 0 Þ¼tyðh 2 ¼ 0 Þ¼
2n 1
n 1 þn 2
ð 2 : 30 Þ
Example 2.8Consider the case when 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 reflection and transmission coefficients?
Solution: From Eq. (2.29) with n 1 =nairand n 2 =ntissueit follows that the
reflection coefficient is
ry¼rx¼ðÞ 1 : 35 1 : 00 =ðÞ¼ 1 : 35 þ 1 : 00 0 : 149
and from Eq. (2.30) the transmission coefficient is
tx¼ty¼ 21 ðÞ: 00 =ðÞ¼ 1 : 35 þ 1 : 00 0 : 851
The change in sign of the reflection coefficient rxmeans that thefield of the
perpendicular component shifts by 180° upon reflection.
Thefield amplitude ratios can be used to calculate thereflectanceR (the ratio of
the reflected to the incidentflux or power) and thetransmittanceT (the ratio of the
transmitted to the incidentflux or power). For linearly polarized light in which the
vibrational plane of the incident light is perpendicular to the interface plane, the
total reflectance and transmittance are
2.4 Reflection and Refraction 41