k i sin θ ¼ k r sin θ
0
¼ k t sin ϕ,
ð8:31Þ
where k r ¼ j k r j and k t ¼ j k t j; θ
0 and ϕ are said to be a reflection angle and a
refraction angle, respectively. Thus, the end points of k i , k r , and k t are connected on a
straight line that parallels the z-axis. Figure 8.3 clearly shows it.
Now, we suppose that a wavelength of the electromagnetic wave in D1 is λ 1 and
that in D2 is λ 2 . Since the incident light and reflected light are propagated in D1, we
have
k i ¼ k r ¼ 2π=λ 1 :
ð8:32Þ
From (8.31) and (8.32), we get
sin θ ¼ sin θ
0
:
ð8:33Þ
Therefore, we have either θ ¼ θ
0 or θ
0
¼ π À θ. since 0 < θ, θ
0
< π/2, we have
θ ¼ θ
0
:
ð8:34Þ
Then, returning back to (8.31), we have
k i sin θ ¼ k r sin θ ¼ k t sin ϕ:
ð8:35Þ
This implies that the components tangential to the interface of k i , k r , and k t are
the same.
Meanwhile, we have
k t ¼ 2π=λ 2 :
ð8:36Þ
Also we have
c ¼ λ 0 ν, v 1 ¼ λ 1 ν, v 2 ¼ λ 2 ν,
ð8:37Þ
where v 1 and v 2 are phase velocities of light in D1 and D2, respectively. Since ν is
common to D1 and D2, we have
c=λ 0 ¼ v 1 =λ 1 ¼ v 2 =λ 2
or
c=v 1 ¼ λ 0 =λ 1 ¼ n 1 , c=v 2 ¼ λ 0 =λ 2 ¼ n 2 ,
ð8:38Þ
302
8 Reflection and Transmission of Electromagnetic Waves in Dielectric Media