Biomedical Engineering Reference
In-Depth Information
X
Ex
Z
Unit area
Hy
Y
FIGURE 17.1
A sinusoidal electromagnetic wave.
2
E
is the partial second derivative of the electric field with respect to position (
2
E
/dx
2
where
r
@
þ
2
E
/dy
2
2
E
/dz
2
),
@
þ @
e
is the electric permittivity of the medium, and
m
is the magnetic permeabil-
ity of themedium. In free space
e ¼ e
0
. Fromsymmetry, there is a similar solution for the
magnetic field (
H
). Since the wave velocity (c
0
) is equal to the known value of the velocity of light,
it can be expressed as follows:
m¼m
0
and
1
=
2
c
0
¼
1
=
(
e
m
0
Þ
ð
17
:
2
Þ
0
Therefore, the change in the electric or magnetic field in space is related to the velocity and
the change of the field with respect to time. Given Cartesian axes O
x
,O
y
, and O
z
, shown in
Figure 17.1, a typical solution to this wave equation is the sinusoidal solution given by
E
x
¼
E
o
exp
½
j(
o
t
kz
Þ
ð
17
:
3
Þ
H
y
¼
H
o
exp
½
j(
o
t
kz
Þ
ð
17
:
4
Þ
which states that the electric field oscillates sinusoidally in the X-Z plane, the magnetic field
oscillates in the Y-Z plane (orthogonal to e-field and in phase), and the wave propagates in
the O
z
direction.
The frequency, f, and wavelength,
l
, of the wave are related to the velocity, c, and given by
f
¼ o=
2
p
ð
17
:
5
Þ
l ¼
2
p=
k
ð
17
:
6
Þ
c
¼
f
l
¼ o=
k
ð
17
:
7
Þ
The ratio of the velocity of the wave in free space, c
o
, to that in the medium through which
the light propagates, c, is known as the index of refraction of the medium, n
¼
c
o
/c.
EXAMPLE PROBLEM 17.1
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.
Continued
