Page 441 - Engineering Electromagnetics, 8th Edition

CHAPTER 12 Plane Wave Reflection and Dispersion 423

Next, we remove the restriction η 1 = η 3 and look for a way to produce zero
reflection. Returning to Eq. (36), suppose we set β 2 l = (2m − 1)π/2, or an odd
multiple of π/2. This means that
2π π
l = (2m − 1) (m = 1, 2, 3,...)
λ 2 2
or

λ 2
l = (2m − 1) (44)
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The thickness is an odd multiple of a quarter-wavelength as measured in region 2.
Under this condition, (36) reduces to
η 2 2
η in = (45)
η 3
Typically, we choose the second region impedance to allow matching between given
impedances η 1 and η 3 .To achieve total transmission, we require that η in = η 1 ,so that
the required second region impedance becomes
√
η 2 = η 1 η 3 (46)

With the conditions given by (44) and (46) satisfied, we have performed quarter-wave
matching. The design of antireflective coatings for optical devices is based on this
principle.

EXAMPLE 12.5
We wish to coat a glass surface with an appropriate dielectric layer to provide total
transmission from air to the glass at a free-space wavelength of 570 nm. The glass
has refractive index n 3 = 1.45. Determine the required index for the coating and its
minimum thickness.
Solution. The known impedances are η 1 = 377
and η 3 = 377/1.45 = 260
.
Using (46) we have

η 2 = (377)(260) = 313
The index of region 2 will then be
377

n 2 = = 1.20
313
The wavelength in region 2 will be
570
λ 2 = = 475 nm
1.20
The minimum thickness of the dielectric layer is then
λ 2
l = = 119 nm = 0.119 µm
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