Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.3.4

Quarter-wave transformer

A transparent dielectric coating is applied to glass ( $ε_{r} = 4$ , $μ_{r} = 1$ ) to eliminate the reflection of red light $λ_{0} = 0.75$ μm for normally incident light.

Determine the required relative permittivity and thickness of the coating.
If violet light $λ_{0} = 0.42$ μm is shone normally on the coated glass, what percentage of the incident power is reflected?

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.8-30.

Try it yourself first.

Strategy

Air, coating and glass are the three media of the three-part problem, with $η = η_{0} / n$ . Dividing the numerator and the denominator of $Γ_{d}$ from Eq. (6.3.39) by $η_{2} η_{3}$ gives a form that shows the quarter-wave limit at a glance,

Γ_{d} = \frac{( 1 - \frac{η _{1}}{η _{3}} ) + j ( \frac{η _{2}}{η _{3}} - \frac{η _{1}}{η _{2}} ) tan β _{2} d}{( 1 + \frac{η _{1}}{η _{3}} ) + j ( \frac{η _{2}}{η _{3}} + \frac{η _{1}}{η _{2}} ) tan β _{2} d} .

Design at $0.75$ μm with a quarter-wave layer, then keep the thickness and evaluate the same formula at $0.42$ μm.

For $d = λ_{2} /4$ , $β_{2} d = π /2$ and $tan β_{2} d \to \infty$ :

Γ_{d} = \frac{η _{2} / η _{3} - η _{1} / η _{2}}{η _{2} / η _{3} + η _{1} / η _{2}} = 0 ⟺ η_{2}^{2} = η_{1} η_{3} ⟺ n_{2} = n_{1} n_{3} = 2,

so $ε_{r 2} = n_{2}^{2} = ε_{r 1} ε_{r 3} = 2$ . With $f = c / λ_{0} = 4 \times 1 0^{14}$ Hz,

λ_{2} d = \frac{c}{ε _{r 2} f} = \frac{0.75 μ m}{2} = 530 nm, = \frac{λ _{2}}{4} = 133 nm .

The layer stays $133$ nm thick, so at $λ_{0} = 0.42$ μm

β_{2} d tan β_{2} d = \frac{π}{2} \times \frac{0.75}{0.42} = 2.805 rad (160. 7^{\circ}), = - 0.350.

In units of $η_{0}$ , $η_{2} = 1/ 2$ and $η_{3} = 1/2$ :

\frac{η _{in}}{η _{0}} = \frac{1}{2} \frac{0.5 - j 0.350/ 2}{1/ 2 - j 0.5 \times 0.350} = 0.529 - j 0.117,

Γ_{d} ∣ Γ_{d} ∣^{2} = \frac{η _{in} - η _{0}}{η _{in} + η _{0}} = \frac{- 0.471 - j 0.117}{1.529 - j 0.117} = - 0.301 - j 0.099 = 0.317 e^{- j 161. 7^{\circ}}, = 10.0 %.

Without the coating the glass reflects $(\frac{2 - 1}{2 + 1})^{2} = 11.1 %$ .

Answer

ε_{r 2} = 2, d = 133 nm, violet: 10.0 % reflected (bare glass 11.1 %)

Check

Two echoes: the air-coating and coating-glass faces reflect equally, $Γ_{12} = \frac{1/ 2 - 1}{1/ 2 + 1} = - 0.172$ and $Γ_{23} = \frac{1/2 - 1/ 2}{1/2 + 1/ 2} = - 0.172$ . At $750$ nm the round trip $2 β_{2} d = 18 0^{\circ}$ makes them cancel. At $420$ nm it is $32 1^{\circ}$ , so they nearly add, and $∣ Γ_{d} ∣ = 0.317$ is close to $2 \times 0.172$ , as found (Figure 6.3.12b).

(a) Reflected power against the free-space wavelength.

(b) The two echoes as phasors. — (a) Reflected power against the free-space wavelength.

Remark

One quarter-wave layer is a narrow-band fix: it works perfectly at the design wavelength and at its odd fractions ( $250$ nm, …), while at the violet end of the visible band ( $380$ nm) the coated glass already reflects $11.1 %$ , as much as bare glass. Broadband optics use two or more layers. The ideal index $2 = 1.41$ for this glass is also hard to find in a durable material, which is why MgF $_{2}$ ( $n = 1.38$ ) is the classic choice.

Problem 6.3.4: Quarter-wave transformer