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

About the further exercises of Section 6.3

Ten further exercises at normal incidence, in the spirit of the problems, the home exercises and the worked examples of this section, each group of five graded from easy to harder. They use only the toolbox at the head of this subsection.

Exercise 6.3.8

Watching a film grow

A film of titanium dioxide ( $n_{f} = 2.3$ ) is evaporated onto a thick glass substrate ( $n_{s} = 1.52$ ). A monitor measures the normal-incidence reflectance at $λ_{0} = 600$ nm while the film grows. Both materials are lossless. (a) What is the reflectance before the deposition starts? (b) Show that the reflectance first rises, and find its first maximum and the film thickness at which it occurs. (c) Find the reflectance when the film is $32.6$ nm thick. (d) The operator stops at the third turning point of the reflectance. How thick is the film? (e) How would the record differ if silica ( $n = 1.46$ ) were evaporated instead?

Try it yourself first.

Strategy

Air, film and glass are the three media of the three-part problem with $η = η_{0} / n$ . The electrical thickness of the film is $β_{f} t = 2 π n_{f} t / λ_{0}$ . The reflectance has turning points whenever the film is a whole number of quarter waves: odd quarters transform the substrate to $n_{f}^{2} / n_{s}$ , even quarters make the film disappear.

$R_{0} = (\frac{1.52 - 1}{1.52 + 1})^{2} = 4.26 %$ .

At $t = λ_{0} / (4 n_{f}) = 600/9.2 = 65.2$ nm the film is a quarter wave and the surface looks like a medium of index

n_{in} Γ = \frac{n _{f}^{2}}{n _{s}} = \frac{5.29}{1.52} = 3.48, = \frac{1 - 3.48}{1 + 3.48} = - 0.554, R = 30.6 %.

Since $n_{in} > n_{s}$ , the reflectance rises from $4.26 %$ to this maximum and falls back to $4.26 %$ at the half-wave thickness $130.4$ nm.

$32.6$ nm is an eighth wave: $β_{f} t = π /4$ , $tan β_{f} t = 1$ . In units of $η_{0}$ , $η_{f} = 1/2.3$ and $η_{s} = 1/1.52$ :

\frac{η _{in}}{η _{0}} Γ = η_{f} \frac{η _{s} + j η _{f}}{η _{f} + j η _{s}} = 0.400 - j 0.170, = - 0.408 - j 0.171, R = 19.6 %.

The turning points lie at $t = m λ_{0} / (4 n_{f})$ : $65.2$ (max), $130.4$ (min), $195.7$ nm (max). The film is $3 λ_{0} / (4 n_{f}) = 195.7$ nm thick, a three-quarter-wave layer with $R = 30.6 %$ .

With $n = 1.46 < n_{s}$ the quarter-wave surface has $n_{in} = 1.4 6^{2} /1.52 = 1.402$ , $Γ = - 0.167$ , $R = 2.8 %$ : the reflectance first falls, to a minimum at $λ_{0} / (4 \times 1.46) = 102.7$ nm.

Answer

R_{0} = 4.26 %, R_{m a x} = 30.6 % at 65.2 nm, R (32.6 nm) = 19.6 %, t = 195.7 nm, silica: minimum 2.8 % at 102.7 nm

Check

At $130.4$ nm the layer formula with $tan β_{f} t = 0$ gives $η_{in} = η_{s}$ and $R = R_{0}$ $✓$ . The curves of Figure 6.3.19, computed from the full layer formula, pass through all the values found.

Exercise 6.3.8. The monitor record: reflectance at 600 nm against the film thickness. A high-index film raises the reflectance, a low-index film lowers it, and every quarter wave is a turning point. — **Figure 6.3.19.** Exercise 6.3.8. The monitor record: reflectance at $600$ nm against the film thickness. A high-index film raises the reflectance, a low-index film lowers it, and every quarter wave is a turning point.

Remark

A film of higher index than the substrate always raises the reflectance, one of lower index lowers it. Counting turning points is how quarter-wave layers are cut accurately during deposition, without knowing the deposition rate. High-reflection mirrors and beam splitters are built from such quarter-wave TiO $_{2}$ layers alternating with low-index ones.

cf. Problem 6.3.4

Exercise 6.3.8: Watching a film grow