Interference from Thin Films — Question 3

Question 3

A quarter-wave coating needs both phase and amplitude matching

At normal incidence, a single lossless layer is intended to eliminate reflection at $\lambda_0=600\,\mathrm{nm}$ between air ($n_0=1$) and a substrate ($n_s=2.25$). The exact amplitude coefficient is $r=(r_{0f}+r_{fs}e^{2i\beta})/(1+r_{0f}r_{fs}e^{2i\beta})$, where $r_{ij}=(n_i-n_j)/(n_i+n_j)$ and $\beta=2\pi n_ft/\lambda_0$.

1. For the thinnest quarter-wave layer, derive the index condition making the numerator vanish.

2. Calculate the required index and thickness, and verify equality of the two interface reflection coefficients.

Question 3 - Solution

A quarter-wave coating needs both phase and amplitude matching

(a) Method and calculation

$$\beta=\pi/2\Longrightarrow e^{2i\beta}=-1;\quad r=0\Longrightarrow\frac{n_0-n_f}{n_0+n_f}=\frac{n_f-n_s}{n_f+n_s}.$$$$(n_0-n_f)(n_f+n_s)=(n_f-n_s)(n_0+n_f)\Longrightarrow\boxed{n_f^2=n_0n_s}.$$

(b) Interpretation and further calculation

$$\boxed{n_f=1.50,\quad t=\lambda_0/(4n_f)=100\,\mathrm{nm}}.$$$$\boxed{r_{0f}=(1-1.5)/2.5=-0.20=r_{fs}=(1.5-2.25)/3.75}.$$

Physical check

A quarter-wave thickness supplies the opposite propagation phase, while the geometric-mean index supplies the required amplitude match.