Two-Layer V-Coat for 550 nm

A single MgF₂ layer on N-BK7 gets the front-surface reflectance down to about 1.25% at 550 nm. A two-layer V-coat can place a much deeper minimum at one wavelength.

This design uses SiO₂ and ZrO₂ on N-BK7. It is meant for normal incidence and a design wavelength of 550 nm.

What a V-coat does

The name comes from the reflectance plot: the curve falls toward a narrow minimum and rises again on either side.

With one layer, only one thickness can move. With two layers, there are two thickness variables. At a single wavelength they can drive both parts of the complex reflection amplitude to zero:

\operatorname{Re}(r) = 0, \qquad \operatorname{Im}(r) = 0

This does not make the coating broadband. It gives a deep minimum around the wavelength used for the calculation.

Material values at 550 nm

Use these specific entries from the coating-material library:

SiO2 (110-800nm, RIT): $n_L = 1.4599$
ZrO2 (190-800nm, RIT): $\tilde n_H = 2.1662 + 0.0049i$
N-BK7 substrate: $n_s = 1.5185$
Air: $n_0 = 1.0000$

The exact catalog entry matters. Another ZrO₂ dataset can have a different index and extinction coefficient, so it will produce different thicknesses.

Why two quarter-wave layers are not enough

A quarter-wave starting thickness for each material would be

d_L^{\mathrm{QW}} = \frac{550}{4 \times 1.4599} \approx 94.18\ \text{nm}

d_H^{\mathrm{QW}} = \frac{550}{4 \times 2.1662} \approx 63.48\ \text{nm}

Quarter-wave phase by itself is not enough. A quarter-wave layer transforms the optical admittance on its far side. For the physical order air / SiO₂ / ZrO₂ / N-BK7, with ZrO₂ touching the substrate, the two ideal lossless quarter-wave transformations give

Y_{\mathrm{in}} = \frac{n_L^2 n_s}{n_H^2}.

There is zero reflection only when the transformed admittance matches air:

Y_{\mathrm{in}} = n_0, \qquad\text{or equivalently}\qquad \frac{n_H}{n_L} = \sqrt{\frac{n_s}{n_0}}.

For these materials,

\frac{n_H}{n_L} = \frac{2.1662}{1.4599} \approx 1.4838, \qquad \sqrt{\frac{n_s}{n_0}} = \sqrt{1.5185} \approx 1.2323.

Those ratios do not match. Using the real indices gives $Y_{\mathrm{in}} \approx 0.6897$ and therefore

R = \left(\frac{n_0-Y_{\mathrm{in}}} {n_0+Y_{\mathrm{in}}}\right)^2 \approx 3.37\%.

The full TMM calculation, including the ZrO₂ extinction coefficient, gives 3.33%. Fixing both layers at one quarter wave fixes both phase thicknesses, so there is no remaining variable to correct this admittance mismatch. Moving the two thicknesses away from quarter wave lets them drive both $\operatorname{Re}(r)$ and $\operatorname{Im}(r)$ to zero.

The 550 nm design

Here is the calculation used to obtain the thicknesses. At normal incidence, the phase thickness and characteristic matrix of layer $j$ are

\delta_j = \frac{2\pi \tilde n_j d_j}{\lambda}

M_j = \begin{bmatrix} \cos\delta_j & -i\sin\delta_j/\tilde n_j \\ -i\tilde n_j\sin\delta_j & \cos\delta_j \end{bmatrix}

The complex index $\tilde n_H = 2.1662 + 0.0049i$ is used directly here; the extinction coefficient is not discarded. For the physical order SiO₂ then ZrO₂, the total matrix is $M = M_L M_H$ . TFStudio then calculates

\begin{bmatrix}B\\C\end{bmatrix} = M\begin{bmatrix}1\\n_s\end{bmatrix}, \qquad r = \frac{n_0B-C}{n_0B+C}, \qquad R = |r|^2

The two unknowns are $d_L$ and $d_H$ . I searched the first half-wave interval for each layer,

0 < d_L < 188.37\ \text{nm}, \qquad 0 < d_H < 126.95\ \text{nm},

using several starting points, then used a two-variable Newton solve for

\operatorname{Re}(r) = 0, \qquad \operatorname{Im}(r) = 0.

The compact root converges to

d_L = 121.9434949763\ \text{nm}, \qquad d_H = 20.8916393494\ \text{nm}.

There is another root in the same search interval, near 66.36 nm of SiO₂ and 105.96 nm of ZrO₂. The solution used here has the smaller total thickness and much less absorbing ZrO₂.

Rounded to practical values, the design is:

d_{\mathrm{SiO_2}} = 121.94\ \text{nm}

d_{\mathrm{ZrO_2}} = 20.89\ \text{nm}

The physical stack is

\text{Air}|\mathrm{SiO_2}\ 121.94\ \text{nm} |\mathrm{ZrO_2}\ 20.89\ \text{nm} |\text{N-BK7}

With the thicknesses rounded to two decimal places, the calculated front-surface reflectance at 550 nm is about $1 \times 10^{-7}$ , which will normally display as zero.

The selected ZrO₂ data has a nonzero extinction coefficient. At the design wavelength the same model gives approximately

T = 99.6817, \qquad A = 0.3183

So the reflected power is nearly zero, but a small amount is absorbed.

Building it in TFStudio

1. Set the design geometry

Reference wavelength: 550 nm
Incident medium: Air
Substrate: N-BK7

2. Add the two layers

The physical order from air to substrate is:

SiO2 (110-800nm, RIT): 121.94 nm
ZrO2 (190-800nm, RIT): 20.89 nm

The Design Editor lists the substrate-adjacent layer first. Enter ZrO₂ first in the table, then SiO₂.

3. Check the spectrum

Run a thickness refinement, then plot reflectance over 400–700 nm. The expected front-surface values are approximately:

500 nm: 0.91%
550 nm: 0%
600 nm: 0.52%
650 nm: 1.54%

The minimum is narrow and the two sides are not perfectly symmetric because the material indices are dispersive.

What to keep in mind

This is a normal-incidence, single-wavelength design.
Changing either catalog entry changes the result.
Zero reflectance does not mean zero absorption.
Thickness errors move the minimum and raise its floor.

For a wider low-reflectance band, two layers are usually not enough; that is a different optimization problem.