Split One Beam into Equal Halves: Design a 45° 50/50 Dielectric Beamsplitter

633 nm, 45° 50/50 dielectric beamsplitter: polarization and angle sweeps

This tutorial designs a 45° dielectric beamsplitter for a 633 nm laser so that the reflected and transmitted powers are close to 50/50. Devices of this kind are used in interferometers, laser metrology, and dual-channel imaging. Intuitively, the coating redistributes reflected and transmitted optical power so that one beam leaves in two directions.

You will express the required reflectance as an optimization objective, then use R/T/A, s/p polarization, and an angle sweep to determine the design's operating range.

Beamsplitter cube mounted on an optical laboratory setup — A cube beamsplitter on an optical tableGeek3 / Wikimedia Commons CC BY-SA 4.0

Diagram of a beamsplitter dividing one incident beam into reflected and transmitted beams — Figure 1 | A beamsplitter produces reflected and transmitted output directionsBenjamin D. Esham / Wikimedia Commons Public Domain

In the diagram, the incident beam divides into reflected and transmitted directions at the splitting surface. In this tutorial, 50/50 means that the two output powers are equal.

Express the device requirement numerically

Item	Teaching specification
Design wavelength	633 nm
Incidence angle	45°
Design polarization	Unpolarized, `pRatio=0.5`
Power target	$R=50\%$ , $T=50\%$
Allowed error	$
Additional acceptance checks	s/p polarization, 0–60° angle sweep, energy conservation

A lossless structure satisfies

R+T+A=1.

Here, $R$ , $T$ , and $A$ are the reflected, transmitted, and absorbed fractions of optical power, respectively. All materials in this tutorial are defined as nonabsorbing, so $A=0$ . Achieving $R=0.5$ at the design point therefore also gives $T=0.5$ .

The software uses $p$ to denote the p-polarized power fraction, or pRatio. The reflectance of a polarization mixture is

R(p)=(1-p)R_s+pR_p.

Here, $R_s$ and $R_p$ are the reflectances for pure s and pure p polarization, respectively, and $p$ ranges from 0 to 1. A value of $p=0.5$ represents an equal-power mixture of the two polarizations. Transmittance is weighted in the same way.

The 50/50 target here constrains only the powers of the two beams. Interferometric applications also require checks of the reflection and transmission phases and the optical path lengths of both arms.

Control the reflected amplitude with three layers

The structure is Air / TiO₂ / SiO₂ / TiO₂ / BK7 / Air. The BK7 substrate is 1 mm thick and is set as an incoherent layer.

Layer	Refractive index	Initial thickness	Status
TiO₂ Top	2.45	30 nm	Optimization variable
SiO₂ Mid	1.46	50 nm	Fixed
TiO₂ Inner	2.45	30 nm	Optimization variable
BK7 Substrate	1.52	1 mm	Incoherent, fixed

Initial Structure page for the three-layer dielectric beamsplitter — Figure 2 | Initial beamsplitter structure

The one-way phase thickness of layer $i$ at oblique incidence is

\delta_i=\frac{2\pi n_id_i\cos\theta_i}{\lambda}.

Here, $\delta_i$ is the phase thickness of layer $i$ , $n_i$ is its refractive index, $d_i$ is its physical thickness, $\theta_i$ is the propagation angle inside the layer, and $\lambda$ is the vacuum wavelength. The two TiO₂ layers occupy different optical environments, so they are assigned separate variables rather than being constrained to the same thickness.

This tutorial uses constant refractive indices to create an easily reproducible model. A practical device requires wavelength-dependent n and k data for the materials.

Run the baseline

In Optics, set 400–900 nm with a 1 nm step, 45° incidence, and pRatio=0.5. Enable Reflectance, Transmittance, and Absorptance. The broad wavelength range shows the spectrum, while the optimization objective uses only 633 nm.

Wavelength range 45 degree incidence and detector settings for the 50 50 beamsplitter — Figure 3 | Optical conditions for the beamsplitter

Predict the result before running: the initial 30/50/30 nm stack is likely to remain transmission-dominant. Even if the unpolarized average reaches 50/50, pure s and pure p polarization may not each be close to 50%.

At 633 nm, the initial structure gives $R=12.883\%$ and $T=87.117\%$ . It behaves more like an antireflection structure and does not yet split the power equally.

Reflectance of the initial three-layer structure near 633 nm — Figure 4 | The initial structure is transmission-dominant

Set 50% reflectance as the optimization objective

In Optimizer, add a target-value objective that makes Reflectance approach 0.5 at 633 nm, 45° incidence, and pRatio=0.5. Adjust only the two TiO₂ layers:

The baseline calculation uses the actual 30 nm thicknesses in Structure. At the optimization stage, set each variable's Initial Value to 60 nm. This value is the local-search starting point when Grid is disabled; it does not rewrite the recorded baseline result. With Grid enabled here, the software uses grid candidates as the starting points instead.

Variable	Minimum	Start	Maximum
TiO₂ Top	15 nm	60 nm	150 nm
TiO₂ Inner	15 nm	60 nm	150 nm

Use TRF with at most 120 evaluations. Enable a 7 × 7 Grid and send the best 3 starting points to local optimization.

633 nm 50 percent reflectance objective and two TiO2 thickness variables — Figure 5 | The 50% reflectance objective and thickness variables

Optimizer settings for a 7 by 7 Grid, three seeds, and the TRF algorithm — Figure 6 | Grid seeds and TRF algorithm settings

This optimization uses 53 objective evaluations and 27 iterations to obtain:

TiO₂ Top: 82.912 nm;
TiO₂ Inner: 60.967 nm;
Unpolarized reflectance at 633 nm: 49.9986%.

Click Apply to Structure, then return to Structure and confirm that both thicknesses have been updated.

Optimum thicknesses and Apply to Structure in the Optimization Report for the 50 50 beamsplitter — Figure 7 | Beamsplitter optimization report

Three-layer dielectric beamsplitter after applying the optimum solution — Figure 8 | Structure after applying the optimum solution

Verify the 50/50 power split

Run the forward calculation again after applying the optimized thicknesses.

Design	$R(633\ \mathrm{nm})$	$T(633\ \mathrm{nm})$	$A$	Deviation from 50%
Initial	12.8832%	87.1168%	0	37.1168 percentage points
Optimized	49.9986%	50.0014%	0	0.0014 percentage points

Reflectance spectrum of the optimized beamsplitter — Figure 9 | The optimized structure reaches 50% reflectance at 633 nm

The design point passes the ±1 percentage-point specification, and the energy-closure error is $1.1\times10^{-16}$ .

Unpolarized 50/50 does not mean polarization-independent

Keep the wavelength at 633 nm and the incidence angle at 45°, then set pRatio to 0 and 1 in turn.

First set pRatio to 0, confirm that the interface shows 100% s polarization, and run the reflectance calculation.

Optics page for the 50 50 beamsplitter showing 633 nm, 45-degree incidence, and pRatio equal to 0 for pure s polarization — Figure 10 | Optics settings for pure s polarization

Real Reflectance result for the optimized 50 50 beamsplitter at 633 nm and 45 degrees with pure s polarization — Figure 11 | Pure s-polarized reflectance is above 50%

Then set pRatio to 1, confirm that the interface shows 100% p polarization, and rerun the same structure.

Optics page for the 50 50 beamsplitter showing 633 nm, 45-degree incidence, and pRatio equal to 1 for pure p polarization — Figure 12 | Optics settings for pure p polarization

Real Reflectance result for the optimized 50 50 beamsplitter at 633 nm and 45 degrees with pure p polarization — Figure 13 | Pure p-polarized reflectance is below 50%

Polarization	$R$	$T$	Deviation from 50%
s	66.9115%	33.0885%	16.9115 percentage points
p	33.0858%	66.9142%	16.9142 percentage points
Unpolarized average	49.9986%	50.0014%	0.0014 percentage points

The unpolarized result is close to 50% because s polarization is above 50% and p polarization is below 50%, so their equal-power average cancels the difference. This structure is a 50/50 beamsplitter only in the unpolarized-average sense; it is not a polarization-independent beamsplitter.

If the application requires any linear polarization to remain close to 50/50, create separate s and p objectives and add layers or release more independent thickness variables.

Determine the operating range with an angle sweep

Fix the wavelength at 633 nm. In Sweep, scan 0–60° in 5° steps while keeping pRatio=0.5.

Incidence-angle sweep from 0 to 60 degrees at 633 nm — Figure 14 | Incidence-angle sweep of the splitting ratio

After running Sweep, reflectance falls gradually from about 57% to about 45% and crosses 50% near 45°.

Real Reflectance result for the optimized beamsplitter during an unpolarized angle sweep from 0 to 60 degrees at 633 nm — Figure 15 | Reflectance versus incidence angle at 633 nm

Incidence angle	$R$	$T$	Deviation from 50%
0°	57.232%	42.768%	7.232 percentage points
30°	54.370%	45.630%	4.370 percentage points
40°	51.697%	48.303%	1.697 percentage points
45°	49.999%	50.001%	0.001 percentage points
50°	48.145%	51.855%	1.855 percentage points
60°	44.801%	55.199%	5.199 percentage points

Unpolarized reflectance and transmittance versus incidence angle at 633 nm — Figure 16 | The 50/50 split occurs only near the design angle

In the coarse 5° sweep, only the 45° sample meets the ±1 percentage-point specification. This result does not define a continuous angular tolerance. To determine the passing interval, repeat the sweep near 45° with a step of 1° or smaller.

Design limits

The current structure achieves a 50/50 power split at 633 nm, 45° incidence, and unpolarized illumination, but it does not simultaneously satisfy broadband, wide-angle, arbitrary-polarization, or phase requirements.

Application requirement	Additional design condition
Polarization-independent splitting	Constrain s and p separately
Wide-angle splitting	Create objectives at multiple incidence angles
Broadband splitting	Replace the single-wavelength objective with a target band
Real glass plate	Use actual dispersion and include the rear surface or a wedge model
Interferometer	Continue by checking reflection and transmission phases and both arm lengths

Variation Exercise

Refine the angle sweep to 40–50° with a 1° step while keeping the structure and wavelength unchanged. Find the continuous interval that satisfies $|R-50\%|\leq1$ percentage point and compare it with the conclusion from the 5° coarse sweep.

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