# Contrast Analysis of Polarization in Three-Beam Interference Lithography
## 2 Contrast in Double-Beam Non-Coplanar Interference Patterns
Under this condition, the azimuth and the polarization angles of the two beams satisfy the following equation:  
\[\phi_{2}=-\alpha+\phi_{1} \tag{9}\]  
In this case, the contrast can simplify to  
\[K\propto\left|\cos^{2}\alpha+\sin^{2}\theta(1-\cos\alpha)\cos\phi_{1}\cos( \alpha-\phi_{1})-\cos\theta\sin^{2}\alpha\right| \tag{10}\]  
From Equation (10), each azimuth angle has a unique polarization co-direction angle corresponding to it, which causes the contrast to obtain a maximum value. We can calculate that the condition for maximum contrast is  
\[\phi_{1}=\frac{\alpha}{2} \tag{11}\]  
This condition is verified in Figure 2. We simulated the contrast of all linear polarization orientations when the azimuth angles were \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\), \(120^{\circ}\), \(150^{\circ}\)and \(180^{\circ}\), and found that the corresponding polarization angle when the maximum contrast is achieved was \(15^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), \(75^{\circ}\), and \(90^{\circ}\). This shows that when the polarization angle is on the intersection of the vibration planes of two waves, the contrast is maximized. By comparing Figure 2a,b, it can be seen that the polarization orientation has a great influence on the contrast, especially under the condition of a large angle of incidence. Figure 1: Wave vectors and the coordinate system for double beam interference. Both incident directions of the two beams are on the cone line of the conical surface. Substituting Equation (11) into Equation (10), the maximum contrast that can be achieved by double-beam interference is  
\[K_{\text{max}}\propto 1-\left(\frac{1}{2}\sin^{2}\theta+\cos^{2}\theta+\cos\theta \right)\sin^{2}\alpha \tag{12}\]  
As shown in Figure 3, only in the case of coplanar interference (\(\alpha=\pi\)) and where two beams are incident in _TE-TE_ mode will the contrast not decrease due to the increase in the incident angle. As the azimuth angle increases, the contrast loss caused by large incident angle interference is considerable until it increases to \(90^{\circ}\), where the contrast loss caused by large incident angle interference reaches the maximum. When the azimuth angle continues to increase beyond \(90^{\circ}\), the contrast loss gradually decreases. Until coplanar interference is reached, the contrast loss caused by the large incident angle completely disappears. As long as the incident angle is not greater than \(35^{\circ}\), the contrast loss of any azimuth non-coplanar interference does not exceed 0.02. However, if the incident angle continues to increase, the contrast sharply decreases. Figure 3: The relationship between the contrast of the two-beam interference light field and the incident angle under the optimal polarization incident condition. In the case of non-coplanar interference of two beams, the contrast loss caused by the increase in the incident angle is inevitable. Figure 2: The relationship between the contrast and the polarization angle combinations with 6 azimuth angles. The interference contrast is most affected by polarization when the azimuth angles are orthogonal. The greater the incident angle, the greater the contrast loss caused by the polarization. For (**a**), the incident angle is \(15^{\circ}\), and for (**b**), the incident angle is \(30^{\circ}\). In addition, the existence of polarization angle deviation will inevitably cause the loss of contrast. To illustrate this impact, we assume that the polarization angle deviations of the two beams are \(\delta_{1}\) and \(\delta_{2}\). In this case, the contrast can be described as  
\[K\propto\left| \begin{array}{c}\sin\alpha\cos\theta\sin(\delta_{1}-\delta_{2} )+\cos\alpha\cos(\delta_{1}-\delta_{2})\\ +\frac{1}{2}(1-\cos\alpha)\sin^{2}\theta\cos(\delta_{1}-\delta_{2})\cos( \alpha+\delta_{1}-\delta_{2})\end{array} \right| \tag{13}\]  
Under the same incident angle, the contrast loss is largest when the azimuth angle is equal to \(90^{\circ}\). The polarization angle deviation at this point further increases the contrast loss.