The transmission path of light in a crystal or some other media is identified mainly by solving a Maxwell equation set. The difference between the solution of the equation set and that of a radiation transfer equation lies in that the phase, amplitude, polarization, etc. of the light must be considered. If the electrical field intensity is used as a dependent variable to characterize the waveform change, then the Maxwell equation set looks like

where k₀ is the wave vector, c₀ is the speed of light in the vacuum, and

The equation is suitable for the single-wavelength incidence scenario, and is mainly used for stead-state analysis and pattern analysis. In this paper, the LiNbO₃ crystal is used as the study subject. The LiNbO₃ crystal is extremely sensitive to the temperature change. The most obvious sensitive parameter is its refractive index. Reference  [24] showed that when the wavelength is in the ultraviolet band or the visible light band, the refractive index dispersion equations for O and E rays can be expressed respectively as follows:

Gaussian energy distribution curves of O-ray and E-ray can be obtained by solving the Maxwell equations through the beam envelope method, and can be used to determine the x coordinates of O-ray and E-ray energy peaks. Their walk-off angles can be obtained through inverse derivation. For specific procedures, see Ref.  [9].

According to Ref.  [25], the LiNbO₃ crystal’ s refraction index vs. temperature derivative relation meets the following formulas when the incident wavelength is within the optical absorption edge range:

The temperature distribution in the crystal can be determined by solving the heat conduction equation

with the boundary condition

where Q is the internal heat source caused by the laser’ s thermal effect, q₀ is the heat diffusion to the surroundings, T_ext is the ambient temperature, k is the LiNbO₃ crystal’ s heat transfer coefficient, C_p is its specific heat capacity, and α is its thermal diffusivity defined as k/(ρ C_p). The LiNbO₃ crystal’ s specific heat capacity and thermal diffusivity vary with the temperature. For their relations with temperature, see Refs.  [26] and [27]. The thermal diffusivity vs. temperature curve can be obtained through numerical fitting.

It is inevitable for the laser to lose energy during transmission. In this study, the elasto-optical effect induced by the heat-induced crystal deformation is neglected and the thermo-optic effect induced by the electromagnetic energy loss is mainly considered. As a heat source in the heat transfer part of the model, it is given by

where the resistive loss is

and the magnetic loss is given by

The electric and magnetic energies are defined respectively as

The time derivatives of these expressions are the electric and magnetic powers

These quantities are related to the resistive and radiative energy or energy losses. According to Poynting’ s theorem, ^[28] we have

The current density J in the above expression is the one satisfying the Maxwell– Ampere law.

On the assumption that the material is linear and isotropic, the following equations are valid:

By interchanging the orders of differentiation and integration, which is justified by the fact that the volume is constant and the fields are continuous in time, the following equation can be obtained:

The integrand on the left side is the total electromagnetic energy density

According to Ref.  [29], the index ellipsoid equation for the LiNbO₃ crystal can be expressed as

when the applied electric field is parallel to the optical axis (x₃). There are no cross terms in Eq.  (21), showing that the major axis of the new index ellipsoid is perfectly coincident with the major axis of the old one, and is still the rotating major axis. So, the new principal refractive index can be directly calculated from Eq.  (21), and given as follows:

In the present paper, only the change of the principal refractive index in the case that the electric field is parallel to the optical axis (x₃) is considered, and those in other cases are not considered.

The incident beam is a Gaussian laser. Its transmission direction is perpendicular to the optical axis (z axis); its polarization direction is perpendicular to the incident direction and at an angle of 45° with respect to the z axis. The physical model and the boundary condition are shown in Fig.  1 (x– y cross section): 1 and 2 denote air zones; 3 refers to the LiNbO₃ crystal whose refractive index is controlled by the temperature and the applied electric field; 4 represents a perfect electrical conductor boundary condition, i.e., the electric field is perpendicular to the boundary interface; 5 represents a virtual receiving screen, which is 10  μ m wide with a vertical distance of 2  μ m from the bottom surface of the crystal, can be moved to the left or right, and is used to record the intensity distribution of the polarized light in different polarization direction at the location. The β and γ , which determine the directions of the line connecting the locations of the incident light and the emitted light and the crystal interface’ s normal line, are respectively the refraction angles of O-ray and E-ray, and their difference is defined as the discrete angle. The simulation is based on the following assumptions: the crystal’ s electro-optical coefficient does not vary with the temperature and its absorption coefficient and thermal expansion coefficient are isotropic. According to Ref.  [29], the electro-optical coefficient is a constant 8.6× 10^{− 6}  μ m/V for O-ray, and it is also a constant 30.8× 10^{− 6}  μ m/V for E-ray. The absorption coefficient of the crystal depends on the incident wavelength. According to Ref.  [30], the absorption coefficient is 0.002  μ m^{− 1}, 0.008  cm^{− 1}, and 0.0019  cm^{− 1} when the incident wavelength is 0.38  μ m, 0.65  μ m, and 1.05  μ m, respectively. The absorption coefficients of O-ray and E-ray have no considerable difference. So, they are assumed to be the same in the present paper. The refractive index of the air is assumed to be 1, i.e., the absorption of light by the air is neglected. The electric field is applied along the z axis. Unless otherwise stated, the electric field’ s intensity is 0  V and the ambient temperature is 293  K. The convective heat transfer coefficient is 10  W/m²· K and the crystal thickness is 8  μ m. In order to observe the wave characteristics in a wavelength range, the grid partition needs to be carried out by wavelength. In this paper, one tenth of the wavelength is regarded as the characteristic length of each grid in the entire area, and all the grids are nonstructural ones. The entire calculation process is shown in Fig.  2, and the origin of the coordinates is set at the center of the simulated area.

Fig.  1.

	Figure Option ViewDownloadNew Window
	Fig.  1. Schematic of model.

Fig.  2.

	Figure Option ViewDownloadNew Window
	Fig.  2. Simulation flowchart.

The solution logic for the coupled thermal and optical effects is as follows: the electromagnetic frequency domain equation is first solved to obtain the optical field distribution on the assumption that the ambient temperature and the heat transfer coefficient are constant; then the heat source term in the energy equation is calculated to obtain the temperature field distribution; the crystal’ s specific heat capacity, the thermal diffusion coefficient, the refractive index, etc. are corrected according to the identified temperature field; new values of the physical parameters are used to solve the light transmission process again; an iteration of the above process is carried out; when the change in temperature or refractive index at the same point in the crystal is less than 0.001  K or 0.00001° , the convergence is considered to be reached, and then the calculation results are outputted.

Figure  3 shows the transmission process of a Gaussian laser in the LiNbO₃ crystal in the case of 0.38  μ m incident wavelength and 40000  V/m incident intensity. The electric field’ s distribution of modulus values, the electric field intensity component distributions in three polarization directions E_x, E_y, and E_z, the temperature field distribution, the total heat source, and the total power dissipation density distribution are shown.

Fig.  3.

	Figure Option ViewDownloadNew Window
	Fig.  3. Distributions of birefringence’ s electromagnetic field and energy: (a) electric field’ s modulus values, (b) electric field intensity distributions Ex, Ey, (c) electric field intensity distribution Ez, (d) temperature distribution, (e) total heat source, and (f) total power dissipation density.

If it is assumed that the O-ray and E-ray waveforms (E_x and E_z) in the case of T = 293  K and h = 10  W/m²· K constitute the rated operating condition and those in the case of T = 283  K and h = 9  W/m²· K constitute the actual operating condition, then their O-ray and E-ray electric field distributions on the receiving screen are shown in Fig.  4. The left peaks and troughs reflect the primary transmitted light’ s energy distribution while the right peaks and troughs reflect the secondary transmitted light’ s energy distribution. The available studies^[6] show that it is feasible to make the O-ray and E-ray electromagnetic waveforms in the case of T = 283  K and h = 9  W/m²· K the same as those in the case of the rated operating condition by using an electric field to modulate the electro-optical crystal’ s index ellipsoid. It is found according to the calculation results that when the applied electrical field’ s intensity is 85.5  V/μ m, its E_z intensity is the same as that in the case of the rated operating condition, and when the applied electrical field’ s intensity is 57.7  V/μ m, its E_x and E_y intensities are the same as those in the case of the rated operating condition.

Fig.  4.

	Figure Option ViewDownloadNew Window
	Fig.  4. The Ex, Ey, and Ez intensity distributions on the receiving screen in the two scenarios (E-ray and O-ray).

Figure  5 shows the changes of the O-ray and E-ray refraction angles with ambient temperature when the heat transfer coefficient is 10  W/m²· K and the incident light intensity is 30000  V/m, 35000  V/m, and 40000  V/m. As shown in the figure, both O-ray and E-ray refraction angles decrease as the incident power increases when the ambient temperature is constant. However, the change of the E-ray refraction angle is faster. This is mainly because the E-ray’ s refractive index temperature coefficient is larger than that of the O-ray (see Eqs.  (5) and (6)). In addition, the refraction angles of O-ray and E-ray decrease as the ambient temperature increases when the incident power is constant, and the decrease of the E-ray’ s refraction angle is faster. This is mainly because the increase of the O-ray’ s refractive index induced by the ambient temperature’ s increase is slower than that of the E-ray’ s refractive index. The calculations show that when the ambient temperature changes from 273  K to 313  K, the refraction angle changes of the O-ray with the incident power are respectively 0.10001° , 0.10453° , and 0.0863° , while those of the E-ray are respectively 0.14959° , 0.15534° , and 0.12737° . It can be seen that the refraction angle changes in the cases of 30000  V/m and 35000  V/m are close and the refraction angle change in the case of 40000  V/m is relatively small. This is mainly because a considerable amount of heat is produced within the crystal in the case of 40000  V/m. The crystal’ s specific heat gradually increases as the temperature increases. When the specific heat reaches up to a certain value, the influence of the specific heat on the temperature change might be greater than that of the power change. Thus, the crystal’ s internal temperature change is the smallest in this case.

Fig.  5.

	Figure Option ViewDownloadNew Window
	Fig.  5. Refraction angle vs. ambient temperature at different incident light intensities.

Figure  6 shows the changes of the O-ray and E-ray discrete angles with ambient temperature when the heat transfer coefficient is 10  W/m²· K and the incident light intensity is 30000  V/m, 35000  V/m, and 40000  V/m respectively. It can be seen that when the temperature is constant, the discrete angle decreases as the incident power increases, and when the incident power is constant, the discrete angle decreases as the ambient temperature increases. This is mainly because the crystal’ s temperature increases as the ambient temperature or incident power increases, thus causing their refractive indices to increase. The refractive index temperature coefficient of the E-ray is higher than that of the O-ray, but when the refractive index of the E-ray is closer to that of the O-ray as the temperature increases, the crystal’ s anisotropism is weaker and thus the discrete angle decreases.

Fig.  6.

	Figure Option ViewDownloadNew Window
	Fig.  6. Discrete angle vs. ambient temperature at different incident light intensities.

Figure  7 shows the changes of O-ray and E-ray refraction angles with heat transfer coefficient at different incident powers when the ambient temperature is 293.15  K. As shown in the figure, both O-ray and E-ray refraction angles decrease as the incident power increases on the premise that the heat transfer coefficient is constant. However, the decrease of the E-ray refraction angle is faster. In addition, when the incident power is constant, both of the refraction angles of O-ray and E-ray increase as the heat transfer coefficient increases. The increase of the E-ray refraction angle is faster. This is mainly because the increased heat transfer coefficient intensifies the heat dissipation and reduces the temperature of the crystal. So both the O-ray and E-ray’ s refractive indices decrease, and the decrease of the O-ray is slower. In addition, when the heat transfer coefficient changes from h = 8  W/m²· K to h = 12  W/m²· K, the refraction angles of the O-ray are respectively 0.19516° , 0.27158° , and 0.32987° as the incident power increases, while the refraction angles of the E-ray are respectively 0.29165° , 0.40322° , and 0.48635° . This indicates that the refraction angle increases as the incident power increases, since the temperature change in the crystal increases as the incident power increases.

Fig.  7.

	Figure Option ViewDownloadNew Window
	Fig.  7. Refraction angle vs. heat transfer coefficient at different incident light intensities.

Figure  8 shows the changes of the O-ray and E-ray discrete angles with the heat transfer coefficient when the ambient temperature is 293.15  K and the incident light intensity is 30000  V/m, 35000  V/m, and 40000  V/m respectively. It can be seen that on the premise that the heat transfer coefficient is constant, the discrete angle decreases as the incident power increases, and on the premise that the incident power is constant, the discrete angle decreases as the heat transfer coefficient increases. In addition, the discrete angle’ s increase rate gradually decreases as the heat transfer coefficient increases and gradually increases as the incident power increases. This is mainly because a higher heat transfer coefficient leads to a lower crystal temperature and refractive index yet higher O-ray and E-ray refraction angles, allowing the E-ray refraction index to gradually deviate from the O-ray refraction index, and the discrete angles to become gradually wider, and as the incident power increases, the temperature change in the crystal induced by the same heat transfer coefficient change increases and thus the discrete angle’ s increase rate increases. When the incident light intensity is constant, the heat dissipation change induced by the optical path change is very small, i.e., the heat source change is very small, and in this case, the crystal’ s temperature is determined mainly by the heat transfer coefficient. When the heat transfer coefficient is very small, the same heat transfer coefficient change causes a larger change in the crystal temperature according to the heat transfer theory. So, the discrete angle change is considerable when the heat transfer coefficient is small.

Fig.  8.

	Figure Option ViewDownloadNew Window
	Fig.  8. Discrete angle vs. heat transfer coefficient at different incident light intensities.

Figure  9 shows the changes of the O-ray and E-ray refraction angles with ambient temperature at different crystal thicknesses when the heat transfer coefficient is 10  W/m²· K. As shown in the figure, both O-ray and E-ray refraction angles decrease as the crystal thickness increases on the premise that the ambient temperature is constant. However, the change of the E-ray refraction angle is faster. This is mainly because a higher crystal thickness leads to a longer optical path in the crystal and a higher internal heat source energy, and meanwhile, the higher crystal thickness weakens the crystal’ s heat dissipation, thus the crystal’ s temperature is higher. In addition, when the ambient temperature changes from 273  K to 313  K, with the crystal thicknesses of 6  μ m, 7  μ m, 8  μ m, 9  μ m, and 10  μ m, the refraction angle changes of the O-ray, are respectively 0.09367° , 0.09812° , 0.0863° , 0.09296° , and 0.09326° while those of the E-ray are respectively 0.13932° , 0.14532° , 0.12737° , 0.13671° , and 0.13662° . It can be seen that their refraction angles in the case of 7  μ m thickness are the widest and their refraction angles in the case of 8  μ m thickness are the narrowest. This is mainly because the crystal has a big temperature change (39.77  K) in the case of 7  μ m thickness and a small temperature change (35.29  K) in the case of 8  μ m thickness.

Fig.  9.

	Figure Option ViewDownloadNew Window
	Fig.  9. Refraction angle vs. ambient temperature at different crystal thicknesses.

Figure  10 shows the discrete angle vs. ambient temperature with different crystal thicknesses. It can be seen that when the ambient temperature is constant, the discrete angle decreases as the crystal thickness increases. This is mainly because the larger crystal thickness leads to a longer total laser path, which causes the heat source term to have a higher value and the crystal’ s internal temperature to sustain a bigger increase, thus making the refraction indices of the O-ray and E-ray higher and their refraction angles narrower. The decrease of the E-ray’ s refraction angle is faster than that of the O-ray’ s refraction angle, so the path of the E-ray is closer to the path of the O-ray. In addition, when the ambient temperature changes from 273  K to 313  K, the discrete angle change in the case of 7  μ m thickness is maximal and the discrete angle change in the case of 8  μ m thickness is minimal.

Fig.  10.

	Figure Option ViewDownloadNew Window
	Fig.  10. Discrete angle vs. ambient temperature at different crystal thicknesses.

Figure  11 shows the changes of the O-ray and E-ray refraction angles with heat transfer coefficient at different crystal thicknesses (6  μ m, 7  μ m, 8  μ m, 9  μ m, and 10  μ m) when the ambient temperature is 293.15  K. As shown in the figure, both O-ray and E-ray refraction angles decrease as the crystal thickness increases on the premise that the heat transfer coefficient is constant. However, the decrease of the E-ray refraction angle is faster. In addition, when the heat transfer coefficient changes from h = 8  W/m²· K to h = 12  W/m²· K, the refraction angles of the O-ray are 0.25967° , 0.30173° , 0.32987° , 0.37092° , and 0.42667° when the crystal thicknesses are 6  μ m, 7  μ m, 8  μ m, 9  μ m, and 10  μ m while the refraction angles of the E-ray are 0.38591° , 0.44667° , 0.48635° , 0.54487° , and 0.62392° . This indicates that the refraction angle increases as the crystal thickness increases when the heat transfer coefficient changes from h = 8  W/m²· K to h = 12  W/m²· K and the ambient temperature is constant. This is mainly because a larger crystal thickness leads to a higher total energy adsorption of the crystal along the optical path, which increases the crystal’ s internal heat source level. According to the energy equation, the temperature change induced by the same Δ h increases as the heat source level increases.

Fig.  11.

	Figure Option ViewDownloadNew Window
	Fig.  11. Refraction angle vs. heat transfer coefficient at different crystal thicknesses.

Figure  12 shows the O-ray and E-ray discrete angles vs. heat transfer coefficient at different crystal thicknesses. It can be found that on the premise that the heat transfer coefficient is constant, the discrete angle decreases as the crystal thickness increases while on the premise that the crystal thickness is constant, the discrete angle increases as the heat transfer coefficient increases. The discrete angle change gradually becomes faster as the crystal thickness increases. On the premise that the crystal thickness is constant, the discrete angle change gradually becomes slower as the heat transfer coefficient increases. This is mainly because the larger crystal thickness causes the crystal’ s internal heat source level to increase and the crystal’ s internal temperature change induced by the same heat transfer coefficient change to increase. According to the heat transfer theory, when the heat transfer coefficient is low, a change in the heat transfer coefficient of the same degree causes a considerable temperature change, and the temperature change gradually decreases as the heat transfer coefficient increases.

Fig.  12.

	Figure Option ViewDownloadNew Window
	Fig.  12. Discrete angle vs. heat transfer coefficient at different crystal thicknesses.

Figure  13 shows the correlations between the applied electric field intensity and the ambient temperature to keep the output E-ray’ s refraction angle and intensity unchanged when the incident power is 40000  V/m, the ambient temperature is 293.15  K, and the heat transfer coefficient is 10  W/m²· K. According to the figure, 293.15  K is the critical point. When the ambient temperature increases, the positive electric field intensity required for offsetting the temperature influence gradually increases; in the case of L = 6  μ m, the required electric field intensity is maximal; in the case of L = 9  μ m, the required electric field intensity is minimal. When the ambient temperature decreases, the negative electric field intensity required for offsetting the temperature influence gradually increases; in the case of L = 6  μ m, the required electric field intensity is minimal; in the case of L = 9  μ m, the required electric field intensity is maximal. This is mainly because the refractive index decreases as the crystal thickness increases when the operating condition is unchanged. The average refractive index in the case of L = 6  μ m is minimal. That is, n_e in Eq.  (23) is minimal. If we assume that T = 293.15  K and h = 10  W/m²· K constitute the rated operating condition, the refractive index to be targeted is n_e1. The n_e1 decreases when the ambient temperature decreases, while n_e1 increases when the ambient temperature increases. Finally, the required applied electric field intensity is determined by n_e and n_e1. The crystal’ s maximum internal temperatures are respectively 560.28  K, 551.97  K, 538.40  K, 530.91  K, and 522.60  K when L = 6  μ m and the ambient temperatures are 273  K, 283  K, 293  K, 303  K, and 313  K. So, in this case, the crystal’ s internal temperature increasing with the increase of the ambient temperature is slightly larger than its internal temperature decreasing with the decrease of the ambient temperature, and thus, in the case of L = 6  μ m, the required positive electric field intensity is higher than the required negative electric field intensity. The crystal’ s maximum internal temperatures are respectively 687.17  K, 676.45  K, 674.40  K, 661.81  K and 648.85  K when L = 9  μ m and the ambient temperatures are 273  K, 283  K, 293  K, 303  K, and 313  K. In this case, the crystal’ s internal temperature increasing with the increase of the ambient temperature is far lower than its internal temperature decreasing with the decrease of the ambient temperature, which makes the required positive electric field intensity lower than the required negative electric field intensity in the case of L = 9  μ m.

Fig.  13.

	Figure Option ViewDownloadNew Window
	Fig.  13. Applied electric field intensity vs. ambient temperature at different crystal thicknesses.

Figure  14 shows the correlations between the applied electric field intensity and the heat transfer coefficient to keep the output E-ray’ s refraction angle and intensity unchanged when the incident power is 40000  V/m, the ambient temperature is 293.15  K, and the heat transfer coefficient is 10  W/m²· K. According to the figure, the critical point is h = 10  W/m²· K. The positive electric field intensity required for offsetting the temperature increase influence gradually increases as the heat transfer coefficient decreases; and the negative electric field intensity required for offsetting the temperature decrease influence gradually increases as the heat transfer coefficient increases. For the same Δ h to be modulated, the required negative electric field intensity is lower than the required positive electric field intensity. This is mainly because the temperature increase induced by a heat transfer coefficient decrease is larger than the temperature decrease induced by the same heat transfer coefficient increase. In addition, when the heat transfer coefficient changes from h = 8  W/m²· K to h = 12  W/m²· K, there is the following order by required applied electric field intensity: L = 10  μ m > L = 9  μ m > L = 8  μ m > L = 7  μ m > L = 6  μ m. This is mainly because higher crystal temperature with larger crystal thickness leads to higher refractive index on the premise that the heat transfer coefficient is constant. The calculation results show that the crystal’ s temperature changes as the thickness increases and the heat transfer coefficient changes from h = 8  W/m²· K to h = 12  W/m²· K are as follows: 104.62  K, 122.59  K, 135.20  K, 153.24  K, and 178.04  K. Thus, the temperature change increases as the thickness increases. The refractive index change induced by the temperature change also increases gradually. Thus, the compensating electric field’ s intensity meets the above pattern.

Fig.  14.

	Figure Option ViewDownloadNew Window
	Fig.  14. Applied electric field intensity vs. heat transfer coefficient at different crystal thicknesses.

In an actual application, there is an electric field intensity threshold for any substance beyond which it will break down. This is the maximum voltage that the substance can sustain. Once the applied electric field intensity exceeds the threshold, the medium will fail to work normally. Currently, there is not any available literature document dealing with the breakdown voltages of thin lithium niobate layers. Their breakdown electric field strengths are approximately a few tens of volts per μ m. In this paper, the thin lithium niobate layer’ s breakdown electric field strength is assumed to be 100  V/μ m. The figure shows that when the heat transfer coefficient change is considerable, such as in the range from h = 10  W/m²· K to h = 8  W/m²· K, a voltage that is higher than the medium’ s breakdown voltage is required for offsetting. This is impossible in fact. Figures  13 and 14 show that the electric field compensation for the influence of the ambient temperature change on the refractive index can be carried out within a considerable ambient temperature range (about 40– 50  K) but the electric field compensation for the influence of the heat transfer coefficient change on the refractive index can be carried out within a narrow range (about 1– 2  W/m²· K). In addition, figure  14 shows that when the thickness decreases, the slope of curve for the applied electric field intensity vs. heat transfer coefficient decreases. If we assume that the index of refraction in the case of h = 10  W/m²· K constitutes the rated condition and the breakdown voltage is 100  V/μ m, the calculation shows that in the case that the thicknesses are 6  μ m, 7  μ m, 8  μ m, 9  μ m, and 10  μ m, the heat transfer coefficient adjustment ranges, which are available through the compensation by an applied electric field for the variation of the E-ray index of refraction incurred by the variation of the heat transfer coefficient, are 8.87– 11.26  W/m²· K, 8.99– 11.07  W/m²· K, 9.02– 10.82  W/m²· K, 8.88– 10.44  W/m²· K, and 8.96– 10.39  W/m²· K, respectively. It can be seen that the electric field-adjustable range of the convection heat transfer coefficient slightly decreases as the thickness increases.

Figure  15 shows the correlations between the applied electric field intensity and the ambient temperature to keep the output E-ray’ s refraction angle and intensity unchanged when the crystal thickness is 8  μ m, the ambient temperature is 293.15  K, and the heat transfer coefficient is 10  W/m²· K, all of which constitute the rated operating condition. This figure shows that the required intensity of the applied positive electric field increases as the ambient temperature increases (> 293.15  K), and the required intensity of the applied negative electric field increases as the ambient temperature decreases (< 293.15  K). It can be seen that the applied electric field intensity vs. temperature is presented in the form of a nonlinear distribution. When the ambient temperature is higher than 283 K and constant, the required positive electric field intensity is maximal and the required negative electric field intensity is minimal in the case ofE = 35000  V/m; on the contrary, in the case of E = 40000  V/m, the required positive electric field intensity is minimal and the required negative electric field intensity is maximal. When the ambient temperature is 273  K, the required negative electric field intensity values follow the following pattern: E = 35000  V/m ≥ E = 40000  V/m ≥ E = 30000  V/m. The above pattern is mainly determined by the crystal’ s final internal temperatures in the case of different ambient temperatures. The final constant crystal temperature is a result of multiple factors such as laser path, crystal– air interface transmission-reflection ratio, and specific heat capacity.

Fig.  15.

	Figure Option ViewDownloadNew Window
	Fig.  15. Electric field intensity vs. ambient temperature at different incident light intensities.

Figure  16 shows the correlations between the applied electric field intensity and heating the transfer coefficient to keep the output E-ray’ s refraction angle and intensity unchanged when the crystal thickness is 8  μ m, the ambient temperature is 293.15  K, and the heat transfer coefficient is 10  W/m²· K, all of which constitute the rated operating condition. According to the figure, the critical point is h = 10  W/m²· K. The positive electric field intensity required for offsetting the temperature increase influence gradually increases when the heat transfer coefficient decreases; and the negative electric field intensity required for offsetting the temperature decrease influence gradually increases when the heat transfer coefficient increases; for offsetting the refraction angle changes induced by the same heat transfer coefficient changes, the required positive electric field intensity is slightly higher than the required negative electric field intensity. In addition, the applied electric field intensity required for the same heat transfer coefficient change increases when the incident power increases. This is mainly because the same heat transfer coefficient change leads to a bigger temperature change when the incident power increases. So, a stronger electric field is required to offset the influence of temperature on the refractive index.

Fig.  16.

	Figure Option ViewDownloadNew Window
	Fig.  16. Applied electric field intensity vs. heat transfer coefficient at different light intensities.

Figures  15 and 16 show that if the breakdown electric field intensity is a certain value, the electric field’ s heat transfer coefficient regulation range in the case of 40000  V/m incident light intensity is widest and the decrease in incident power leads to the narrowing of the heat transfer coefficient regulation range. The calculation shows that if the incident electric field intensities are 30000  V/m, 35000  V/m, and 40000  V/m, the heat transfer coefficient adjustment ranges, which are available through the compensation by an applied electric field for the variation of the E-ray index of refraction incurred by the variation of the heat transfer coefficient, are 8.60– 11.68  W/m²· K, 8.84– 11.15  W/m²· K, and 9.02– 10.82  W/m²· K, respectively. It can be seen that the electric field-adjustable range of the convection heat transfer coefficient decreases as the power of the incident light increases. This is mainly because the slope of curve for the applied electric field intensity vs. heat transfer coefficient gradually increases as the incident light intensity increases. The offsetting regulation patterns of the applied electric field to temperature change influence are relatively complex. This is mainly because the relation between the applied electric field intensity and the temperature is in the form of a nonlinear distribution.

The crystal does not adsorb any incident light in the window area but the light refractive index in the area is still subjected to the temperature influence. So, for some wide-spectrum lasers (the light wavelength contains both adsorption band and transmission band), it is possible to consider the modulation of the window area’ s light refractive index by changing the ambient temperature or heat transfer coefficient. Figure  17 shows two scenarios: 1) vertical incidence of one single beam (1.06  μ m); 2) vertical incidence of two beams (1.06  μ m and 0.38  μ m) into the lithium niobate crystal; on the receiving screen, the 1.06  μ m beam is distributed in the polarization direction. The visible waveforms shift toward the left, mainly because the 0.38  μ m beam is easy to be adsorbed by the crystal, thus resulting in higher crystal temperature and making the 1.06  μ m beam’ s refractive index lower. It has been proved in this paper that when the external environment changes, both refractive index and heat source loss of light within the adsorption band change. The resulting crystal temperature change affects the refractive index of the light in the window area. This indicates that for a laser with a wide spectrum containing the adsorption band, it is theoretically feasible to modulate the refractive index of light within the window band by changing the ambient boundary conditions.

Fig.  17.

	Figure Option ViewDownloadNew Window
	Fig.  17. Electric field intensity distributions for O-ray and E-ray on receiving screen in the cases of single and dual incident wavelengths.