Method and measuring system for solving birefringence phenomenon double-light track of uniaxial crystal

Abstract

The invention discloses a method and a measuring system for solving birefringence phenomenon double-light tracks of a uniaxial crystal, and the method comprises the steps: defining a front surface normal vector, a rear surface normal vector and an optical axis direction vector of the crystal, and representing the spatial orientation of the crystal through the rotation operation around an x axis and a y axis; calculating a normal vector and an optical axis direction vector after rotation according to the rotation angle of the crystal around the z axis; based on the refraction law, the Snell law and the Fresnel equation, the propagation directions of the o light and the e light in the crystal are deduced; considering the refraction behavior of the rear surface of the crystal, and respectively calculating direction vectors after the o light and the e light are emitted; and determining the coordinates of the o light and the e light on the observation screen by utilizing a point equation and combining a ray tracing method, and rotating the crystal for one circle to generate a dual-light trajectory. According to the method, any crystal rotation operation is processed through a unified coordinate system, and the calculation complexity of a complex scene is reduced; the systematic intersection solution algorithm provided by the invention is combined with the e light wave vector / light separation calculation model, so that the light spot positioning efficiency is improved.

Description

Technical Field

[0001] This invention relates to the fields of optical measurement and crystal physics, and in particular to a method and measurement system for solving the birefringence phenomenon of a uniaxial crystal and its two light trajectories. Background Technology

[0002] Birefringence is a fundamental optical effect in anisotropic crystals. Its ideal model typically assumes that the front and back surfaces of the crystal are strictly parallel and that the optical axis is fixed. Traditional theories, based on Snell's law and the wave vector equation, derive the propagation paths of the ordinary (o) and extraordinary (e) rays within the crystal. However, in actual fabrication, crystals often exhibit surface non-parallelism and optical axis orientation deviations, causing the ideal model to fail to accurately describe the experimentally observed light spot trajectory.

[0003] In 1989, Yariv et al. proposed a classic birefringence theory model, which, under the assumption of perfectly parallel crystal surfaces, could preliminarily predict the separation of o-rays and e-rays. In 2005, Born and Wolf further refined the model in "Principles of Optics," but still did not solve the problems caused by actual crystal processing errors.

[0004] In addition, Shen Weimin et al. proposed a uniaxial crystal model with parallel front and rear surfaces in their paper entitled “Discretion between ordinary and extraordinary rays when the optical axis is arbitrarily oriented in a uniaxial crystal” (Acta Optica Sinica, 2002, (06): 765-768). In this model, when the crystal is rotated, the o-ray appears as a fixed point, which is significantly different from the circular trajectory observed in the experiment. Summary of the Invention

[0005] Purpose of the invention: The purpose of this invention is to propose a method and measurement system for solving the birefringence phenomenon of uniaxial crystals and the two-light trajectory. By introducing the crystal surface tilt angle into the model, the accuracy of the two-light trajectory prediction is significantly improved.

[0006] Technical solution: This invention includes the following steps:

[0007] S1. Define the crystal's front surface normal vector, back surface normal vector, and optical axis direction vector, and represent its spatial orientation through rotation operations around the x-axis and y-axis;

[0008] S2. Calculate the normal vector and optical axis direction vector after rotation based on the rotation angle of the crystal around the z-axis;

[0009] S3. Based on the law of refraction, Snell's law and Fresnel's equation, derive the propagation directions of o-ray and e-ray within the crystal;

[0010] S4. Considering the refractive behavior of the back surface of the crystal, calculate the direction vectors of the o-ray and e-ray after they are emitted;

[0011] S5. Using the point-direction equation and the ray tracing method, determine the coordinates of the o-ray and e-ray on the observation screen, rotate the crystal once, and generate the double ray trajectory.

[0012] Specifically, S1 is: Let the normal vector of the front surface of the crystal be... The normal vector of the back surface is The optical axis direction vector is , by the direction vector of the rotation axis Rotate about the x-axis respectively Then rotate around the y-axis The result was:

[0013] in, Let be the rotation matrix about the x-axis; Let be the rotation matrix about the y-axis.

[0014] Specifically, S2 is:

[0015] Given that the rotation matrix of a crystal about the z-axis is The normal vector of the front surface of the rotated crystal , back surface normal vector and optical axis direction vector for:

[0016] .

[0017] In S3, the propagation direction vector of the o-light within the crystal is... The calculation steps include: First, calculating the angle of incidence. cosine = According to Snell's Law Calculate the angle of refraction Finally, using the vector form of the law of refraction, the direction vector of the o-ray within the crystal can be obtained as:

[0018] .

[0019] In S3, when calculating the propagation direction of the e-ray within the crystal, its wave vector direction is also considered. and the direction of light The wave vector direction satisfies the boundary conditions: ,in, It is the effective refractive index of the e-ray, relative to the wave vector direction. and optical axis direction The included angle Regarding this, the effective refractive index satisfies the wave vector equation:

[0020] Wave vector direction It can be decomposed into a component perpendicular to the interface and a component parallel to the interface. The component parallel to the interface is determined by Snell's law, and its magnitude is the same as that of the incident light. The component perpendicular to the interface is determined by the wave vector surface equation and the boundary conditions, and its expression is:

[0021]

[0022] After obtaining the wave vector direction, the direction of the e-ray within the crystal is used. From the relationship with the wave vector direction, we can obtain the corresponding propagation direction of the e-ray within the crystal:

[0023] .

[0024] In S4, the direction vector of the emitted o-light The calculation process is as follows: First, calculate the incident angle between the inside of the crystal and the back surface. cosine According to Snell's Law Calculate the angle of refraction of the o-ray Finally, using the vector form of the law of refraction, the direction vector of the o-ray propagation outside the crystal is determined:

[0025] .

[0026] In step S4, when the e-ray exits from the rear surface of the crystal, it is refracted again. Outside the crystal, the wave vector direction... =Direction of light At this point, the incident medium is a crystal, and the exit medium is an isotropic medium. Using the condition that the wave vector is continuous in the tangential component at the interface, we can obtain... The effective refractive index This can be obtained by solving the wave vector surface equation:

[0027]

[0028] in, It is Wave Arrow and optical axis The angle between the two sides, the direction of the wave vector Represented as:

[0029] .

[0030] S5 specifically includes:

[0031] (1) Calculate the coordinates of the intersection point between the e-ray and the back surface.

[0032] Using point-to-point formulas, the point is... , The three components After combining the surface equation and the ray equation:

[0033]

[0034] The intersection point between the emitted e-beam and the rear surface can be obtained. ;

[0035] (2) Calculate the coordinates of the intersection point between the o-ray and the back surface.

[0036] Using point-to-point formulas, the point is... , The three components After combining the surface equation and the ray equation:

[0037]

[0038] The intersection point between the emitted o-beam and the rear surface can be obtained.

[0039] (3) Calculate the coordinates of the intersection point between the e-ray emitted and the receiving screen.

[0040] Given that d1 is the length of the crystal along the z-axis, L is the distance between the rear exit surface of the crystal and the receiving screen along the z-axis, and the position of the receiving screen is the position of the screen. Using point-to-point formulas, the point is... , The three components Solve the equations of the receiving screen plane and the e-ray emitted simultaneously:

[0041]

[0042] The coordinates of the center of the incident point of the e-ray emitted on the screen can be obtained. ;

[0043] (4) Calculate the coordinates of the intersection point between the outgoing o-ray and the screen.

[0044] Using point-to-point formulas, the point is... , The three components Solve the equations of the receiving screen plane and the emitted ray simultaneously:

[0045]

[0046] The coordinates of the center of the incident point of the emitted light on the screen can be obtained. .

[0047] A measurement system for solving the birefringence phenomenon of a uniaxial crystal and its two light trajectories includes a laser source, a tilted uniaxial crystal, a rotating stage, and a receiving screen. The rotating stage is perpendicular to the rotation axis, and the uniaxial crystal is placed on the rotating stage. The laser spot on the receiving screen is collected once every time the crystal is rotated by a set angle.

[0048] The uniaxial crystal is placed at an angle.

[0049] Beneficial effects: This invention establishes an improved model for systematically solving the trajectories of o-rays and e-rays emitted from birefringence in uniaxial crystals. By using a unified coordinate system to handle arbitrary crystal rotation operations, the computational complexity of complex scenarios is significantly reduced. The systematic intersection point solving algorithm proposed in this invention, combined with the e-ray wave vector / ray separation calculation model, improves the efficiency of spot positioning. In addition, the improved model proposed in this invention fully considers the influence of the non-parallelism of the front and rear surfaces of the crystal. It is a reasonable correction based on experimental conditions, which is more in line with the actual experimental environment and has a lower dependence on experimental adjustments. The simulation results obtained are in high agreement with the experimental data, providing reliable theoretical support for high-precision inversion of crystal optical axis direction and other parameters. Attached Figure Description

[0050] Figure 1 This is a schematic diagram of the theoretical model of the present invention;

[0051] Figure 2 This is a schematic diagram of the measurement system of the present invention;

[0052] Figure 3 This is a diagram of dual-light trajectory data collected in the experiment of this invention;

[0053] Figure 4 This is a dual-light trajectory data diagram based on the improved model of this invention. Detailed Implementation

[0054] The invention will now be further described with reference to the accompanying drawings.

[0055] Example 1

[0056] The method for solving the birefringence phenomenon of uniaxial crystals in this embodiment, by introducing corrections for the crystal surface tilt angle and the spatial orientation of the optical axis, not only simplifies the experimental operation but also more accurately fits the actual behavior of o-ray and e-ray, significantly improving the accuracy and practicality of birefringence prediction.

[0057] like Figure 1The diagram shown illustrates the theoretical model of the experiment in this embodiment, including incident light from a laser source, a uniaxial birefringent crystal, and a receiving screen. The front and rear surfaces of the birefringent crystal are not parallel, and its rotation around the z-axis can be controlled. The incident light strikes the intersection of the crystal's incident surface and the rotation axis at a certain angle, undergoing a first refraction on the front surface to produce o-rays and e-rays. These propagate within the crystal and are refracted again on the rear surface, propagating in different directions. After traveling a distance in a straight line through the air, they illuminate different positions on the screen. As the crystal rotates, the o-ray and e-ray illuminate different positions on the screen. After one complete rotation of the crystal, the trajectories of the o-ray and e-ray can be recorded on the screen.

[0058] The method for solving the two-light trajectory of birefringence in a uniaxial crystal in this embodiment specifically includes the following steps:

[0059] S1. Let the normal vector of the front surface of the crystal be... The normal vector of the back surface is The optical axis direction vector is These vectors can be obtained through vector rotation operations. Specifically, they are derived from the rotation axis (z-axis) direction vector. Rotate about the x-axis respectively Then rotate around the y-axis The result was:

[0060] in, and This is a rotation matrix for rotations about the x-axis and y-axis.

[0061] S2, when the crystal is wrapped Axis rotation When rotating the crystal around the z-axis, the orientations of the front and back surfaces and the optical axis must be rotated accordingly. Therefore, the aforementioned vectors also need to be transformed accordingly. The rotation matrix of the crystal around the z-axis is given as... The normal vector of the front surface of the rotated crystal , back surface normal vector and optical axis direction vector for:

[0062]

[0063] S3, when light travels from a direction with a refractive index of _____ The medium incident on the refractive index When the surface of a crystal is exposed, according to the law of refraction, the direction vector of the o-ray within the crystal is... The calculation steps include: First, calculating the angle of incidence. cosine = According to Snell's Law Calculate the angle of refraction Finally, using the vector form of the law of refraction, the direction vector of the o-ray within the crystal can be obtained as follows:

[0064]

[0065] For e-rays, their refraction behavior is more complex, and the direction of their wave vector must also be considered. And the direction of energy flux density (i.e., the direction of light rays). The wave vector direction satisfies the boundary condition, meaning that the incident wave vector and the refracted wave vector are continuous in their tangential components at the interface: .in, It is the effective refractive index of the e-ray, relative to the wave vector direction. and optical axis direction The included angle Relevant. The effective refractive index satisfies the wave vector equation:

[0066] Wave vector direction It can be decomposed into a component perpendicular to the interface and a component parallel to the interface. The component parallel to the interface is determined by Snell's law, and its magnitude is the same as that of the incident light. The component perpendicular to the interface is determined by the wave vector surface equation and the boundary conditions. Its expression is:

[0067]

[0068] After obtaining the wave vector direction, the direction of the e-ray within the crystal is used. The relationship between the energy flow direction and the wave vector can be used to obtain the propagation direction of the e-ray within the crystal:

[0069]

[0070] S4, o-ray and e-ray from crystal (refractive index) ) ejected into the external medium (refractive index) When the light exits, refraction will occur again. The vector of the light exit direction... The derivation process is similar to that of the first refraction. First, the angles of incidence inside the crystal and at the back surface are calculated. cosine According to Snell's Law Calculate the angle of refraction of the o-ray Finally, using the vector form of the law of refraction, the direction vector of the o-ray propagation outside the crystal is determined:

[0071]

[0072] When the e-ray exits from the back surface of the crystal, it is refracted again. Outside the crystal, the wave vector direction... With respect to the direction of light Becoming the same again ( The derivation process is similar to that for incident on a crystal, the difference being that the incident medium is a crystal and the exit medium is an isotropic medium. Using the continuity condition of the tangential component of the wave vector at the interface, we can obtain... The effective refractive index This can be obtained by solving the wave vector surface equation:

[0073]

[0074] in, It is Wave Arrow and optical axis The included angle. Wave vector direction. It can be represented as:

[0075]

[0076] S5. Calculate the coordinates of the intersection points of the light ray with each plane using the point-to-point method, where the coordinates of the incident point between the light source and the front surface are: Specifically, it includes:

[0077] (1) Calculate the coordinates of the intersection point between the e-ray and the back surface.

[0078] Using point-to-point (point is) , The three components After combining the surface equation and the ray equation:

[0079]

[0080] The intersection point between the emitted e-beam and the rear surface can be obtained. .

[0081] (2) Calculate the coordinates of the intersection point between the o-ray and the back surface.

[0082] Using point-to-point (point is) , The three components After combining the surface equation and the ray equation:

[0083]

[0084] The intersection point between the emitted o-beam and the rear surface can be obtained. .

[0085] (3) Calculate the coordinates of the intersection point between the e-ray emitted and the receiving screen.

[0086] Given that d1 is the length of the crystal along the z-axis, and L is the distance between the rear exit surface of the crystal and the receiving screen along the z-axis, then the position of the receiving screen is the position of the screen. Using point-to-point (point is) , The three components Solve the equations of the receiving screen plane and the e-ray emitted simultaneously:

[0087] The coordinates of the center of the incident point of the e-ray emitted on the screen can be obtained. .

[0088] (4) Calculate the coordinates of the intersection point between the outgoing o-ray and the screen.

[0089] Using point-to-point (point is) , The three components Solve the equations of the receiving screen plane and the equations of the emitted ray:

[0090]

[0091] The coordinates of the center of the incident point of the emitted light on the screen can be obtained. .

[0092] Based on the above model, the crystal is rotated once, and the coordinates of the incident points of the o-ray and e-ray emitted light on the receiving screen are calculated at different rotation angles, thus obtaining the dual-light trajectory.

[0093] Example 2

[0094] This embodiment is an experimental example, based on... Figure 1 The given model diagram was used to build... Figure 2 The measurement system shown includes a laser source 1, a uniaxial crystal 2 (whose front and rear surfaces are not necessarily parallel and are placed at an angle), a rotating stage 3, and a receiving screen 4 (CCD camera). During the experiment, it is necessary to measure the thickness of the uniaxial crystal 2 along the rotation axis and the distance from the receiving screen 4 to the rear surface of the uniaxial crystal 2 along the rotation axis. The uniaxial crystal 2 is placed on the rotating stage 3 perpendicular to the rotation axis. The emitted e-beam and o-beam laser spots on the receiving screen 4 are collected every 30 degrees of rotation. The center coordinates of the spots are calculated using the centroid method. After one rotation, the centroid coordinates of the o-beam and e-beam spots are extracted and plotted to obtain the corresponding dual-beam trajectory. The experimentally collected dual-beam trajectory data and its Bspine interpolation are shown below. Figure 3 As shown.

[0095] The parameters of other main instruments used in the experiment were as follows: the CCD resolution was 5496×3672, and the single pixel size was... The laser source is a helium-neon laser with a wavelength of 632.8 nm. The uniaxial crystal is a high-quality Iceland spar crystal from calcite crystals, with a thickness of 25 mm. Its corresponding o-ray and e-ray refractive indices are respectively... = 1.6557, =1.4852. The crystal is placed on a rotating stage and rotates synchronously with it.

[0096] Using the improved model of this invention, the incident polar angle can be calculated by following these steps. = -0.0187°, rotation angle of the front surface about the x-axis =14.1102°, rotation angle of the rear surface about the x-axis =13.0705°, the angle of rotation of the optical axis around the x-axis = -70.5184°, incident light azimuth angle = -180.0000°, rotation angle of the front surface about the y-axis = -5.5154°, rotation angle of the rear surface about the y-axis = -5.3647°, the angle of rotation of the optical axis around the y-axis = 101.2265°, crystal rotation angle Iceland spar crystals under certain parameters (i.e.) ( ), d=25mm, L= Under these 12 parameters (mm), the coordinates of the incident points of the o-ray and e-ray emitted light on the screen are:

[0097] S1. Calculate the normal vectors of the front and back surfaces of the crystal. , and optical axis direction vector .

[0098] S2, Calculation of crystal rotation The normal vectors of its front and back surfaces at that time and and optical axis direction vector .

[0099] S3, Calculate the refractive index of light. A medium with a refractive index of 1 is incident on a medium with a refractive index of 1. = or When the crystal surface is exposed, the direction of the wave vector of the o-ray inside the crystal is... and the wave vector direction of e-ray And by utilizing the direction of the e-ray within the crystal... The relationship between the energy flow direction and the wave vector square allows us to obtain the propagation direction of the e-ray within the crystal. .

[0100] S4. Calculate the wave vector directions of the outgoing o-ray and e-ray from the crystal, where the wave vector direction of the o-ray is... The direction of the e-wave vector is .

[0101] S5, the coordinates of the light source and the incident point on the front surface are: By using the point-to-point method to calculate the coordinates of the intersection points of the ray with each plane, the coordinates of the intersection point between the e-ray emitted and the receiving screen can be obtained. And the coordinates of the intersection point of the emitted light and the screen. .

[0102] Repeating the above steps, the remaining 11 crystal rotation angles can be obtained ( The coordinates of the intersection points of the o-ray and e-ray of the Iceland spar crystal on the receiving screen. These coordinates together constitute the required dual-ray trajectory data, and can be further used to obtain the simulated trajectory when the rotation angle sampling rate is increased, such as... Figure 4 As shown. For ease of comparison, in Figure 4 The experiment trajectory data is also presented. The results show that the model results are consistent with... Figure 3 The experimental data showed high consistency, with a root mean square error of 0.1620 mm.

[0103] This invention can effectively solve the outgoing light trajectories of o-rays and e-rays in uniaxial crystal birefringence. The model exhibits superior experimental consistency and accuracy, providing a more reliable theoretical tool for related research and applications, and possesses greater applicability and broader application potential.

Claims

1. A method for solving the two-ray trajectory of birefringence in a uniaxial crystal, characterized in that, Includes the following steps: S1. Define the crystal's front surface normal vector, back surface normal vector, and optical axis direction vector, and represent its spatial orientation through rotation operations around the x-axis and y-axis; S2. Calculate the normal vector and optical axis direction vector after rotation based on the rotation angle of the crystal around the z-axis; S3. Based on the law of refraction, Snell's law and Fresnel's equation, derive the propagation directions of o-ray and e-ray within the crystal; S4. Considering the refractive behavior of the back surface of the crystal, calculate the direction vectors of the o-ray and e-ray after they are emitted; S5. Using the point-direction equation and the ray tracing method, determine the coordinates of the o-ray and e-ray on the observation screen, rotate the crystal once, and generate the double ray trajectory.

2. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 1, characterized in that, Specifically, S1 is: Let the normal vector of the front surface of the crystal be... The normal vector of the back surface is The optical axis direction vector is , by the direction vector of the rotation axis Rotate about the x-axis respectively Then rotate around the y-axis The result was: in, Let be the rotation matrix about the x-axis; Let be the rotation matrix about the y-axis.

3. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 2, characterized in that, Specifically, S2 is: Given that the rotation matrix of a crystal about the z-axis is The normal vector of the front surface of the rotated crystal , back surface normal vector and optical axis direction vector for: 。 4. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 3, characterized in that, In S3, the propagation direction vector of the o-light within the crystal is... The calculation steps include: First, calculating the angle of incidence. cosine = According to Snell's Law Calculate the angle of refraction Finally, using the vector form of the law of refraction, the direction vector of the o-ray within the crystal can be obtained as: 。 5. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 4, characterized in that, In S3, when calculating the propagation direction of the e-ray within the crystal, its wave vector direction is also considered. and the direction of light The wave vector direction satisfies the boundary conditions: ,in, It is the effective refractive index of the e-ray, relative to the wave vector direction. and optical axis direction The included angle Regarding this, the effective refractive index satisfies the wave vector equation: Wave vector direction It can be decomposed into a component perpendicular to the interface and a component parallel to the interface. The component parallel to the interface is determined by Snell's law, and its magnitude is the same as that of the incident light. The component perpendicular to the interface is determined by the wave vector surface equation and the boundary conditions, and its expression is: After obtaining the wave vector direction, the direction of the e-ray within the crystal is used. From the relationship with the wave vector direction, we can obtain the corresponding propagation direction of the e-ray within the crystal: 。 6. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 5, characterized in that, In S4, the direction vector of the emitted o-light The calculation process is as follows: First, calculate the incident angle between the inside of the crystal and the back surface. cosine According to Snell's Law Calculate the angle of refraction of the o-ray Finally, using the vector form of the law of refraction, the direction vector of the o-ray propagation outside the crystal is determined: 。 7. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 6, characterized in that, In step S4, when the e-ray exits from the rear surface of the crystal, it is refracted again. Outside the crystal, the wave vector direction... =Direction of light At this point, the incident medium is a crystal, and the exit medium is an isotropic medium. Using the condition that the wave vector is continuous in the tangential component at the interface, we can obtain... The effective refractive index This can be obtained by solving the wave vector surface equation: in, It is Wave Arrow and optical axis The angle between the two sides, the direction of the wave vector Represented as: 。 8. The method for solving the two-ray trajectory of birefringence in a uniaxial crystal according to claim 7, characterized in that, S5 specifically includes: (1) Calculate the coordinates of the intersection point between the e-ray and the back surface. Using point-to-point formulas, the point is... , The three components After combining the surface equation and the ray equation: The intersection point between the emitted e-beam and the rear surface can be obtained. ; (2) Calculate the coordinates of the intersection point between the o-ray and the back surface. Using point-to-point formulas, the point is... , The three components After combining the surface equation and the ray equation: The intersection point between the emitted o-beam and the rear surface can be obtained. (3) Calculate the coordinates of the intersection point between the e-ray emitted and the receiving screen. Given that d1 is the length of the crystal along the z-axis, L is the distance between the rear exit surface of the crystal and the receiving screen along the z-axis, and the position of the receiving screen is the position of the screen. Using point-to-point formulas, the point is... , The three components Solve the equations of the receiving screen plane and the e-ray emitted simultaneously: The coordinates of the center of the incident point of the e-ray emitted on the screen can be obtained. ; (4) Calculate the coordinates of the intersection point between the outgoing o-ray and the screen. Using point-to-point formulas, the point is... , The three components Solve the equations of the receiving screen plane and the emitted ray simultaneously: The coordinates of the center of the incident point of the emitted light on the screen can be obtained. .

9. A measurement system for solving the two-ray trajectory of birefringence in a uniaxial crystal, the measurement system being capable of implementing the method for solving the two-ray trajectory of birefringence in a uniaxial crystal as described in any one of claims 1 to 8, characterized in that, It includes a laser source, a single-axis crystal, a rotating stage, and a receiving screen. The rotating stage is perpendicular to the rotation axis, and the single-axis crystal is placed on the rotating stage. Every time the stage rotates by a set angle, it collects the laser spot of the emitted e-light and o-light on the receiving screen.

10. The measurement system for solving the birefringence phenomenon of a uniaxial crystal and its two light trails according to claim 9, characterized in that, The uniaxial crystal is placed at an angle, and its front and back surfaces are not parallel.

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