Tutorial Zemax 9: Physical optical modelling I 2012-11-04 9 Physical optical modelling I 1 9.1 Gaussian Beams... 1 9.2 Physical Beam Propagation... 3 9.3 Polarization... 7 9.4 Polarization II... 11 9 Physical optical modelling I 9.1 Gaussian Beams Consider a simple model of a basic mode (TEMoo) laser resonator at a wavelength of = 632.8 nm. The retroreflecting mirror is plane and here the gaussian beam waist has a size of w o = 0.3 mm. After a distance of 50 mm, a thin ideal lens with focal length f = 150 mm is located inside the resonator. The outcoupling mirror has a radius of curvature of R 2 = -100 mm (concave). a) Determine the distance between the lens and the outcoupling mirror to get a stable gaussian fundamental mode. Prove the result with the gaussian beam transformation menue. What is the 2w-diameter of the beam at the outcoupling mirror? b) If the outcoupling component is a lens with 5 mm thickness and made of quarz, determine the outer radius to get a divergence of the outgoing laser beam of = 10 mrad. c) What is the distance behind the resonator, where the beam diameter is 5 mm? Solution: The resonator data ar established as follows: The thickness after the lens is set as variable. In the merit function, the operator GBPR is set at surface 4 with a target value of -100 mm. An additional virtual surface with distance 0 to the outcoupling mirror must be introduced here, because the radius of curvature is measured behind the surface (incorporating the refraction at the surface). The optimization delivers the desired distance d = 75.1126 mm. The gaussian beam transformation confirmes the radius of curvature of -100 mm. The beam diameter at the outcoupling is 2w = 0.323 mm.
2 b) Now the outer radius of the lens is set as variable while the distance d 3 is now fixed. In the merit function, the operator GBPD is used with the corresponding parameters. The result is a radius of R = 9.539 mm. c) Now the operator GBPW is used with target 2.5 mm and the last thickness is taken as variable. Since there are always two possible solutions for a quadratic gaussian beam matching problem, a positive starting value for the distance is necessary here. The desired distance is d = 266 mm.
3 9.2 Physical Beam Propagation Take the system of the previous exercise to evaluate the data by a more rigorous diffraction beam propagation. a) First check the beam diameter at the outcoupling mirror. b) Check the beam curvature and the final divergence of the outgoing beam. c) If a plano convex lens is with focal length f = 30 mm, thickness 3 mm and of BK7 is used in a distance of 400 mm behind the resonator, calculate the obtained focus width. Compare the result with the ideal case of an unperturbed gaussian beam. d) The last transition is a far field propagation. Try to improve the resolution be resampling the POP at the last surface or by using another operator. Solution: The data are as follows: For the beam data, a x-width of 4 mm and 1024 sampling points are used to get a proper resolution of the profile. The result of the beam profile is quite near to a gaussian beam.
4 For the peak irradiance of 23.625 W/mm -2, the corresponding intensity at 1/e 2 is 3.197 W/mm -2. The beam diameter is interpolated from the text window 2w = 0.32832 mm and is identical to within 5 digits with the result of the ideal gaussian beam. Unfortunatelly, there is no automatic performance evaluation option in Zemax for the numerically obtained results. b) If the phase function is calculated at the surface 4, the following figure is obtained: From the text window, we get an absolute phase on axis of o = -0.928908 rad. For a radius of r = 0.277115 we get = -3.059039 rad. Corresponding to the parabolic phase formula we get e i e 2 i r 2 R 2 r, R 89.47mm 2 which is considerably smaller than the paraxial result. o To find the divergence angle, a large distance of d = 10000 mm is set behind the resonator and the beam intensity is calculated again. From the 1/e 2 intensity we find the divergence angle w 99.617 0.0012734*1.12/ 0.00601 9.985mrad z 10000 which again is very close to the perfect value. If the starting peak intensity 110782.7 W/mm -2 is set, the peak is normalized to 1 at this distance.
5 c) The lens is established as required with a solve of the focal power F = 0.02 mm -1. The best focal distance of 18.9747 mm is determined by the quick focus option. The POP calculation than gives the following profile: It is seen, that the phase aberration of the beam has perturbed the gaussian profile. Therefore the focus diameter is not defined properly in this case. An estimation of the central peak alone gives for a 1/e 2 threshold the diameter 55.8 m. The ideal beam radius is 52.54 m. d) If it is tried to improve the resolution by resampling the last propagation with the following data
6 The result seems to be critical. The resampling don't work well. If the angulat spectrum operator is used, the result is completely nonsense:
7 9.3 Polarization Establish a system with only Hoya glasses. the incoming light is a collimated beam with 10 mm diameter at the wavelength 632.8 nm. There are 3 surfaces with the radii 30 / 5 / -30 mm. The first lens has the thickness 2 mm and is of BACL3, the second cemented lens with thickness 6 mm is of E-C3. a) Show, that the refractive indices of the two glasses are very close together. Determine the incidence angle of the marginal ray at the cemented surface. b) The refractive index of the glass is quite near to that of BK7. Therefore for all surfaces the coating ZEC- V633_BK7 is to be inserted in the corresponding surface property menue. Calculate the pupil map of the system, if the incoming collimated light is linear polarized in y-direction. What is the largest local transmission loss? What is the mean transmission loss? c) Investigate the results, if a wrong coating is used at the surface No. 3. Try and compare the results for the wrong substrate (ZEC_V633_SF11) and the wrong wavelength (ZEC_V1550_BK7). d) Analyse the properties especially of the cemented surface due to the large incidence angle by inspection of the coating here for the correct coating ZEC_V633_BK7. Solution: a) The system data looks as follows: The menue Prescription Data' shows, that the difference in the indices is only 0.000156.
8 If a ray trace is performed, the incidence angle of the marginal ray is obtained to be 67.4. b) The pupil map shows only a minor rotation of the linear polarized field component. The transmission fan shows a maximal decrease of the intensity on a value of approximately 80%. The mean value is approximately 95%.
9 c) The corresponding charts for SF11 as substrate (right column) and the wavelength 1550 nm (left column) have the following behavior. The false substrate has only a minor influence, the wrong wavelength enhanced the transmission loss considerably. The effect on the pupil phase map is only small in both cases. d) We get the following properties at the surface 3, where the maximum incidence angle is set to 67.4
10 It is seen, that the intensity drop towards the rim is in the range of 10% for the transmitted light, the phase changes by approximately 70. The strong effect onto the phase results in circular polarized light in the outer zone of the aperture cone.
11 9.4 Polarization II Establish an optical diode by ideal Jones matrix components and demonstrate their function by rotating the /4-plate. a) First set up a system with polarisator in y-direction, rotatable /4-plate and crossed analysator plate in a collimated beam. b) Find the rotation angle of the /4-plate, where the transmission is 10%. c) Establish a universal plot for rotating the /4-plate and demonstrate the Malus-law. What is the largest transmission of the setup? Now use as input a circular polarization. What is changed if the plate is rotated? Solution: a) Data of the system with ideal Jones-matrix surfaces: b) the rotation angle around z is set as variable und the operator CODA is used in the merit function. As a result, an angle of 19.62 is obatined. The Transmission menue confirms the optimization result. c) The universal plot shows the expected sin-variation with 90 period. The maximum transmission is 25%.
12 If now circular polarization is chosen as input, the curve reduces to a maximum transmission of 12.5%, the relative behavior is the same