1

2019-02-27 Prof. Herbert Gross Uwe Lippmann, Yi Zhong Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena

Exercise Solutions 16:

Optical Design with Zemax for PhD - Advanced

Exercise 16-1: Gaussian beam I

a) A resonator which works at the wavelength = 1.064 m with a length of 200 mm is given. The waist of the fundamental mode has a radius of 0.2 mm and is located in a distance of 70 mm from the first mirror. Determine the radii of curvature for both mirrors. b) Now assume the outcoupling component to be a thin lens made of BK7 with a plane outer radius. What is the divergence angle of the outgoing beam? Determine the beam diameter at a distance of 500 mm. c) Now add two cylindrical lenses with the focal lengths -25 mm and +100 mm out of the Thorlabs catalog behind the resonator working in the y-section. Adjust the intermediate distance to get the minimum value of divergence behind the beam train. What are the beam radii in x and y at the final lens? What are the divergence angles in both cross sections? Is the telescope adjusted to be afocal now? d) Now adjust the telescope to be ideally afocal. Is now the transform factor of the divergence exactly 4? Solution: a)

2

The radius of curvature of the Gaussian beam wave front is measured relative to the waist, therefore the signs of the wavefront radii are opposite to the sign of the required surface curvature. The radii of curvature are +269.268 mm and -237.298 mm for the two mirrors. It must be noticed, that in the convention of the gaussian beams table, the next medium to the outcoupling mirror must be air. In the definition of the beam, the distance 70 mm is defined to be positive (waist to the right of surface 1), in the gaussian beam parameter table, the distance is indicated to be negative (surface relative to the waist).

3

b)

The divergence angle outside of the resonator is 2.205 mrad, the beam diameter in a distance of 500 mm is 2w = 2 x 1.369 mm = 2.738 mm.

c)

4

The beam data behind the system is:

beam radius [mm] divergence [mrad]

x-section 4.888 2.205

y-section 1.484 0.228

The telescope is not afocal, in this case the divergence angles should have the ratio 1:4 according to the ideal telescope magnification.

d) In this case, a paraxial ray trace is used for the adjustment. Make sure to disable surface 2 prior to optimization to have both sides of the telescope collimated at the same time. After adjusting the telescope for afocal, re-enable surface 2:

5

The ratio is now 2.205 / 0.549 = 4.02.

6

Exercise 16-2: Anamorphic Gaussian beam Real semiconductor lasers usually exhibit an anamorphic behavior. The source point of the wave in the slow axis cross section lies inside the chip, the fast axis source point is located at the surface.

x :slow axis

y :fast axischip

ideal

cylinder lens

Due to this so called intrinsic astigmatism, the collimation with one lens cannot be obtained in both cross sections. This is demonstrated in the following exercise.

a) Establish an anamorphic semiconductor laser model at a wavelength of = 700 nm with the waist sizes of 0.001099 mm in x and 0.000527 mm in the y-cross section. What are the corresponding divergence angles of the beam? Establish a cylindrical perfect lens that acts only in the y-section and generates an intrinsic astigmatism of 0.1 mm at the same divergence. b) Now collimate the fast y-axis by an ideal lens of focal length f = 10 mm. What is the remaining divergence and waist location of the x-section behind the lens? c) Now insert a cylindrical lens working in the x-section with focal length of -20 mm. At the location where both beam cross sections have the same size, a second cylindrical lens shall collimate the x-section. Then the beam has a circular cross-section. What is the necessary focal length of this second lens? Solution: a) The gaussian beam initial data show, that the divergence values are 0.2 rad and 0.4 rad respectively. The cylindrical lens should work as a 4f-system. Therefore the focal length is 0.025 mm corresponding to a focal power of 40 mm-1. The positions of the waist at the surface 3 are now zty = 0 mm and ztx = -0.1 mm.

7

b) For the exact optimization of the distance between focal point and lens, the parameter GBPD ist used, to minimize the fast axis divergence. The resulting distance of the lens is very near to 10 mm corresponding to the focal length. Since we have a small Rayleigh length we are quite close to the geometric case.

The remaining divergence in x is 0.002029 and the waist is located at -1007.69, which means, the waist is located behind the lens in +z direction and the beam is converging. c) In the first step, the beam size in x and y is determined by the operator GBPS. In case of the X cross-section, the UseX flag must be set. It should also be noticed, that the waist radii have to be explicitly set in merit function again. Then the distance behind the first cylindrical lens is varied to optimize for minimum difference between the beam sizes.

8

The second target in the merit function is a minimum beam divergence in x. The parameters for y are included in the merit function for information. The resulting focal length is f = +42.266 mm.

9

Exercise 16-3: Gaussian beam waist transform

a) A Gaussian beam with the wavelength 550 nm and a waist size of w0 = 2 m should be

transformed by an ideal lens to have a waist of w0' = 4 m in the distance of exactly 200 mm. Find the lens focal length and the corresponding position of the lens to perform this requirement.

b) Now the initial waist has a size of 8 m and only a lens of focal length f = 100 mm is available.

What are the two possible positions of the lens to get a final waist of 4 m independent of the location? Solution: a) The Gaussian beam is established from the left with the corresponding data. The lens is movable, its position and strength is variable. The waist location and the waist position are required by the operators GBPP and GBPW. The result is a lens with focal length 44.4 mm at a distance 66.67 mm from the initial waist.

b) The real solution gives the distances 300 and 100 mm.

In principle there is also a second solution existent, which has a virtual position for the lens.

10

A Universal Plot of the waist size over the first distance shows both solutions:

11

Exercise 16-4: Gaussian Beam II

Consider a simple model of a basic mode (TEM00) laser resonator at a wavelength of = 632.8 nm. The retro-reflecting mirror is plane and here the Gaussian beam waist has a size of wo = 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 R2 = 100 mm (convex). 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 propagation analysis. 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 quartz, 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 is 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 (remember: the Gaussian wavefront radius has the opposite sign of the surface radius for the same curvature). The beam parameters are calculated after refraction at a surface. To get the parameters at the position of the mirror but before the interaction with the mirror we would need to introduce a dummy surface with thickness 0 right before the mirror. In our case, we don’t need it since the mirror is to be placed at the retro-reflecting position where no change in wave front radius occurs on the surface. The optimization delivers the desired distance d = 75.113 mm. The gaussian beam transformation confirmes the radius of curvature of -100 mm. The beam diameter at the outcoupling is 2w = 0.328 mm.

b) Now the outer radius of the lens is set as variable while the distance d3 is now fixed. In the merit function, the operator GBPD is used with the corresponding parameters. There are two possible solutions: One gives a positive lens and an external waist and one gives a negative lens with a virtual waist inside the resonator. Since we have convergent light in our starting system the optimization will converge towards the positive solution. To get the negative

12

solution we have to give the back side of the lens a rather strong concave curvature (e.g. +10 mm) to get diverging light. Then the result of the optimization is a 10 mrad diverging beam. The radius of the back surface of the lens is then R = 6.817 mm.

c) Now the operator GBPS 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 = 234.5 mm.

13

Exercise 16-5: Physical Beam Propagation I Take the system of the previous exercise (16-4) 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 = 20 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 using another propagator. Solution: The system data is 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.

14

For the peak irradiance of 23.625 W/mm2, the corresponding intensity at 1/e2 is 3.197 W/mm2. 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. Unfortunately, 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 mm we get = -3.059039 rad. Corresponding to the parabolic phase formula we get

mm

rRee

o

R

ri

i 47.892

,2

2

2

which is considerably smaller than the paraxial result.

15

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/e2 intensity we find the divergence angle 10 mrad. c) The lens is established as required with a solve of the focal power F = 0.05 mm-1. The focal distance of 19.031 mm is the position of the unperturbed Gaussian beam waist.

The POP calculation then 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/e2 threshold a radius of 5.7 m.

The ideal beam radius is 1.0 m.

16

d) If the angular spectrum operator is used, the result is completely nonsense:

17

Exercise 16-6: Physical Beam Propagation II

a) Establish a collimated input beam at the wavelength 0.7 m, a top hat profile and diameter 6 mm. This beam is focused by a plano-convex lens with radius of curvature 25 mm made of BK7. Determine the plane of best focusing, the residual wave aberration and the beam profile in this plane. Determine the radius of the first dark ring. b) Now introduce a circular stop to clip the outer diffraction rings in the focal plane. After re-collimating the beam with an achromate of focal length f = 100 mm out of a catalog inspect the intensity profile in the far field. c) Place a thin glass plate behind the system with a Zernike Fringe phase plate on one side.

Select a normalization radius of 10 mm and introduce an astigmatic coefficient of 5 . Show the phase after the plate and the intensity profile in a distance of 5000 mm. Solution: a)

The wave aberration has a peak valley value of 0.124 .

The POP propagation is performed with 1024 pixels, a guard range of 100 and a zoomfactor of 4.

The intensity profile shows some rings, the first zero is located at a radisu of 6.8 m.

18

b)

c)

19

The phase is toric for the Zernike coefficient c5 and the profile is elliptical.