1. The Field of View of a Thin Lens Interferometer Baseline=2B F F=range from array center to detector  ’’  Nulled here. B B 2Bsin  Bsin  2 Channels in Phase here Pathlengths from “in phase” positions: Top Channel 2Bsin  +(F-Bsin  )/cos  ’ Bottom Channel (F+Bsin  )/cos 

2. The Field of View of a Thin Lens Interferometer: Approximations Baseline=2B F F=range from array center to detector  ’’  Nulled here. Pathlengths from “in phase” positions: Top Channel 2Bsin  +(F-Bsin  )/cos  ’ Bottom Channel (F+Bsin  )/cos  APPROXIMATIONS: DeflectionAngles of each Channel:  ’~  Cos Approximation: cos  ~ 1-  2 /2 Taylor Expansion: q/(1+dx) ~ q(1-dx) Also Assumed: Internal workings of both lenses are identical! OPD between top and bottom channels: OPD ~B  2 sin 

3. OPD between top and bottom channels: OPD ~B  2 sin   FOV =  2 B)  = F 2 /(10B 3) Q: Given the OPD between the channels, what is the condition to NOT seriously distort a fringe on the detector? A: OPD <  Note that this condition also defines the field of View for an interferometer of this type. NOTE: This FOV calculation makes no assumptions about maximum graze Angles. Therefore, at most, the FOV will be ~0.5 degrees due to graze restrictions.

4. Designing a Mission 1: Baseline, Focal Length, FOV, and Formation Tolerances are derived. What angular resolution at what wavelength do you want?  res  What is the smallest X-ray pixel size(  m) you can imagine? s The baseline (2B) needs to be: 2B =  res The Focal Length (F) needs to be: F =  s/  res The FOV will be: FOV = 4/5  (s/ ) 2  res Tightest Formation Flying Tolerance between optics s/c = s. “Lateral” Longitudal Formation Flying Tolerance between optics s/c = 4/5  (s/ ) 2  res B

5. Some Typical Numbers 1 “MAXIM Pathfinder” 100  as Science (5x10 -10 radians)  Angstroms S=10  m  F=20 km  2B= 2 m  FOV=2.5 arcseconds  Long. Control=0.7 mm “Full MAXIM” 1  as Science (5x10 -12 radians)  Angstroms S=10  m  F=2000 km  2B= 200 m  FOV=25 mas  Long. Control=7  m

6. Designing a Mission 2: What angular resolution at what wavelength do you want?  res  What is the smallest X-ray pixel size(  m) you can imagine? s The baseline (2B) needs to be: 2B =  res The Focal Length (F) needs to be: F =  s/  res The FOV will be: FOV = 80  (s/ ) 2  res Tightest Formation Flying Tolerance between optics s/c = s. “Lateral” Longitudal Formation Flying Tolerance between optics s/c = 80  (s/ ) 2  res B The Difference Here is That we will have fringes 10x bigger than the CCD pixel Size.

7. Some Typical Numbers 2 “MAXIM Pathfinder” 100  as Science (5x10 -10 radians)  Angstroms S=10  m  F=200 km  2B= 2 m  FOV=250 arcseconds  Long. Control=7 cm “Full MAXIM” 1  as Science (5x10 -12 radians)  Angstroms S=10  m  F=20000 km  2B= 200 m  FOV=2.5 arcseconds  Long. Control=700  m

8. Mirror Module Dimensions The Mirror modules are pairs of flat (better than /100) mirrors. One mirror is fixed, the other has pitch (~mas) and yaw (arcminute) control. The module also has the ability to adjust the spacing of the mirrors at the nm level to introduce ~ angstrom pathlength control. Thermal control consistent to maintain optical figure (~0.1 degrees). There is structure to hold the module together.

9. The Mass of Glass 1: If a mirror length is “m”, and the graze angle is “g”, then the width of the mirror is m*sin(g)- in order to have square effective areas for each module. The effective area of one module will be: A module = (r*m*sin(g)) 2 Where r is the reflectivity off one mirror. m msin(g) Some Numbers: r~0.8 g~2 degrees sin(g)~1/30 => A module ~ m 2 /1400

10. The Mass of Glass 2: The “1/6” rule suggests that the thickness of the mirror be about 1/6th the length in order to preserve figure. If the density of the mirror is , then the mass of the glass of one module is: M glass =  m 3 sin(g)/3 m msin(g) Some Numbers:  ~2.5 g/cc sin(g)~1/30  M glass ~ m 3 /16 grams  (m in cm) m/6

11. The Mass of Glass 3: The ratio of mass to effective area becomes: Mass/area = r 2 m  /(3sin(g)) m msin(g) Some Numbers:  ~2.5 g/cc sin(g)~1/30 m/6 Mirror Length (cm) Area/Module (cm 2) Glass Mass Per Module (kg) Number of Modules to get 1000 cm 2 area. Total mass of glass to get 1000 cm 2 (kg) 100.0710.062514000875 300.6431.68815552624 1007.14362.51408750 NOTE- this mass estimate is for glass only. There may be some scaling of masses for structure and actuators- but that is not considered here.

12. Actuator Requirements: The pitch control should be to the some fraction of the diffraction spot size.  (m*sin(g))~30 /m ~ 6 mas for m=100 cm, 62 mas for m=10cm ~ 30 nm of control for anysize mirror. The range of pitch control should be able to accommodate the range of baselines over the range of focal lengths.  max ~ B/F =  s) ~ 1 arcsecond of range ~ 5x10 -6 m of linear range for a mirror of length m. where s=CCD pixel size

13. What would one of these modules look like? m msin(g) m/6 Gap~msin(g) Yaw ontrol Pitch Control msin(g) 3/2m+d m/3 + msin(g) 2(w+gap)+msin(g) By 2(w+gap)+msin(g)+m/3+actuator+encoder ASSUME: w+gap~5 cm Encoder+encoder~5cm Sin(g)~1/30 -->(10cm+m/30)x(15cm+m/3+m/30) -->m=30cm-> 13cmx26cm

14. Packing into a Rocket