1 / 14

The Field of View of a Thin Lens Interferometer

The Field of View of a Thin Lens Interferometer. Pathlengths from “in phase” positions: Top Channel 2Bsin q +(F-Bsin q )/cos f ’ Bottom Channel (F+Bsin q )/cos f. 2Bsin q. B. 2 Channels in Phase here. B. Bsin q. f ’. F. q. f. Nulled here. Baseline=2B.

abedi
Download Presentation

The Field of View of a Thin Lens Interferometer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Pathlengths from “in phase” positions: Top Channel 2Bsinq+(F-Bsinq)/cosf’ Bottom Channel (F+Bsinq)/cosf The Field of View of a Thin Lens Interferometer: Approximations APPROXIMATIONS: DeflectionAngles of each Channel:f’~f Cos Approximation: cosf ~ 1-f2/2 Taylor Expansion: q/(1+dx) ~ q(1-dx) Also Assumed: Internal workings of both lenses are identical! OPD between top and bottom channels: OPD ~Bf2sinq f’ F q f Nulled here. Baseline=2B F=range from array center to detector

  3. Q: Given the OPD between the channels, what is the condition to NOT seriously distort a fringe on the detector? A: OPD < l/10 Note that this condition also defines the field of View for an interferometer of this type. • OPD between top and bottom channels: • OPD ~Bf2sinq • qFOV=l/(10f2B) • = lF2/(10B3) 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? qres, l Tightest Formation Flying Tolerance between optics s/c = s. “Lateral” What is the smallest X-ray pixel size(mm) you can imagine? s The baseline (2B) needs to be: 2B = l/ qres The Focal Length (F) needs to be: F = s/qres Longitudal Formation Flying Tolerance between optics s/c = 4/5(s/l)2 qresB The FOV will be: FOV = 4/5(s/l)2 qres

  5. Some Typical Numbers 1

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

  7. Some Typical Numbers 2

  8. Mirror Module Dimensions • The Mirror modules are pairs of flat (better than l/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: msin(g) 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: Amodule= (r*m*sin(g))2 Where r is the reflectivity off one mirror. m Some Numbers: r~0.8 g~2 degrees sin(g)~1/30 => Amodule~ m2/1400

  10. The Mass of Glass 2: msin(g) 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 s, then the mass of the glass of one module is: Mglass= sm3sin(g)/3 m m/6 • Some Numbers: • s~2.5 g/cc • sin(g)~1/30 • Mglass~ m3/16 grams • (m in cm)

  11. The Mass of Glass 3: msin(g) Some Numbers: s~2.5 g/cc sin(g)~1/30 The ratio of mass to effective area becomes: Mass/area = r2ms/(3sin(g)) m m/6 • 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. dq~l/(m*sin(g))~30l/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. qmax~ B/F = l/(20s) ~ 1 arcsecond of range ~ 5x10-6m of linear range for a mirror of length m. where s=CCD pixel size

  13. What would one of these modules look like? msin(g) msin(g) m m/3 + msin(g) 3/2m+d 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 m/6 Gap~msin(g) Yaw ontrol Pitch Control

  14. Packing into a Rocket

More Related