# Presentation: 448927

## 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

## 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2sin

## OPD between top and bottom channels:

- 	OPD ~B2sin
- FOV=2B)
-              = F2/(10B3)

- 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.**

## 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 resB

## 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
 |

## 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 resB

- The Difference Here is
- That we will have fringes 10x bigger than the CCD pixel
- Size.

## 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
 |

## 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.

## 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:
- Amodule= (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
- =>  Amodule~ m2/1400

## 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:
- Mglass= m3sin(g)/3

- m

- msin(g)

**Some Numbers:**
- 	~2.5 g/cc
- 	sin(g)~1/30
- Mglass~ m3/16 grams
- (m in cm)

- m/6

## The Mass of Glass 3:

- The ratio of mass to effective area becomes:
- Mass/area = r2m/(3sin(g))

- m

- msin(g)

**Some Numbers:**
- 	~2.5 g/cc
- 	sin(g)~1/30

- m/6

| Mirror Length (cm) | Area/Module
(cm2) | Glass Mass Per Module
(kg) | Number of Modules to get 1000 cm2 area. | Total mass of glass to get 1000 cm2
(kg) |
| --- | --- | --- | --- | --- |
| 10 | 0.071 | 0.0625 | 14000 | 875 |
| 30 | 0.643 | 1.688 | 1555 | 2624 |
| 100 | 7.143 | 62.5 | 140 | 8750 |

- 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.

## 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
- 	      ~ 5x10-6m of linear range for a mirror of length m.
- where s=CCD pixel size

## 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

## Packing into a Rocket