Practice Questions

Question 1

M(obj) =

Answer 1

I(obj)(max)-I(obj)(min) / I(obj)(max)+I(obj)(min)

Question 2

I(obj)(max)

Answer 2

b0 + b1

Question 3

I(obj)(min) =

Answer 3

b0 - b1

Question 4

M(img) =

Answer 4

I(img)(max) - I(img)(min) / I(img)(max) + I(img)(min)

Question 5

I(img)(max) =

Answer 5

c0 + c1

Question 6

I(img)(min) =

Answer 6

c0 - c1

Question 7

M =

Answer 7

M(img)/M(obj)

Question 8

What happens to the image contrast and the modulation transfer function if the value of c1 is halved

Answer 8

contrast is halved

harder to distinguish edge of image

the peak of the PSF is reduced

the image is dimmer, less contrast

Question 9

<ε(iso)^2> =

Answer 9

(θ/θ(0))^5/3

Question 10

θ(0) =

Answer 10

r(0)/h

Question 11

Seeing =

Answer 11

β = 0.98 λ/r(0)

Question 12

Strehl ratio =

Answer 12

S = (r(0)/D)^2

Question 13

<ε(fit)^2> =

Answer 13

k/N^(5/6) (D/r(0))^(5/3)

Question 14

Marechal approximation

Answer 14

S ~ e^(<ε^2>)

Question 15

fewer actuators means

Answer 15

design and operation of AO system is easier

Question 16

angular resolution

Answer 16

α = 1.22 λ/D

Question 17

cut off frequency of the modulation transfer function

Answer 17

f(c) = D/λF

most cases F = 1

Question 18

phase structure function is

Answer 18

a measure of the phase variance between two locations on the phase front separated by r

Question 19

phase structure function =

Answer 19

D(φ)(r) = <|φ(x) - φ(x+r)|^2>

Question 20

coherence function is

Answer 20

an indication of the correlation of the phase front between two locations on the phase front

Question 21

coherence function =

Answer 21

B(φ)(r) = <|φ(x) φ(x+r)|>

Question 22

qualitatively explain this expression

D(φ)(r) = 2[B(φ)(0) - B(φ)(r)]

Answer 22

phase structure function eq and definition

coherence function eq and definition

B(φ)(0) is the correlation of the phase value with itself at the same location (since the shift parameter r is 0).

2[B(φ)(0) - B(φ)(r)] is then equivalent to the evaluation of the phase structure function.

Question 23

Show that

D(φ)(r) = k^2δh (∞ ∫ -∞) [D(N)(r,z) - D(N)(0,z)] dz

Answer 23

B(φ)(r) = k^2δh (∞ ∫ -∞) B(N)(r,z)dz

substitue into

D(φ)(r) = 2[B(φ)(0) - B(φ)(r)]

inserting B(N)(0,0)-B(N)(0,0)

Question 24

re-express the fried parameter in terms of temperature T and its corresponding structure constant C(T)

Answer 24

r(0) ∝ [ ∫ C^2(N)(h) dh]^(-3/5)

n - 1 ∝ P/T

dn/dT ∝ -P/T^2

C^2(N) = C^2(T) |dn/dT| ∝ C^2(T) P/T^2

=> r(0) ∝ [ ∫ C^2(T)(h) P/T^2dh]^(-3/5)

Question 25

Should the Fried parameter vary with zenith angle

Answer 25

r(0) ∝ [0.423k^2(cosγ)^-2 ∫ C^2(N)(h) dh]^(-3/5) and r(0) ∝ (cosγ)^(3/5) where γ is the zenith angle. We see that at large zenith angles, cosγ goes to 0. So the fried paramter is smaller which is expected for stronger turbulence

Question 26

Show that the phase structure function can be expressed as D(φ)(r) = 6.88 (τ/τ(0))^(5/3)

Answer 26

D(φ)(r) = 6.88 (r/r(0))^(5/3) r = v(bar)τ and τ(0) = r(0)/v(bar) substituting into D(φ)(r)

Question 27

<ε(servo)^2> =

Answer 27

(τ(0) f(3dB)^(-5/3)

Question 28

τ(3dB) =

Answer 28

1/f(3dB) where f(3dB) is the bandwidth of the servo

Question 29

Littrow configuration

Answer 29

θ(i) = θ(B) θ(m) = -θ(B) θ(m) = -θ(i)

Question 30

how should the FSR be considered when designing a spectrograph

Answer 30

can adjust the free spectral range by changing the spacing between grating rules

Question 31

groove separation

Answer 31

a measured in m

Question 32

optical path difference

Answer 32

OPD = Δnl

Question 33

refractive index =

Answer 33

n = 1+7.8x10^-7 P/T

Question 34

turbulence occurs around the

Answer 34

tropopause at an altitude of 10km with pressure 100mbar and temperature 200K

Question 35

number of grooves

Answer 35

N = 2π/Δφ where Δφ = Δl k

Question 36

in terms of 1 rad <ε(fit)^2> =

Answer 36

(N(1 rad)/N)^(5/6)

Question 37

<ε(total)^2> =

Answer 37

sum of different <ε^2> =

Question 38

for a loseless system

Answer 38

G(s) = G(r)

Question 39

Kolmogorov Model

Answer 39

describes how the Kelvin-Helmholtz instabilities are broken up into smaller eddies

Question 40

The main points of the Kolmogorov Model

Answer 40

1. The kinetic energy is added to the system at the outer scale 2. The energy then cascades down to the smaller scales 3. when the scale is sufficiently small, the air viscosity dissipates the kinetic energy and the formation of new eddies stops. The average diameter at the smaller eddies is 1mm.

Question 41

Shack-Hartmann wavefront sensor scheme

Answer 41

1. an array of lenslets to focus the incoming wavefront onto an array of CCDs. 2. Senses the position by measuring the gradient of wavefronts and is therefore, achromatic 3. a plane wavefront would be focused to a grid of equally spaced images 4. any distortion in wavefront will cause image to be focused away from central position

Question 42

Shack-Hartmann WS diagram

Answer 42

see notes

Question 43

Re-express the Fourier transform U in polar coordinates and evaluate the Fourier transform.

Answer 43

X = pcosθ and Y = psinθ dXdY= |J|dpdθ = pdpdθ x = ωcosψ and y = ωsinψ substitute into U Bessel function identify Bessel function in U n=0 , sigma = (θ-ψ) and x = kpω use Bessel function identities U(P) = U(0) = [2J(1)(ka)/(ka)]

Question 44

How does the fourier transform U change with a

Answer 44

the intensity takes the form of an Airy function -> resulting image an Airy disc for an ideal telescope the PSF is described by an airy function as a changes the height of the central peak of the airy function will increase and its width decreases a is the radius of the primary mirror. as a increases the width of the central spot of the Airy disc will decrease and the central spot will be brighter

Question 45

Fried parameter describes

Answer 45

the strength of the wave front distortion due to atmospheric turbulence

Question 46

the value of the Fried parameter is

Answer 46

the equivalent diameter of an ideal telescope for observations limited by random inhomogeneities in the atmosphere's refractive index

Question 47

Strehl ratio

Answer 47

the ratio between the intensity of the central spot of the point spread function with that expected in the ideal case

Question 48

the Strehl ratio compares

Answer 48

the observed intensity peak to that for a telescope working at the diffraction limit

Question 49

For telescopes with large primary mirrors, the observed image will be limited by

Answer 49

the length scale of atmospheric turbulence which is characterized by the Fried parameter

Question 50

for a wavefront sensing scheme that uses the same wavelength as the observation

Answer 50

α(SA) = λ/d

Question 51

The fried parameter is a measure

Answer 51

of the characteristic coherence length of the distorted wavefront

Question 52

isoplanatic angle is

Answer 52

the greatest angular distance between the guide star and the observing target. Where light from both pass through the same turbulent region.

Question 53

Why does the isoplanatic angle vary with zenith

Answer 53

The isoplanatic angle is smaller with increasing zenith angle because light passes through more turbulent atmosphere at larger zenith angles.

Question 54

Sodium Laser Guide Stars

Answer 54

There is a layer of sodium at altitudes of about 90km above sea level. Laser light is used to excite an atomic resonance line in this layer of sodium to produce an artificial guide star.

Question 55

How do laser guide stars improve the sky coverage

Answer 55

laser guide stars improve the sky coverage of AO systems because observations no longer need to be close to natural guide stars

Question 56

The cone effect is

Answer 56

the result of laser guide stars being much closer to the observer than their targets. As a result, light from the laser guide star probe all the turbulent air that distorts the observed wavefront for light from a distant star

Question 57

cone effect diagram

Answer 57

see notes

Question 58

As a increases the PSF

Answer 58

becomes narrower tending towards a delta function

Question 59

S =

Answer 59

I(0)PSF/I(0)airy

Question 60

as a increases the height

Answer 60

of the PSF increases thus the Strehl function decreases as a increases

Question 61

how to obtain an image from the point spread function

Answer 61

I(x,y) = O(x,y) * P(x,y)

Question 62

DΦ(r) = <|Φ(x) - Φ(x+r)|^2> =

Answer 62

(∞ ∫ -∞) |Φ(x) - Φ(x+r)|^2 dx

Question 63

<ε(iso)^2> =

Answer 63

(∞ ∫ h(0)) DΦ(r) dh

Question 64

PSF tells us

Answer 64

how the light from a source spreads out in an image

Question 65

Modulation transfer function

Answer 65

measure of how contrast is lost in the image

Question 66

Optical transfer function can be expressed

Answer 66

in terms of a phase structure function

Question 67

turbulence generates

Answer 67

temperature cells

Question 68

the long exposure OTF is given by

Answer 68

the phase structure parameter C^2(N) with height

Question 69

the seeing is

Answer 69

the FWHM of the atmospheric PSF

Question 70

atmospheric time constant

Answer 70

is the timescale over which a turbulence cell moves by its own size

Question 71

adaptive optics compensate

Answer 71

the wavefront by sensing the wavefront then actuating on a deformable mirror

Question 72

fitting error comes from

Answer 72

finite actuator spacing

Question 73

bandwidth error comes from

Answer 73

finite servo bandwidth

Question 74

free spectral range

Answer 74

is the wavelength difference at which orders start to overlap

Question 75

a blazed grating

Answer 75

concentrates light into the desired order and makes efficient use of power

Question 76

wavefront correction

Answer 76

is carried out by segmented or deformable mirrors

Question 77

Curvature WFS is

Answer 77

a simple alternative to SHWS

Question 78

doppler shift can be resolved by

Answer 78

using a grating in High order downside is that orders overlap contaminating the spectrum an echelle spectrograph overcomes this by using echelle grating followed by a cross disperser to create a 2D echellogram

Question 79

the etendue of a system is set by

Answer 79

the stop

Question 80

the bandpass of a spectrometer is

Answer 80

the minimum wavelength resolution achieved by matching diffraction from the entrance slit with that from the grating