Download Second Year Problem sheet 1

Second Year Problem sheet 2
14th February 2003
Course ASTR2B15: Techniques in modern astronomy
(Answers to be handed in on Friday 21rd February at end of 1pm lecture)
1. Using aperture photometry on a V-band image it is found that a target star gives 13451
counts in a one arcsecond diameter circular aperture centred on the star (these counts
include the background of course). The sky background gives a total of 3564 counts in a
two arcsecond diameter circular aperture. What is the signal to noise of the observation?
A standard calibration star gives 21562 counts using a one arcsecond circular aperture
centred on the star (again including background). If the V-band magnitude of the standard
calibration star is known to be 15.0 (with negligible error), what is the magnitude (with
errors) of the target star? (Assume that there is no detector noise and that Poisson
statistics hold).
NB. Can use that if f 
 dI b2
df  f  2 
 I b
Ib
Ia
the error df is given by
1
dI a2  2
2.5 I
and magnitude error m  

ln 10 I
I a 2 
[10]
2. Give a description, including a diagram, of an optical astronomical imaging camera.
Explain the function of each component and what is meant by a pupil conjugate.
If the diameter of the telescope is 4.2m and the f-ratio of the beam at the telescope focus
is 11, what is the image scale in arcseconds per millimetre at the telescope focal plane?
(consider the telescope a simple lens) If the detector in the camera has pixels of size
30m by 30m and the size of the detector array is 2000 by 2000 pixels. What should the
magnification of the camera optics be to give a field of view on the detector with a
diameter of 2 arcminutes.
[10]
3. What are the main design constraints that determine the resolving power of a grating
spectrograph?
A reflection grating spectrograph (consisting of slit, collimating lens, main dispersion
grating, cross dispersion element, imaging lens and detector array) has a slit width of
200microns. The size of the main dispersion grating is 0.2x0.2m square, with a line
density of 100 lines per millimetre and is fully illuminated by the collimated beam. The
angle of diffraction off this grating for the 25th order at a wavelength of 500nm is 45
degrees. The pixel size of the detector array is 30x30 microns. The total magnification of
the spectrograph optical system is 0.1 and the focal length of the final imaging lens is
0.5m. What is the spectral resolving power of the spectrograph at 500nm in the 25 th
order? (show how you have decided on what determines the resolving power in this case)
[10]
1. Using aperture photometry on a V-band image it is found that a target star gives
13451 counts in a one arcsecond diameter circular aperture centred on the star
(these counts include the background of course). The sky background gives a total of
3564 counts in a two arcsecond diameter circular aperture. What is the signal to
noise of the observation?
A standard calibration star gives 21562 counts using a one arcsecond circular
aperture centred on the star (again including background). If the V-band magnitude
of the standard calibration star is known to be 15.0 (with negligible error), what is
the magnitude (with errors) of the target star? (Assume that there is no detector
noise and that Poisson statistics hold).
NB. Can use that if f 
df  f
Ib
Ia
 dI b2
 2
 I b
the error df is given by

1
2 2
dI a

I a 2 
Assuming poisson statistics, error on signal N   N
Therefore signal from target star = 13451 13451
Signal from background in one arcsecond = 3564 / 4  3564 / 4
= 891 891
Signal for star alone = (13451  891)  13451  891
= 12560  14342
Signal to noise ratio =
S
Ns  NB
12560


 104.9
N
NS  NB
14342
Signal from calibration star = 21562  21562
Signal for calibration star alone = (21562  891)  21562  891
= 20671  22453
[5]
Magnitude relation
ma  mb  2.5 log 10
Ib
Ia
Therefore
I b 20671

=1.646
I a 12560
ma  15  2.5 log 10 1.646
If f 
ma  15  0.541  15.541
1
 dI 2 dI 2  2
and the error df is given by df  f  b2  a2 
I a 
 I b
Thus
1
df  22453
14342  2


  0.012
2
f
(12560) 2 
 (20671)
The error on magnitude given by
m  
2.5 I
2.5

0.012  0.013
ln 10 I
ln 10
Students can put maximum or minimum values into above equation to get errors or use
this formula.
[5]
ma  15.541  0.013
2. Give a description, including a diagram, of an optical astronomical imaging camera.
Explain the function of each component and what is meant by a pupil conjugate.
If the diameter of the telescope is 4.2m and the f-ratio of the beam at the telescope
focus is 11, what is the image scale in arcseconds per millimetre at the telescope focal
plane? (consider the telescope a simple lens) If the detector in the camera has pixels
of size 30m by 30m and the size of the detector array is 2000 by 2000 pixels. What
should the magnification of the camera optics be to give a field of view on the
detector with a diameter of 2 arcminutes.
Simple camera design
1)
Collimating lens – produces a parallel beam of light.
2)
Aperture stop – a circular aperture placed at a pupil conjugate that cuts off stray light
from getting into the detector.
3)
Filter – filters the light to let through only a selection of wavelengths. Filters are
normally termed broad band or narrow band.
4)
Imaging lens – produces the final image on the detector. With the collimating lens
decides size of image on the detector (the plate scale – arcsec/mm).
5)
Detector – records the image and converts it into digital data. Usually has a shutter in
front of it
By selecting the focal lengths of the collimator and imaging lens one can either magnify or
reduce the plate scale.
[4]
A pupil conjugate is where an image of the pupil is formed by the subsidiary optics. At this
point rays from all sky angles converge
[2]
If the focal ratio = 11 and the telescope diameter D=4.2
1 arcsecond angle subtends the following distance at the telescope focus
=
2
x11x 4.2  2.24x10-4m
360 x60 x60
Therefore the plate (image) scale = 4.46 arcsec/mm
[2]
Size of pixel array is 2000x30microns = 60mm
If a 2 arcminute field is required (120 arcseconds), this requires an image scale of
120
 2 arcsec/mm
60
Therefore a demagnification of 2/4.46 is required
=0.448
[2]
3. What are the main design constraints that determine the resolving power of a
grating spectrograph?
A reflection grating spectrograph (consisting of slit, collimating lens, main
dispersion grating, cross dispersion element, imaging lens and detector array) has a
slit width of 200microns. The size of the main dispersion grating is 0.2x0.2m square,
with a line density of 100 lines per millimetre and is fully illuminated by the
collimated beam. The angle of diffraction off this grating for the 25 th order at a
wavelength of 500nm is 45 degrees. The pixel size of the detector array is 30x30
microns. The total magnification of the spectrograph optical system is 0.1 and the
focal length of the final imaging lens is 0.5m. What is the spectral resolving power of
the spectrograph at 500nm in the 25th order? (show how you have decided on what
determines the resolving power in this case)
Three factors determine a spectrograph's resolving power (spectral resolution).
a)
Slit size
Essentially the spectrograph forms an image of the slit on the detector for all
wavelengths. The size of this image on the detector will impose a resolution
limit.
b)
Resolvance of the grating
The grating has a fundamental limit to its resolution defined by it’s defraction
properties (Rayleigh limit).
R  Nm
where N is the number of grating lines and m is the order number.
c)
Detector pixel size
The size of the detector pixels also defines the spectral resolution as it also
imposes a limit on the  that can be detected.
If the slit size is 200microns and the magnification of the optics is 0.1 the size of the
slit on the detector is 20microns. This is smaller than the pixel size so this is not the
dominant effect on the resolution.
The resolvance of the grating is
R  Nm
If grating has 100 lines/mm, then N= 100*0.2*1000=20,000 and m=25
Therefore
R  500,000
Now consider linear dispersion of grating together with detector pixel size
dx
fm

d
a cos
  x
a cos
fm
If dx=30microns (pixel size), =45o a=1/100000=1.0x10-5m, f=0.5m and m=25
  30 x10 6
1.0 x105 cos 45
 1.697 x1011m
0.5 x 25
Therefore spectral resolution at 500nm is
500 x109
R
 29463
1.697 x1011
This is what decides the resolution of the detector.