AST 443/PHY 517
Homework 2 Solutions
Consider a 2-mirror telescope with a 4m diameter primary and a 0.5m diameter secondary.
The primary has a focal ratio f/2, that is, F1 =2.
1. What fraction of the light that passes through the aperture is blocked by the secondary
mirror?
(0.5/4)2 = 1/64
2. What is the focal length of the primary f1 ?
f1 = D1 F1 = 8m
3. In a Cassegrain telescope, the focus lies behind the primary mirror, so the back focal
length f1 β must be >0. What constraint does this give for the magnification of the
secondary m?
β > 0 therefore k(m+1)−1 > 0. k = D2 /D1 = 1/8, so m> 7
4. Let the magnification m=9. What is β, and how far behind the primary mirror is the
focus?
β = k(m+1)−1 = 10/8 - 1 = 0.25. The focus is f1 β = 2 meters behind the
primary.
5. What is the radius of curvature of the secondary R2 (for m=9)?
ρ=mk/(m-1) = 9/64 = R2 /R1 . R2 = 9/8 meters.
6. Explain why R2 is not the same as the distance from the secondary to the focus.
The light rays reflected by the primary are not parallel. It is a converging
beam.
7. For this system, verify that a Ritchey-Chrétien telescope with K1 = −1.02 has K2 =
−2.44. What would K2 have to be if the primary is spherical?
The equation that must be satisfied for an aberration free design is
1+K1 = k4 /p3 [K2 +(m+1)2 /(m-1)2 ]
where k is the ratio of the off-axis distances for a ray incident on the secondary to that of the primary, and p = R2 /R1 . k = p. Plug in. The
numbers are close, but not exact.
A spherical primary has K1 =0. K2 = 6.634
8. For imaging faint objects, would you prefer to use the Palomar 5m telescope with
photographic plates (2% quantumn efficiency), or the CTIO 0.9m with a CCD (70%
quantum efficiency). Why? Assume everything else is equal.
The grasp G is the product of the quantum efficiency and the telescope
area.
G5m
G0.9m
=
π(5/2)2
π(0.9/2)2
= 0.88
The 0.9m telescope is about 10% better.
9. What is the diffraction-limited resolution of the Hubble Space Telescope (diameter=2.4m)
at wavelengths of 1500Å, 5500Å (V-band), and 2.2µm (K-band). State your answers
in arcseconds. What is the maximum size of a diffraction-limited (not seeing-limited)
telescope on the Earth at those wavelengths?
The Rayleigh Criterion R=1.22λ/D rad = 251643λ/D arcsec. At 1550Å,
5500Å, and 2.2µm, the resolution is 16, 59, and 258 milliarcsec, respectively.
The ratio of the seeing disk to R is S/R = D/24(5500/λ)1/2 . This ratio =
10 at 5500Å, or 1.9 at 2.2µm. For a ground-based telescope the size of the
HST, seeing dominates over the diffraction limit at V (and barely so at K).
The comparison is irrelevant in the UV.
At what size does the diffraction limit equal the seeing disk? At 5500Å the
seeing disk is 0.59 arcsec. This is the Rayleigh limit for a 23 cm telescope.
Anything smaller is diffraction-limited. The limit is 1.13m at K.
10. Show that at a given speed, the number of photons detected in a 24µm pixel of a CCD
detector is independent of the size of the telescope (for a source large compared to the
pixel size),
Speed is a measure of the amount of flux collected on a pixel in a given time.
Speed ∝ (D/f )2 , where D is the aperture diameter and f is the focal length.
The plate scale is ∝ 1/f; the area subtended by a large source is ∝f−2 ; this
is the dilution of the source brightness. The amount of light collected ∝D2 .
Therefore, the flux per pixel increases as D2 and decreases as f2 . Since the
speed is fixed, so is the brightness (incident energy) per fixed pixel size.
(See Bradt, page 103.)
11. (PHY 517; extra credit for AST 443)
You want to observe an anonymous quasar with the HST for the purpose of looking for
its host galaxy. Let the V magnitude of the quasar be 20. Assume diffraction limited
optics at 5500Å. Assume a total system efficiency of 50% (half the photons that enter
the telescope are detected), and 0.05 arcsecond CCD pixels. If you can integrate for
one hour, what is the brightness (in magnitudes per square arcsecond) of the faintest
galaxy you can detect at distances of 0.1, 0.5, and 1.0 arcsec from the quasar? You can
assume a black background and negligible detector read noise. State clearly all other
assumptions.
A source with V=0 emits 996 photons/cm2 /s/Å. The HST has a 2.4m
diameter. The V band has a width of about 1000Å. The number of counts
detected in time T from a source of magnitude M in a telescope of diameter
D is 999*1000.*T*( D2 )2 . Divide by 2 to account for the system efficiency
(detector QE, filter throughput, secondary blockage, large-angle scattering,
etc). In 1 hour you should detect about 258,000 counts. These will be
in a 2-dimensional Gaussian with σ=59 mas. At this point, I generated
this 2-D Gaussian numerically with 50 mas pixels.The peak pixel detects
about 29,500 counts. at 0.1 arcsec (2 pixels from peak), there are about
7000 counts. By 0.5 arcsec there is less than 1 count detected from the
source. 7000 counts corresponds to about 16.4 mag, so this is a reasonable
estimate of the minimum detectable brightness of a source at V∼16.4. But
this is in a 50 mas pixel, which is 0.0025 square arcsec. A full square
arcsec would encompass 400 times the flux, so the surface brightness is
about V=8.4 mag/arcsec2 . At the greater distances, you would need to set
a detection threshold, since we have unrealistically set the background to
zero. A 3 count threshold corresponds to about 16.8 mag/sq arcsec.
You can also do this analytically.
The upshot is that
A.) it is hard to see faint extended objects near a point source, and that
B.) it is hard to detect low surface brightness objects at high spatial resolution.