Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.stsci.edu/ftp/instrument_news/FOC/Foc_handbook/v5/handbook_v5_ch07.ps Äàòà èçìåíåíèÿ: Fri Dec 9 19:44:32 1994 Äàòà èíäåêñèðîâàíèÿ: Sun Dec 23 20:15:10 2007 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: zodiacal light

FOC Instrument Handbook Version 5.0 73
7.0 OBSERVER'S GUIDE(PRESCRIPTION FOR ESTIMATING EXPOSURE TIMES)
The first step consists in specifying the required signal to noise ratio S/N or the
relative accuracy ffiN=N = (S=N) \Gamma1 of the measurement. Then, the exposure time required
to attain that accuracy is given, in general, for Poisson statistics, by:
t = (S=N) 2 (R S + 2R B )R \Gamma2
S (1)
where R S is the source rate and R B the background rate in an appropriate resolution element
in counts sec \Gamma1 . The problem then simply reduces to properly estimating R S and R B .
For a point source in the FOC field of view and for a count rate per pixel much less
than N MAX calculated from Table 9, the source rate is given by:
R S = ú
4 D 2 (1 \Gamma p)
Z 1
o
ffl(–)F (–)Q(–)T (–)d– (2)
where:
D = diameter of the ST primary = 2.4 meters
p = ratio of obscured area to total area of primary mirror = 0.138
ffl(–) = fraction of energy intercepted by the appropriate resolution element
F(–) = source flux at ST in photons cm \Gamma2 sec \Gamma1 š A \Gamma1
Q(–) = FOC+OTA response function for T(–) =1 in counts photon \Gamma1
T(–) = transmission of filters or efficiency of dispersing elements
The terms in eq. (2) can be assumed to be appropriate averages over the pixel to pixel
variations in the instrument response function. Q(–) and T(–) are plotted in Figures 28 and
11 through 15.
The background rate, on the other hand, can be expressed, in general, as:
R B = nz
Ÿ
B p + ú
4 D 2 (1 \Gamma
p)\Omega p
Z 1
o
I B (–)Q(–)T (–)d–
–
(3)
where:
n = number of normal (z=1) or zoomed (z=2) pixels in appropriate resolution element
B p = inherent detector background count rate per normal pixel
I B (–) = specific intensity of diffuse background at ST in photons cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1
\Omega p = solid angle subtended by a normal FOC pixel in steradians.
Equations (2) and (3) can be evaluated numerically or by approximating them by
assuming that the spectral passband is sufficiently narrow. This permits the following sim­
plifications:
R S ' 3:9 \Theta 10 4 ffl(– 0 )F (– 0 )Q(– 0 )T (– 0 )\Delta– (4)
R B ' nz
''
B p + 6:9 \Theta 10
K
\Gamma10
I B (– 0 )Q(– 0 )T (– 0 )\Delta–
#
(5)
where all the relevant functions are evaluated at wavelength – 0 of peak response and \Delta– is
the FWHM bandpass of the instrument in Angstroms. The latter two parameters are listed
in Table 3. K takes on the numerical values 1 and 4 for the new F/48 and the new F/96
relays, respectively.

74 FOC Instrument Handbook Version 5.0
For an extended source, the size of the resolution element nz is determined by the user
according to his application. For a point source, the encircled energy tabulated in Table
8 should be used to determine ffl(–) and nz for each specific case. The precise area to be
used depends in general on the S/N ratio. If it is very high, one can afford to increase the
size of the resolution element nz to collect more photons, if it is low, nz should be kept as
small as possible. For any particular situation, there is an optimum nz at which the S/N is
maximum for a given exposure time t or at which t is minimum for a given S=N ratio. A
few quick calculations should be enough to locate this condition once the background has
been properly defined as indicated in the next paragraphs.
At least two sources of diffuse background have to be considered in estimating I B (– 0 )
in eq. (5). The first is residual airglow above the ST altitude of 500­600 km. For the FOC
bandpass of 1200--6000 š A only two features need to be considered: the HI, Lyman ff line at
1216 š A and the OI, 1304 š A triplet. The latter feature need only be considered for daytime
observations. Their contribution to R B can be evaluated via the graphs shown in Figure
39. In this graph, the second term in the brackets in eq. (5) is evaluated for the three FOC
relays for the condition T(– 0 ) = 1 as a function of spacecraft position in the orbit and for
a zenith oriented line of sight. Solar zenith angle 0 ffi corresponds to local noon, 180 ffi local
midnight. Lyman ff intensities can be expected to increase approximately a factor of 40%
if the line of sight drops to the horizon. R B can be determined by multiplying the data on
Figure 39 by the appropriate T(1216 š A) or T(1304 š A) and nz and adding to B p .
The second source of background is zodiacal light, which can be an important con­
tributor to R B in the 3000--6000 š A range. This contribution as a function of wavelength is
plotted in Figure 40 for the three relays. An intensity of 90 S10 units (' 3 \Theta 10 \Gamma4 photons
cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1 ) and a standard solar spectrum is assumed in these calculations. This
corresponds to a line of sight direction of ecliptic latitude fi = 40 ffi and helioecliptic longitude
– \Gamma – \Theta =85 ffi . Thus, R B for the zodiacal light can be computed by multiplying the results
shown in Figure 40 by the appropriate nz T(– 0 )\Delta– and by the factor S/90 where S can be
computed for any target position by means of the data tabulated by Levasseur­Regourd and
Dumont (Astr. Ap., 84, 277, 1980) and reprinted here for convenience as Table 12.
7.1 POINT SOURCES
How all this works in practice is best illustrated by some examples.
7.1.1 Imaging
How all this works in practice is best illustrated by some examples. Say that you are
interested in observing an unreddened A0V star of mu =24 with an accuracy of 10% using the
F342W filter, and the new F/96 relay with normal sized (unzoomed) pixels for which z = 1.
From Table 3 for the F346M filter, you find that – 0 =3480 š A, \Delta– =434 š A, QT(– 0 ) =0.045.
The stellar flux F (– 0 ) = 8:0 \Theta 10 2 \Theta 10 \Gamma0:4\Theta24 = 2:0 \Theta 10 \Gamma7 photons cm \Gamma2 sec \Gamma1 š A \Gamma1 .
Inserting these values into eq. (4), you get R S = 0:15 \Theta ffl(– 0 ) counts sec \Gamma1 . This is the total
count rate from that star spread out over a certain number of pixels corresponding to the
ffl(– 0 ) chosen from Table 8. If the star is reddened by a given total extinction A V , you should
use a standard or average reddening curve (see Savage and Mathis, Ann. Rev. Astr. Ap.,
17, 73, 1979 for an example) to deduce the appropriate A –0 . Then, R S can be multiplied
by 10 \Gamma0:4A –0 to take this effect into account in the simplest possible way. The possible

FOC Instrument Handbook Version 5.0 75
Table 12: Zodiacal Light Intensities in S10 Units
b
l - l O .
0š 10š 20š 30š 40š 50š 60š 70š 80š 90š
180 180 152 127 105 89 76 66 59 58 63
170 161 147 123 104 89 76 66 59 58 63
160 147 134 113 98 86 75 65 59 59 63
150 140 129 107 91 80 71 63 58 59 63
140 139 129 105 87 75 67 62 58 60 63
130 141 132 105 86 74 65 61 59 60 63
120 147 138 108 88 75 66 61 59 61 63
110 158 148 113 91 78 68 63 61 62 63
100 175 160 120 96 82 72 65 62 62 63
90 202 176 130 103 87 76 68 64 63 63
80 239 197 144 113 94 82 72 67 64 63
70 296 228 162 124 103 88 77 69 65 63
60 394 275 190 143 116 96 82 72 66 63
50 572 355 238 173 135 108 89 76 67 63
40 920 510 316 220 160 123 95 80 68 63
30 1930 825 475 285 194 140 103 84 70 63
20 355 226 157 111 88 73 63
10 260 167 117 90 74 63
0 275 170 118 90 74 63
inaccuracies introduced by this method are probably not worse than the uncertainties on
the validity of the reddening curve itself and/or the prediction of the continuous flux to be
observed.
Next, calculate R B from eq. (5) using the data in Figures 39 and 40. Far ultraviolet
airglow is not going to be a factor in the B bandpass. The zodiacal light background per
pixel can be estimated by means of the data graphed in Figure 40 and Table 12. Suppose
the star is viewed at fi = 15 ffi and – \Gamma – fi = 120 ffi for which S=120 S10 units. Then, this
contribution is 2:3 \Theta 10 \Gamma7 \Theta 120=90 \Theta 0:58 \Theta 434 = 8 \Theta 10 \Gamma5 counts sec \Gamma1 per normal pixel
so that, assuming B p = 7 \Theta 10 \Gamma4 counts sec \Gamma1 per normal pixel, eq. (5) can be written as:
R B =n ' 7 \Theta 10 \Gamma4 + 8 \Theta 10 \Gamma5 = 8:0 \Theta 10 \Gamma4 counts s \Gamma1 pix \Gamma1
Then, the required exposure time can be easily computed from eq. (1) and the data in
Table 8 for ffl(– o ). Using the data in the column marked F346M, one obtains:
n e(3400å) R S (cs ­1 ) R B (cs ­1 ) t(sec)
1 0.087 0.013 0.0008 8639
9 0.409 0.061 0.007 2016
21 0.528 0.079 0.017 1811
37 0.601 0.090 0.030 1852
69 0.722 0.108 0.055 1869
97 0.773 0.116 0.078 2021

76 FOC Instrument Handbook Version 5.0
Figure 39. Residual 1216 and 1304 š A airglow contribution to the FOC background counting
rate with no filters in place in counts sec \Gamma1 per normal pixel as a function of
the solar zenith angle at the spacecraft at 500km altitude. The line of sight is
assumed to be oriented towards the zenith.

FOC Instrument Handbook Version 5.0 77
2000 3000 4000 5000 6000
Figure 40. Zodiacal light contribution to the FOC background counting rate with no filters
in place in counts sec \Gamma1 š A \Gamma1 per normal pixel as a function of wavelength. The
zodiacal light intensity is assumed to be 90 S10 units.
Thus, integrating under the PSF out to a radius of 0.036 arcseconds at n = 21 provides
enough flux for the required S/N to be achieved in a minimum exposure time of 1800 seconds.
If the background rate for some reason had been 5 times higher, the minimum exposure time
would have been 3500 seconds at n = 9.
The accuracy of this approximation is, of course, a sensitive function of the shape of the
instrument bandpass and is, therefore, expected to be highest for the narrow, well­defined
passband filters with negligible red and/or blue leaks. It will certainly only give rough order
of magnitude estimates for the wide band pass filters for which a numerical integration of
eq. (2) is required for higher confidence predictions. If there are other point sources within
a few Airy radii of the primary source, their contribution to the background R B must be
evaluated by means of the appropriate system point spread function. It should also be kept

78 FOC Instrument Handbook Version 5.0
in mind that some background sources may vary in intensity during an exposure. This will
be the case for the airglow or scattered light emission sources for exposures lasting a good
fraction of an ST orbit (see Section 6.5). In this situation, it is advisable to pick the worse
case intensity to evaluate the required exposure time.
Particular attention has to be paid, in any case, to the expected count rate since it may
violate the assumption that R S n \Gamma1 Ü N MAX (see Table 9 and Section 6.2.2). If it does for
the particular format chosen as indicated in Table 6, either the format must be changed or a
neutral density filter inserted in order to drop the expected rate below the threshold. This,
of course, will also result in an increase in the exposure time required to reach the required
S/N ratio.
7.1.2 Spectroscopy
Similar computations can be carried out for a point source in the slit of the spectrograph
for the new F/48 relay except that, of course, the long slit spectrograph efficiencies plotted
in Figure 28 have to be used in equations (4) and (5). The bandpass \Delta– is now naturally
limited by the projected slit width of 0.06 arcseconds corresponding to 4, 2, 1.3 and 1 š A
for first, second, third and fourth order, respectively. The transmission of the order sorting
filter also has to be taken into account with special attention devoted to possible higher
order confusion if the filter has an appreciable near uv and visible leak and the source has
appreciable emission in these regions. This confusion can be eliminated completely for point
or pseudo point objects with the use of the objective prism FOPCD as the cross disperser.
In this case, the transmission and the dispersion of the prism given in Table 5 have to be
factored into the calculations.
For the case of the objective prisms, eq. (4) can be rewritten in the form:
R S (–) = 3:9 \Theta 10 4 ''(–)F (–)Q(–)T op (–)T (–)ffi–
where T op is the transmission of the prism tabulated in Table 5 and ffi– is the wavelength
interval in š A corresponding to the FOC spatial resolution. This interval can be expressed
simply as:
ffi– = 2r(–)D(–)PS \Gamma1
where D(–) is the prism dispersion in š A mm \Gamma1 tabulated in Table 5, r(–) is the radius of
the circle enclosing the required energy ffl(–) in arcseconds given in Table 8 and PS is the
plate scale of the appropriate relay in arcseconds mm \Gamma1 given in Section 3. Then, the source
count rate around – is:
R S (–) = 7:8 \Theta 10 4 ffl(–)r(–)D(–)F (–)Q(–)T op (–)T (–)PS \Gamma1 (7)
Equations (3) and (5) for the noise calculations remain the same except that some simpli­
fication can be introduced due to the fact that the overwhelming sources of background in
the case T(–) = 1 are the system integrated zodiacal light and the geocoronal Lyman ff line.
Thus, in this case, eq. (5) can be written as:
R B = nz
''
B p + 2:7 \Theta 10 \Gamma5 S
K
+ c
b
(7:7 \Theta 10 \Gamma4 )I kR
#
(8)

FOC Instrument Handbook Version 5.0 79
where c = 0 for the NUVOPs and the FUVOP and FOPCD in the new F/48 relay, c = 1
for the FUVOPs on the new F/96 relay, b = 1 for the new F/48 relay, and b = 2 for the new
F/96 relay. S is the intensity of the zodiacal light in S10 units and I kR is the intensity of
the Lyman ff airglow in kilorayleighs.
To see how this works, suppose you want to observe a 20th visual magnitude QSO with
a š \Gamma2 spectrum and you want to compute the required exposure time to obtain a S/N=10
at 1700 š A with the FUVOP of the new F/96 relay. In this case, F(1700 š A)=10 \Gamma5 photons
cm \Gamma2 sec \Gamma1 š A \Gamma1 . From the data tabulated in Table 8, you find r(1700 š A)=0.08 arcseconds
for ffl(1700 š A) = 0:7 and, from the data in Table 5, D(1700 š A) = 11:72=24 \Theta 10 \Gamma3 = 488 š A
mm \Gamma1 and T op (1700 š A) = 0:88 while Q(1700 š A)=0.014 from Table 10. This means that the
source rate from eq. (7) at 1700 š A is 0.46 counts sec \Gamma1 . The count rate is spread over n = 97
pixels for z = 1 from Table (8). Assuming that S = 120 S10 for the zodiacal light, I kR = 5
kilorayleighs, B p = 7 \Theta 10 \Gamma4 counts sec \Gamma1 pixel \Gamma1 , and K=4, c=1, b=2, eq. (8) gives:
R B =97
''
7 \Theta 10 \Gamma4 + 2:7 \Theta 10 \Gamma5 \Theta 120
4
+ 7:7 \Theta 10 \Gamma4 \Theta 5
2
#
=
97
i
7 \Theta 10 \Gamma4 + 8:1 \Theta 10 \Gamma4 + 1:9 \Theta 10 \Gamma3 j
= 0:33 counts sec \Gamma1
Finally, the required exposure time is:
t =
100(0:46 + 2 \Theta 0:33)
0:46 2 = 529 seconds:
7.2 EXTENDED SOURCES
The prescription for an extended source deviates only slightly from the formulation
discussed so far provided R S is redefined as:
R S = nz
ú
4 D 2 (1 \Gamma
p)\Omega p
Z 1
o
I S (–)Q(–)T (–)d– (9)
= nz
6:9 \Delta 10 \Gamma10
K
Z 1
o
I S (–)Q(–)T (–)d–
where n is now the chosen number of normal (z = 1) or zoomed (z = 2) pixels in the required
resolution element and I S (–) is the specific intensity of the extended source in photons
cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1 . Equations (3) and (5) for R B need not be modified. Conversion
of other specific intensity units into photons cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1 can be executed via the
following relations:
Units Photons cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1
U magnitudes per arcseconds squared =3.2\Theta10 13 \Theta 10 \Gamma0:4U at 3600 š A
B magnitudes per arcseconds squared = 6.4\Theta10 13 \Theta 10 \Gamma0:4B at 4470 š A
V magnitudes per arcseconds squared = 4.3\Theta10 13 \Theta 10 \Gamma0:4V at 5560 š A
1 Rayleigh š A \Gamma1 = 8.1\Theta10 4
1 erg cm \Gamma2 sec \Gamma1 sr \Gamma1 š A \Gamma1 = 5 \Theta10 7 – ( š A)
1 Wm \Gamma2 Hz \Gamma1 sr \Gamma1 =1.5\Theta10 29 [–( š A)] \Gamma1

80 FOC Instrument Handbook Version 5.0
1 S10 =333
Suppose, for example, you want to observe a Lyman ff aurora above the limb of Jupiter
of intensity 20 kiloRayleighs with a spatial resolution of 0.28 arcseconds with a S/N = 10
with the new F/96 relay. You will be using 400 new F/96 pixels for this purpose. You
should use the F120M filter because it has the highest transmission at Lyman ff and the
lowest transmission at the longer wavelengths where the disk Rayleigh scattering spectrum
may overwhelm any far uv auroral features.
From Figure 13, you find that at – =1216š A, the F120M filter has T=0.1 and from
Table 10 you deduce that Q(1216 š A)=0.008. Then, since in this case the Jovian emission
line of width Ÿ1š A is much narrower than the instrumental bandpass of 86 š A, eq. (9) can be
written simply as:
R S ' 400 \Theta 6:9 \Theta 10
4
\Gamma10
\Theta 2 \Theta 10 4 \Theta 8:1 \Theta 10 4 \Theta 0:008 \Theta 0:1 = 0:09
counts sec \Gamma1 per resolution element
The background rate R B will be dominated by the detector background and the geo­
coronal Lyman ff airglow if the observation is carried out at night. From the curve marked
the new F/96, 1216 š A in Figure 39 for a typical observing configuration of 150 ffi local solar
zenith angle, you obtain 2.2 \Theta 10 \Gamma3 counts sec \Gamma1 pixel \Gamma1 looking towards the zenith. This
implies that, for B p = 7 \Theta 10 \Gamma4 counts sec \Gamma1 pixel \Gamma1 , you have:
R B = 400
h
7 \Theta 10 \Gamma4 + 2:2 \Theta 10 \Gamma3 \Theta 0:1
i
= 0:37
counts sec \Gamma1 per resolution element
This means that S/N=10 for this Jovian aurora and resolution can be reached in:
t = 75(0:09 + 2 \Theta 0:37)
0:092 2 = 2562 seconds
Observations at higher spatial resolution would require correspondingly longer exposure
times.
If this same aurora is to be observed against a planetary disk background of Lyman ff
emission of 15 kilorayleighs with the same accuracy, the relevant background rate becomes:
R B = 400
''
7 \Theta 10 \Gamma4 + 2:2 \Theta 10 \Gamma4 + 6:9 \Theta 10
4
\Gamma10
1:5 \Theta 10 4 \Theta 8:1 \Theta 10 4 \Theta 0:008 \Theta 0:1
#
= 400
h
7 \Theta 10 \Gamma4 + 2:2 \Theta 10 \Gamma4 + 1:7 \Theta 10 \Gamma4 i
= 0:44 counts sec \Gamma1 per res: el:
so that:
t = 25(0:09 + 2 \Theta 0:44)
0:092 2 = 2994 seconds

FOC Instrument Handbook Version 5.0 81
In this case, however, you might be looking onto the visible disk of the planet and the
visible leak will dominate the count rate. To estimate the visible leak contribution notice
that at '5000 š A, the F120M filter has a residual transmission of 10 \Gamma4 and assume the Jovian
spectrum to be solar with an intensity of ' 2 \Theta 10 6 Rayleighs š A \Gamma1 at 5000 š A. Thus, you can
approximate the effect by spreading this intensity over '1500 š A where Q(–) '0.03. Then,
with these assumptions:
R B ' 0:44 + 400
''
6:9 \Theta 10 \Gamma10
4 2 \Theta 10 6 \Theta 8:1 \Theta 10 4 \Theta 0:03 \Theta 10 \Gamma4 \Theta 1500
#
' 0:44 + 50 counts sec \Gamma1 per resolution element:
A solution to this problem would be to insert another filter into the beam to suppress
the visible contamination. A good choice would be F140W for which T(1216 š A) = 0.05 and
T(5000 š A) = 3 \Theta 10 \Gamma4 and:
R S = 0:09 \Theta 0:05 = 4:5 \Theta 10 \Gamma3 count sec \Gamma1 per resolution element
R B = 0:44 \Theta 0:05 + 400
''
6:9 \Theta 10 \Gamma10
4
\Theta 2 \Theta 10 6 \Theta 8:1 \Theta 10 4 \Theta 0:03 \Theta 10 \Gamma4 \Theta 3 \Theta 10 \Gamma4 \Theta 1500
#
= 0:02 + 0:015 = 0:035 counts sec \Gamma1 per resolution element
t =
235(4:5 \Theta 10 \Gamma3 + 2 \Theta 0:035)
(4:5 \Theta 10 \Gamma3 ) 2 = 9:2 \Theta 10 4 seconds = 26 hours
Obviously, this hypothetical program cannot be accomplished with the FOC. To reduce
the exposure time to physically realistic levels one needs to, say, reduce the required accuracy
and/or spatial resolution. For example, halving both the accuracy and the resolution yields
a more acceptable exposure time of 1.6 hours.
Finally, suppose you wish to image an extended object (a planetary nebula, for exam­
ple) with the new F/96 relay at the highest possible resolution in the zoomed configuration
for the biggest possible field of view. Suppose the object exhibits a line spectrum with a
surface brightness at Hfi of 5 \Theta 10 \Gamma13 ergs cm \Gamma2 sec \Gamma1 arcsec \Gamma2 and you wish to use the
F486N interference filter to isolate the line to an accuracy of 10%. From the data shown
in Figures 12 and 28 you find that at 4861 š A, T = 0:6, and Q = 0:03. From the conversion
relations on page 78, you note that I s (4861) = 5 \Theta 10 \Gamma13 \Theta 4:25 \Theta 10 10 \Theta 5 \Theta 10 7 \Theta 4861 =
5:16 \Theta 10 9 photons cm \Gamma2 sec \Gamma1 sr \Gamma1 . Thus, eq. (9) becomes for n = 1, z = 2:
R S = 2 \Theta 6:9 \Theta 10 \Gamma10
4 \Theta 5:16 \Theta 10 9 \Theta 0:6 \Theta 0:03 = 0:03 counts s \Gamma1 per zoomed pixel
From the data shown in Figure 40 and a zodiacal light brightness of 90 S10 and B p = 7 \Theta 10 \Gamma4
counts s \Gamma1 per normal pixel, eq. (5) becomes:
R B = 2[7 \Theta 10 \Gamma4 + 3 \Theta 10 \Gamma7 \Theta 0:63 \Theta 34] = 1:4 \Theta 10 \Gamma3 counts s \Gamma1 per zoomed pixel

82 FOC Instrument Handbook Version 5.0
because T (– o ) = 0:63 and \Delta– = 34 š A for the F486N filter from the data in Table 3. In
consequence, finally:
t = 100(0:03 + 2 \Theta 1:4 \Theta 10 \Gamma3 )
0:03 2 = 3644 seconds: