A star with V = 21.0 is known to produce a count rate of 10 electrons per second
ID: 1502918 • Letter: A
Question
A star with V = 21.0 is known to produce a count rate of 10 electrons per second for a
certain telescope/detector combination. The detector read noise is 4 electrons per
pixel, and the dark rate is zero. Compute the exposure time needed to reach a SNR
= 10 under the following conditions:
(a) dark sky and good seeing: aperture radius = 3.5 pixels, sky brightness = 1.4
electrons per pixel per second;
(b) moonlit sky and poor seeing: aperture radius = 5.0 pixels, sky brightness = 4
electrons per pixel per second.
Explanation / Answer
The signal-to-noise ratio (SNR) is calculated as follows. First, the signal is simply the count rate (e/sec) from the star N measured through some aperture times the exposure time t
The noise is make up of three components, which are added in quadrature:
The photon noise from the star itself sqrt(N*t)
The photon noise from the sky sqrt(S*p*t),
The read-noise from the device sqrt(p)*R in the measuring aperture
Put this all together and we have:
SNR = N*t / sqrt(N*t + S*p*t + p*R^2)
where, N = count rate = 10
S is the count rate per pixel (e/sec/pixel)
p is the number of pixels in the measuring aperture
R is the read-noise per pixel
a) SNR = 10, N = 10, S = 1.4, p = 3.5, R = 4
10 = 10t/ sqrt(10t + 4.9t + 56)
=> t^2 - 14.9t - 56 = 0
Solving the above eqation, we get
The exposure time = t = 18 sec
b) SNR = 10, N = 10, S = 4, p = 5, R = 4
10 = 10t/ sqrt(10t + 20t + 80)
=> t^2 - 30t - 80 = 0
Solving the above eqation, we get
The exposure time = t = 32 sec
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