查看“2020年GeCAA理论第1题-天文摄影”的源代码

==英文题目==

'''1 Astrophotography'''

An astrophotographer, based on the Equator, uses a good digital camera on a tripod,
but with no tracking. The camera has a telescopic lens with focal length of 150mm and
aperture (objective diameter) of 84mm. The sensor has effective light collecting diameter
of 22.5mm. The photographic target is a star field at the observer’s Zenith.

(a) (2 points) Calculate the field of view (the angular width of the image) which can
be captured on the sensor using this equipment.

(b) (5 points) The pixels in the camera’s sensor are separated by a distance of 3.23 μm.
What is the maximum possible exposure time for a single frame, so that no star trails
appear on the exposed image?

(c) (3 points) For a better-quality image of the star field, the photographer decides to
take multiple shots at the exposure time calculated in b) and then to stack them
together. The total time for all these shots is 30 s (ignore any time taken to write
data to the memory card) What proportion of the total field of view is possible at
the higher signal to noise ratio?

==中文翻译==

==官方解答==
''Solution''

(a) By simple right angle triangle,
$$FOV = 2 \times tan^{−1}
( \dfrac{sensor width}
{2 × Focal Length})
= 2 × tan^{−1}
(
\dfrac{22.5}
{2 × 150})$$

$$α ≈ 8.58°$$                                                                                                                                                             (''2.0 points'')

<nowiki>(b) Number of pixels on the sensor will be given by,
$$N =
\dfrac{Sensor width}{pixel width}=
\dfrac{22.5}{3.23 × 10^{−3}}
= 6965.94$$  (''1.0 point'')</nowiki>

As the number of pixels have to be integer, we take $$N = 6966$$                                                                           (''1.0 point'')

Angular coverage of each pixel will be,
$$Pixel view =
\dfrac{8.578°}
{6966}
= 0.001 23°/pixel
≈ 4.43′′/pixel$$   (''1.0 point'')

As the stars complete one full circle in 23.9344 hours,                                                                                                (''1.0 point'')

''student loses this one mark if 15° per hour is used''

$$t_1 = 1.23 × 10^{−3}\dfrac{23.9344h}
{360}
= 0.294 s
≈ 0.3 s$$                                                                           (''1.0 point'')

(in reality this is probably a factor of 10 smaller than the eye would detect)

(c) In 30 seconds, the sky will move by,
$$Δα =
\dfrac{360°}
{23.9344 × 120}
= 0.125 342°$$ (''1.0point'')

$$\therefore \dfrac{ α − Δα}
{α}
=
\dfrac{8.578 − 0.125342}
{8.578}
= 98.5%$$ (''2.0 points'')

(or 8.453° is total field of view in high resolution images from the stack)

Only penalise 15°/hr once in the whole solution.