2024年IOAA理论第6题-星团摄影
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英文题目
T6. Cluster Photography (20 points)
An astronomer takes pictures, in the $$V$$-band, of a faint celestial target, from a place with no light pollution. The selected target is the globular cluster Palomar 4, which has an angular diameter of $$\theta = 72.0''$$ and a uniform surface brightness in the $$V$$-band of $$m_V = 20.6\ \text{mag/arcsec}^2$$. The observation equipment consists of one reflector telescope, with diameter $$D = 305\ \text{mm}$$ and $$f/5$$, and a prime focus CCD with quantum efficiency $$\eta = 80\%$$ and square pixels with size $$\ell = 3.80\ \mu\text{m}$$.
Given data:
- $$V$$-band central wavelength: $$\lambda_V = 550\ \text{nm}$$
- $$V$$-band bandwidth: $$\Delta\lambda_V = 88.0\ \text{nm}$$
- Photon flux for a 0-magnitude object in the $$V$$-band: 10000 counts/mm/cm$$^2$$/s
(a) (3 points) Calculate the plate scale (the angle of sky projected per unit length of the sensor) of the observation equipment in arcmin/mm.
(b) (4 points) Estimate the number of pixels, $$n_p$$, covered by the cluster image on the CCD.
(c) (13 points) With an exposure time of $$t = 15\ \text{seconds}$$, the astronomer obtains a signal-to-noise ratio of $$S/N = 225$$. Compute the brightness of the sky at the observation site, knowing that the CCD has a readout noise (standard deviation) of 5 counts per pixel and dark noise of 6 counts per pixel per minute. Give your answer in mag/arcsec$$^2$$. You may find useful: $$\sigma_{RON}^2 = n_p \cdot 1 \cdot RON^2$$ and $$\sigma_{DN}^2 = n_p \cdot DN \cdot t$$.
中文翻译
T6. 星团摄影（20分）
一位天文学家在无光污染地区使用$$V$$波段对暗淡天体目标进行拍摄。目标为球状星团帕洛玛4，其角直径为$$\theta = 72.0''$$，$$V$$波段表面亮度均匀，为$$m_V = 20.6\ \text{mag/arcsec}^2$$。观测设备包括一台口径$$D = 305\ \text{毫米}$$、焦比$$f/5$$的反射望远镜，以及量子效率$$\eta = 80\%$$、像元尺寸$$\ell = 3.80\ \mu\text{m}$$的方像素CCD。
给定数据：
- $$V$$波段中心波长：$$\lambda_V = 550\ \text{纳米}$$
- $$V$$波段带宽：$$\Delta\lambda_V = 88.0\ \text{纳米}$$
- $$V$$波段0等天体的光子流量：10000 计数/毫米/平方厘米$$^2$$/秒
(a) (3分) 计算观测设备的底片比例（传感器单位长度对应的天空角度），单位为角分/毫米。
(b) (4分) 估算星团图像在CCD上覆盖的像素数$$n_p$$。
(c) (13分) 在曝光时间$$t = 15\ \text{秒}$$下，信噪比$$S/N = 225$$。已知CCD的读出噪声（标准差）为5计数/像素，暗噪声为6计数/像素/分钟，计算观测地点天空亮度（以mag/arcsec$$^2$$为单位）。可用公式：$$\sigma_{RON}^2 = n_p \cdot 1 \cdot RON^2$$ 和 $$\sigma_{DN}^2 = n_p \cdot DN \cdot t$$。
官方解答
(a) **解答：**
底片比例计算公式为：
$$\text{Plate Scale} = \frac{1\ \text{mm}}{f} \cdot \frac{180 \times 60}{\pi}\ \text{arcmin/mm}$$
其中焦距$$f = D \times f/\text{ratio} = 305\ \text{mm} \times 5 = 1525\ \text{mm}$$。代入得：
$$\text{Plate Scale} = \frac{1}{1525} \times \frac{10800}{\pi} \approx 2.25\ \text{arcmin/mm}.$$
(b) **解答：**
星团直径在CCD上的投影长度为：
$$d_{\text{GC}} = \theta \times \text{Plate Scale}^{-1} = 72.0'' \times \frac{1\ \text{mm}}{2.25\ \text{arcmin}} \times \frac{1\ \text{arcmin}}{60''} \approx 0.532\ \text{mm}.$$
覆盖像素数为：
$$n_p = \frac{\pi}{4} \left( \frac{d_{\text{GC}}}{\ell} \right)^2 \approx \frac{\pi}{4} \left( \frac{0.532\ \text{mm}}{3.80 \times 10^{-3}\ \text{mm}} \right)^2 \approx 15,400.$$
(c) **解答：** 总信号$$N_{\text{GC}}$$计算如下：
$$N_{\text{GC}} = \eta \cdot \Phi_0 \cdot 10^{-0.4m_V} \cdot \frac{\pi D^2}{4} \cdot \Delta\lambda_V \cdot t \cdot \Omega_{\text{GC}},$$
其中$$\Omega_{\text{GC}} = \pi (\theta/2)^2 = 4071.5\ \text{arcsec}^2$$。代入数据得：
$$N_{\text{GC}} \approx 180,540\ \text{counts}.$$
噪声项为：
$$\sigma_{\text{total}}^2 = N_{\text{GC}} + N_{\text{sky}} + n_p \cdot RON^2 + n_p \cdot DN \cdot t.$$
由$$S/N = 225$$得：
$$N_{\text{sky}} = \left( \frac{180540}{225} \right)^2 - 180540 - 385000 - 23100 \approx 55,250\ \text{counts}.$$
天空表面亮度计算为：
$$m_{\text{sky}} = m_V + 2.5 \log_{10} \left( \frac{N_{\text{sky}}}{N_{\text{GC}}} \right) \approx 21.9\ \text{mag/arcsec}^2.$$
