2015IOAA理论短问题第6题-大气光深
英文题目
At the start of every observation, a radio telescope is pointed at a point-source calibrator that has a known flux density of 21.86 Jy outside the Earth’s atmosphere. However, on a certain date, the measured flux density of the calibrator source was 14.27 Jy. If the calibrator source was at an altitude of 35 degrees, estimate the zenith atmospheric optical depth, $$\tau_{z}$$.
中文翻译
每次观测开始时，射电望远镜都会对准一个点源校准源；已知该校准源在地球大气层外的流量密度为 21.86 央斯基（Jy）。但在某一日，测得该校准源的流量密度仅为 14.27 央斯基。若该校准源的地平高度为 35°，试估算天顶方向大气光学深度$$\tau_{z}$$
解答
官方答案
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官方答案 |
(50%)Atmospheric optical depth $${tau}_A$$ can be determined from the measured flux, that is: \[S_{meas} = exp(-{tau}_A) S_{real}\] \[14.27 = exp(-{tau}_A) \times 21.86 \implies exp(-{tau}_A) = 14.27 / 21.86 = 0.65\] Then \[{tau}_A = -\ln(0.65) = 0.43\]
(50%)Now we have:
\[{tau}_A = {tau}_z \sec z\]
\[{tau}_z = {tau}_A \cos z\]
\[= 0.43 \times \cos (90 - 35)^\circ = 0.25\]
where $$z$$ is zenith angle.
翻译
AI翻译
大气光学深度 $${tau}_A$$ 可以从测量流量密度确定，即： \[S_{meas} = exp(-{tau}_A) S_{real}\] \[14.27 = exp(-{tau}_A) \times 21.86 \implies exp(-{tau}_A) = 14.27 / 21.86 = 0.65\] 然后 \[{tau}_A = -\ln(0.65) = 0.43\]
现在我们有： \[{tau}_A = {tau}_z \sec z\] \[{tau}_z = {tau}_A \cos z\] \[= 0.43 \times \cos (90 - 35)^\circ = 0.25\] 其中 $$z$$ 是天顶角。
