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IOAA 2016, Теория, английски — Задача 8. U-Band photometry

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Всички материали от IOAA 2016

Theoretical Examination · 20 т.

Внимание: Бележка към темата

Both archived English question editions print T12 AstroSat, but both 27-page English worked keys substitute T12 Exoplanets. No matching AstroSat solution was found in the inspected full year inventory. Summary Answer Sheets and Cover Sheet named by the instructions were not found in this archived source pack. Instructions specify 13 questions totalling 300 points; both question editions contain 12 questions totalling 250 points. Printed numerical/formula discrepancies are retained: T8 uses c/lambda_max for nu_max and prints 1233 pi then 3813; T9 numerical velocity denominator lacks a power of ten and the stated semimajor axis is 20933 km; T11.4 prints a direct rather than inverse mass ratio in the radius-cubed relation; T11.7 prose says 166.67 Hz while its mass calculation uses 142.86 Hz. No inferred corrections are substituted.

Instructions to the Contestants

(1) Total time duration of this examination is five hours.

(2) This examination consists of

Questions	Marks each	Total
Question 1 to 5	10	10 x 5 = 50 marks
Question 6 to 10	20	20 x 5 = 100 marks
Question 11 to 13	50	50 x 3 = 150 marks
	Total	300 marks

(3) Inside the envelope you will find

a. Question paper in English

b. Question paper in your native language (if applicable)

c. Table of Constants in English

d. Table of Constants in your native language (if applicable)

e. A set of Summary Answersheets

f. Cover Sheet

(4) Use only black or blue pen for writing. For figures, you may use pencils.

(5) There is no negative marking.

(6) Some marks will be deducted if the final answer is given without detailed solution.

(7) Some marks will be deducted if the final answers have inappropriate number of significant digits, no units or wrong units.

(8) The necessary values of fundamental and astronomical constants should be taken from the Table of Constants provided to you.

(9) Instructions regarding using the answersheets

a. For each question a separate Summary Answersheet has been provided. Final answer(s) for each question / each part of the question must be written in the corresponding box in the Summary Answersheet. Write your contestant code and page number on each Summary Answersheet.

b. You should also show detailed solution in the blank space provided on each Summary Answersheet. If necessary, you can ask for extra blank sheets from the invigilators.

c. Use separate blank sheets for each question.

d. Write your contestant code, question number and page number on FRONT side of each blank sheet. Write your contestant code and page number on the BACK side of each blank sheet.

e. The page numbers should be continuous, i.e., if you use 20 sheets for the entire examination (including the Summary Answersheets), the page numbers should run from 1 to 40.

f. Write only inside the boxed area.

g. For the work that you do not want to be evaluated, cross that part out.

(10) At the end of the examination

a. On the Cover Sheet, clearly write page numbers for each question.

b. Check that you have written your contestant code on all pages.

c. Put the Cover Sheet, all Summary Answersheets, blank sheets and rough sheets inside the envelope. You may keep question paper and Table of Constants with you.

Table of Constants

Fundamental Constants

Name	Symbol	Value
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$1.602\times10^{-19}\ \mathrm{C}$
Gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Gas constant	$R$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement constant	$\lambda_mT$	$2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of electron	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}$
Mass of proton	$m_p$	$1.673\times10^{-27}\ \mathrm{kg}$
Mass of neutron	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}$
Atomic mass unit	$u$	$1.661\times10^{-27}\ \mathrm{kg}$

Astronomical Constants

Name	Symbol	Value
Parsec	pc	$3.086\times10^{16}\ \mathrm{m}$
Astronomical unit	$a_\oplus$	$1.496\times10^{11}\ \mathrm{m}$
Mass of the Sun	$M_\odot$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of the Sun	$R_\odot$	$6.955\times10^8\ \mathrm{m}$
Luminosity of the Sun	$L_\odot$	$3.826\times10^{26}\ \mathrm{W}$
Apparent magnitude of the Sun at midday		$-26.72\ \mathrm{mag}$
Solar constant		$1366\ \mathrm{W\,m^{-2}}$
Angular diameter of the Sun		$30'$
Mass of the Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of the Earth	$R_\oplus$	$6.371\times10^6\ \mathrm{m}$
Tropical year		$365.242$ solar days $=3.156\times10^7\ \mathrm{s}$

Условие

A star has an apparent magnitude $m_{\mathrm{U}} = 15.0$ in the $U$ -band. The $U$ -band filter is ideal, i.e., it has perfect (100%) transmission within the band and is completely opaque (0% transmission) outside the band. The filter is centered at 360 nm, and has a width of 80 nm. It is assumed that the star also has a flat energy spectrum with respect to frequency. The conversion between magnitude, $m$ , in any band and flux density, $f$ , of a star in Jansky ( $1\ \mathrm{Jy} = 1 \times 10^{-26}\ \mathrm{W\ Hz^{-1}\ m^{-2}}$ ) is given by

f = 3631 \times 10^{-0.4m}\ \mathrm{Jy}

This star is being observed in the $U$ -band using a ground based telescope, whose primary mirror has a diameter of 2.0 m. Atmospheric extinction in $U$ -band during the observation is 50%. You may assume that the seeing is diffraction limited. Average surface brightness of night sky in $U$ -band was measured to be 22.0 mag/arcsec $^2$ .

T8.1 Approximately how many $U$ -band photons, $N_0$ , from this star will be incident normally on a 1 m $^2$ area at the top of the Earth's atmosphere every second? [8 т.]

T8.2 What is the ratio, $R$ , of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter 2''? [8 т.]

T8.3 In practice, only 20% of U-band photons falling on the primary mirror are detected. How many photons, $N_{\mathrm{t}}$ , from the star are detected per second? [4 т.]

Решение

T8.1

Approximately how many $U$ -band photons, $N_0$ , from this star will be incident normally on a $1\ \mathrm{m^2}$ area at the top of the Earth's atmosphere every second? [8]

Solution:
The $U$ -band is defined as $(360 \pm 40)\ \mathrm{nm}$ . Thus, the maximum, minimum and average frequencies of the band will be,

\nu_{\max} = \frac{c}{\lambda_{\max}} = 9.369 \times 10^{14}\ \mathrm{Hz}

\nu_{\min} = 7.495 \times 10^{14}\ \mathrm{Hz}

\nu_{\mathrm{avg}} = 8.432 \times 10^{14}\ \mathrm{Hz}

\Delta\nu = \nu_{\max} - \nu_{\min}

= 1.874 \times 10^{14}\ \mathrm{Hz} \quad \textbf{[2.0]}

f_{\mathrm{st1}} = 3631 \times 10^{-0.4 \times 15}

= 3.631\ \mathrm{mJy} = 3.631 \times 10^{-29}\ \mathrm{W\ Hz^{-1}\ m^{-2}} \quad \textbf{[2.0]}

Now, $N_0 \times h\nu_{\mathrm{avg}} = \Delta\nu \times f_{\mathrm{st1}} \times A \times \Delta t$ [2.0]

where, $A = 1\ \mathrm{m^2}$ & $\Delta t = 1\ \mathrm{s}$

\therefore\ N_0 = \frac{1.874 \times 10^{14} \times 3.631 \times 10^{-29}}{6.626 \times 10^{-34} \times 8.432 \times 10^{14}}

\simeq \boxed{12180} \quad \textbf{[2.0]}

Exact calculation including integration is accepted with full credit (exact answer: 12190).
Accepted range: $12180 \pm 200$ .
Using flat spectrum for $\Delta\lambda$ instead of $\Delta\nu$ is considered a major conceptual error, and will incur penalty of 2.0 marks.

T8.2

What is the ratio, $R$ , of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter $2''$ ? [8]

Solution:
Let us call sky flux per square arcsec as $\Phi$ and total sky flux for the given aperture as $\phi_{\mathrm{sky}}$ . Let total star flux be $\phi_{st}$ .

\phi_{\mathrm{sky}} = A\Phi = \pi \times (1\ \mathrm{arcsec})^2 \times \Phi = \pi\Phi \quad \textbf{[3.0]}

\therefore\ m_{\mathrm{sky}} = 22.0 + 2.5\log_{10}\left(\frac{\Phi}{\phi_{\mathrm{sky}}}\right) \quad \textbf{[1.0]}

= 22.0 + 2.5\log_{10}\left(\frac{\Phi}{\pi\Phi}\right)

= 22.0 - 2.5\log_{10}(\pi)

m_{\mathrm{sky}} = 20.76\ \mathrm{mag} \quad \textbf{[1.0]}

As extinction is 50% [1.0]

R = \frac{\phi_{\mathrm{st2}}}{\phi_{\mathrm{sky}}} = \frac{0.5\phi_{\mathrm{st1}}}{\phi_{\mathrm{sky}}} = 0.5 \times 10^{(20.76-15)/2.5}

\simeq \boxed{100} \quad \textbf{[2.0]}

Accepted range: $\pm 5$ .
A student may also calculate number of photons incident per second per metre square in this case and then compare it with the answer in the first case to get the correct ratio.

T8.3

In practice, only 20% of $U$ -band photons falling on the primary mirror are detected. How many photons, $N_{\mathrm{t}}$ , from the star are detected per second? [4]

Solution:

N_{\mathrm{t}} \times 1\ \mathrm{m^2} = N_0 \times 0.5 \times 0.2 \times A_{\mathrm{t}} \quad \textbf{[2.0]}

N_{\mathrm{t}} = 12180 \times 0.5 \times 0.2 \times \pi\left(\frac{2.0}{2}\right)^2 = 1233\pi \quad \textbf{[1.0]}

N_{\mathrm{t}} \simeq \boxed{3813} \quad \textbf{[1.0]}

Accepted range $(3813 \pm 50)$ .

Оригинал в Архива: theory-qp.pdf · официални решения: theory_sol.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: theory.pdf · допълнителни данни: theory.pdf