Question

At an observing site the brightness of the night sky background per `` is equivalent to a 21st magnitude star. In the search for a faint object a telescope scans the sky with a field of view limited to 200 ``. Calculate the equivalentmagnitude of the star matching the total effective brightness of the sky background.

Answer 1

Answer on Question #54960, Physics / Astronomy | Astrophysics

The ratio of recorded energy in the experiment to that from $1\ \psi$ is 200:1. Using equation:

$$m_1 - m_2 = -2.5 \log \left(\frac{B_1}{B_2}\right)$$

Hence:

$$m_{200} - 21 = -2.5 \log \left(\frac{200}{1}\right)$$ $$m_{200} = 21 - 5.25 = 15.25$$

Answer: $m_{200} = 15.25$

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