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I have the following problem for an astrophysics course:

A star is seen through a rather dusty region of space has its brightness dimmed by +1 magnitude/kpc, which makes it seem further away than it actually is. If the observed apparent magnitude of the star is $m_v = +4.0$ and absolute magnitude $M_v = -4.5$, determine the distance using the extinction coefficient ("A" in the distance modulus equation).

This seems to be a relatively simple problem using the following equation which was given to us in class:

$$ m - M = 5\log{\frac{d}{10}} + A(\lambda) $$

The issue comes in that our professor did not say anything about the $A(\lambda)$ term. I figure that I could go through deriving it in some way, but that is something that this course should not require. How do I find this term?

Also, our professor is not the greatest, so does anyone know of any good resources for astrophysics that they have used? More complex resources are actually preferred since I am a physics major as well.

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  • $\begingroup$ Extinction coefficients are usually given as a value for a particular wavelength. The wavelength isn't specified, but you are given a rate of extinction ($1mag/kpc$). The formula you have works in parsecs, so substituting $A(\lambda) = 0.1d$, then you might be able to solve it. This is a bit of a stab. $\endgroup$
    – Carl
    Oct 13 '13 at 5:53
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The extinction coefficient is usually given as a value for a particular wavelength. The wavelength isn't specified for you, but you are given a rate of extinction ($1mag/kpc$).

Start by calculating the distance without extinction:

$$ d = 10^{\big(\frac{m_v - M_v + 5}{5}\big)} \\ d = 10^{2.7} \\ d = 501 = 0.5kpc $$

Using this, we can now determine how much the $1mag/kpc$ impacts. $$ m_v = 4 - (0.5kpc * 1mag/kpc) \\ m_v = 3.5 \\ $$

With the new apparent magnitude $m_v=3.5$ we can now calculate the distance using the same formula as previously: $$ d = 10^{\big(\frac{m_v - M_v + 5}{5}\big)} \\ d = 10^{2.6} \\ d = 398 pc = 0.39 kpc $$

N.B. There may be a more elegant way to achieve this.

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