Clarification on calculating IR Dust Extinction $A_K$/$A_V$ So I'm calculating V_Alf Ori's Luminosity (or Betelgeuse) using Davies et al. (2013) SED fitted formula from Near-Infrared/NIR K-Band magnitudes and it requires the subtraction of 'extinction'. However, the conventional prescriptions for extinction in the NIR K-regime is a little way off for a massive star with a very dusty environment (likely thick as well due to stellar mass loss and winds enriching it). So instead of using those, I may need to do a quick approximation of it.
This maybe a rookie stuff but as I scour in, which needs some quick clarification or validation, I kept seeing this &quot;$A_K$/$A_V$&quot; which I presume came from this one?
$\frac{A_λ}{A_V}$ = $\frac{R_λE_{(λ-V)}}{R_VE_{(B-V)}}$
where;

$R_V$ is the canonical Cardelli redenning value of 3.1.
$A_V$ is optical extinction.
E(B-V) is conventional color excess.
While the remaining upper variables are 'almost' the same depending on the bandpass of choice λ.

Edit (1): Is that 'correct'/applicable or there's something else on that like applying power law etc.?
Edit (2): Already got the idea of $\frac{A_λ}{A_V}$. Then how about $A_K$?how can I get that?
The extinction $A$ of a medium (e.g. dust) is defined as the number of magnitudes an object is increased (because higher magnitude means less light), as the light from that object travels through the medium. It is usually (but not always) a function of wavelength $\lambda$, and is thus written $A_\lambda$ (less often $A(\lambda)$).
The wavelength can be a single wavelength such as the Lyman $\alpha$ line at $\lambda=1216\,\mathrm{Å}$, in which case we may write that exact wavelength as a subscript, $A_{1216}$. Alternatively it can be the total, average extinction in a wavelength band such as the $K$ band, i.e. the near-infrared wavelength region roughly between $\lambda = 2\,\mu\mathrm{m}$ and $\lambda = 2.5\,\mu\mathrm{m}$ (depending on the manufacturer of the filter).
In this case we write $A_K$.
I can't seem to find the paper you're referring to, but you ask &quot;How can I get that?&quot;, and the answer is that you measure it. To measure an extinction, you must know how much light you'd expect in the case that some object would not have been obscured (it doesn't necessarily have to be at the particular wavelength you're interested in; it could also just be at a set of other wavelengths, and then you fit an extinction law to these points, under certain assumptions of the properties of the medium). Or you could simply not measure anything, and the assume that the extinction is &quot;similar&quot; to some &quot;standard&quot; extinction law, such as the Cardelli (1989) law, or the Calzetti (2000) law.
Relative extinction
Because extinction depends not only on the dust properties, but also one the distance traveled, it makes sense to measured it relative to a fixed wavelength, e.g. the optical band $V$ centered on $\lambda \sim 5500\,\mathrm{Å}$, thus factoring out distance. Depending on available data, sometimes the $B$ band at $\lambda \sim 4400\,\mathrm{Å}$ is used as reference, but since $A_B$ is usually larger than $A_V$, the flux is larger in $V$ and hence more precise. From the definition of the color excess
$$
E(B-V) \equiv A_B - A_V
$$
and the absolute-to-selective extinction
$$
R_V \equiv \frac{A_V}{E(B-V)},
$$
the relation between the two ratios is
$$
\frac{A_\lambda}{A_B} = \left(\!\frac{R_V}{1+R_V}\!\right) \frac{A_\lambda}{A_V}
$$
You could call the function $A_\lambda$ the extinction law, but I think it's more common to use the term for the quantity
$$
k(\lambda) = \frac{A_\lambda}{E(B-V)}.
$$