Estimating galactic dust extinction for medium band filters

I have a set of medium-band filters and I would like to compute $A_{\lambda}/E(B-V)$ for each filter which are not reported in the literatures. The magnitudes of the objects in the photometry catalogue should be corrected for the galactic dust extinction using Cardelli et al. 1989 law.

Update

In appendix B of D.J. Schlegel, D.P. Finkbeiner, & M. Davis (1998, ApJ, 500, 525) estimating the $A_{\lambda}/E(B-V)$ for filters is obtained by using this equation $$\Delta m_f=-2.5\frac{\int d\lambda W_f(\lambda)F(\lambda)10^{\frac{-A(\lambda)\Delta m_V}{2.5}}}{\int d\lambda W_f(\lambda)F(\lambda)}$$ $\Delta m_V$ is the reddening magnitude of an elliptical galaxies and $F(\lambda)$ is the flux of source and $W_f(\lambda)$ is the response function of a filter. I am wondering how I can use Cardelli et al. 1989 law to estimate these quantities for different filters or there is an alternative approach?

• Just out of curiosity, what filters are you talking about? Apr 13, 2016 at 17:51
• @skytux optical filters. Apr 13, 2016 at 18:10
• I asked about the name of the filters to be more exactly. Apr 13, 2016 at 20:58
• @skytux here you can find the info about the set of filters I am using. Apr 13, 2016 at 22:09

The quantity you want is basically the extinction law, and is usually called $k(\lambda)$. An extinction law is a fit to several measurements of the extinction $A_\lambda$ in some direction (or an average of several directions).

Cardelli et al. (1989) provides different functional forms for the mean extinction law, parametrized in their Eq. 1 as $$\frac{A_\lambda}{A_V} = a(x) + \frac{b(x)}{R_V},$$ where $x$ is the inverse wavelength in $\mu\mathrm{m}^{-1}$, and the coefficients are given separately for IR, optical, UV, and FUV in Eqs. 2, 3, 4, and 5, respectively. The total-to-selective extinction $R_V\equiv A_V/E(B-V)$ takes different values for different lines of sight, but usually lies in the range 2.5 to 6, with 3.1 being a typical value in the Milky Way.

To get the quantity you're interested in, simply convert your favorite wavelength to $x$, stick into Eq. 1, and multiply by $R_V$: $$k(\lambda) \equiv \frac{A_\lambda}{E(B-V)} = \frac{A_\lambda}{A_V} R_V.$$

• Thanks for your answer. I have updated my question. I have been well aware of this equation, I am wondering how I can get appropriate values for my filter set? Apr 13, 2016 at 22:05
• Okay, I thought using the central wavelength would be fine for you. If you want the exact extinction, you give your formula in the updated question, right? The $A(\lambda)$ in the formula is then given by Cardelli's Eq. 1, and as far as I can see, $\Delta m_V$ is $A(V)$.
– pela
Apr 13, 2016 at 22:59
• @Dalek: $A_V$ is also dimensionless, so I think you're good. Did you try doing the calculation and see if you get meaningful values? If you know Python, I have a script that can do the Cardelli law for you.
– pela
Apr 14, 2016 at 8:46
• I code with python mostly. I will appreciate if I can get your code. If I only multiply $d=A_{\lambda}/A(V)$ by $\Delta m_V \equiv A(V)$ then I will get a reasonable value. Apr 14, 2016 at 11:40
• I guess an $A(V)$ in the denominator is missing in the original equation of Schlegel et al. 1998 paper. Apr 14, 2016 at 11:50