<p>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).</p>

<p>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.</p>

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

CC BY-SA 3.0
2016-04-13T22:05:44.317Z
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?
Dalek
CC BY-SA 3.0
2016-04-13T22:59:10.877Z
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
CC BY-SA 3.0
2016-04-14T08:46:28.610Z
@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
CC BY-SA 3.0
2016-04-14T11:40:49.543Z
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.
Dalek
CC BY-SA 3.0
2016-04-14T11:50:25.730Z
I guess an $A(V)$ in the denominator is missing in the original equation of Schlegel et al. 1998 paper.
Dalek
CC BY-SA 3.0
2016-04-14T12:25:17.657Z
The extinction depends on the exact filter profile *and* on the intrinsic spectrum of the source.
ProfRob
CC BY-SA 3.0
2016-04-14T13:03:05.943Z
@RobJeffries Well, I have optical data therefore I don't have access to the intrinsic spectrum of my objects. The process described in Schlegel et al. 1998 they have used a spectrum of an elliptical galaxy and they obtained the values for $A(V)$ but I suspect the equation is not quite right as I mentioned before.
Dalek
CC BY-SA 3.0
2016-04-14T18:06:07.573Z
@Dalek Well if you can't do it properly then you just interpolate the extinction curve to the central wavelength of your filters. The filter profile is only relevant if you are going to fold a spectrum through it.
ProfRob
CC BY-SA 3.0
2016-04-13T21:12:58.207Z
2016-04-13T21:12:58.207Z
pela
<p>I have a set of <strong>medium-band filters</strong> 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 <strong><em>Cardelli et al. 1989</em></strong> law.</p>

<p><strong>Update</strong></p>

<p>In appendix B of <a href="http://iopscience.iop.org/article/10.1086/305772/pdf" rel="nofollow">D.J. Schlegel, D.P. Finkbeiner, &amp; M. Davis (1998, ApJ, 500, 525)</a> 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  <strong><em>Cardelli et al. 1989</em></strong> law to estimate these quantities for different filters or there is an alternative approach?</p>

CC BY-SA 3.0
2016-04-13T17:51:46.417Z
Just out of curiosity, what filters are you talking about?
skytux
CC BY-SA 3.0
2016-04-13T18:10:38.367Z
@skytux [optical filters](https://en.wikipedia.org/wiki/Photometric_system).
Dalek
CC BY-SA 3.0
2016-04-13T20:58:35.093Z
I asked about the name of the filters to be more exactly.
skytux
CC BY-SA 3.0
2016-04-13T22:09:57.747Z
@skytux [here](http://www.mpia.de/COMBO/combo_filters.html) you can find the info about the set of filters I am using.
Dalek
<galaxy><redshift><dust><photometry><extinction>
Estimating galactic dust extinction for medium band filters
Dalek
# Estimating galactic dust extinction for medium band filters

(Asked by: username_0 on 2016-04-13)

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)](http://iopscience.iop.org/article/10.1086/305772/pdf) 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?

**Tags:** galaxy, redshift, dust, photometry, extinction
> **Comments:**
> **username_1**: Just out of curiosity, what filters are you talking about?  
> **username_0**: @skytux [optical filters](https://en.wikipedia.org/wiki/Photometric_system).  
> **username_1**: I asked about the name of the filters to be more exactly.  
> **username_0**: @skytux [here](http://www.mpia.de/COMBO/combo_filters.html) you can find the info about the set of filters I am using.  

---

## 1 answers

### Answer by username_2 on 2016-04-13. Score: 4

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 $R\_V$:
$$
k(\lambda) \equiv \frac{A\_\lambda}{E(B-V)} = \frac{A\_\lambda}{A\_V} R\_V.
$$

> **Comments:**
> **username_0**: 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?  
> **username_2**: 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 $A(V)$.  
> **username_2**: @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.  
> **username_0**: 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.  
> **username_0**: I guess an $A(V)$ in the denominator is missing in the original equation of Schlegel et al. 1998 paper.  
> **username_3**: The extinction depends on the exact filter profile *and* on the intrinsic spectrum of the source.  
> **username_0**: @RobJeffries Well, I have optical data therefore I don't have access to the intrinsic spectrum of my objects. The process described in Schlegel et al. 1998 they have used a spectrum of an elliptical galaxy and they obtained the values for $A(V)$ but I suspect the equation is not quite right as I mentioned before.  
> **username_3**: @Dalek Well if you can't do it properly then you just interpolate the extinction curve to the central wavelength of your filters. The filter profile is only relevant if you are going to fold a spectrum through it.  

---

