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Apr 14, 2016 at 18:06 comment added ProfRob @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.
Apr 14, 2016 at 13:03 comment added Dalek @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.
Apr 14, 2016 at 12:25 comment added ProfRob The extinction depends on the exact filter profile and on the intrinsic spectrum of the source.
Apr 14, 2016 at 11:50 comment added Dalek I guess an $A(V)$ in the denominator is missing in the original equation of Schlegel et al. 1998 paper.
Apr 14, 2016 at 11:40 comment added Dalek 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 8:46 comment added pela @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.
Apr 13, 2016 at 22:59 comment added pela 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 $A(V)$.
Apr 13, 2016 at 22:05 comment added Dalek 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 21:12 history answered pela CC BY-SA 3.0