# How much does the equivalent width of a line change by the introduction of 5% scattered light?

How much does the equivalent width of a line change by the introduction of 5% scattered light? We know the equivalent width is defined as $$W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$$; where $$F_{\nu}$$ represents the flux in the line and $$F_c$$ represents the flux in the continuum.

The measured equivalent width is $$W_m = \int_{-\lambda_o}^{\lambda_o} \frac{I(\lambda)*(F_c - F_\nu)}{D_c} \, d\nu$$ in which $$\lambda_o$$ is the spectral range over which the profile can be traced, $$I(\lambda)$$ is the instrumental profile, and $$D_c$$ is the apparent continuum.

If we choose $$\lambda$$ to be 0 at the center of the line, and the range spans 200 Angstroms, then does the equivalent width of the line change by $$200 A \cdot 0.05$$ = 10 A? So is the equivalent width of a line changed depend on our range? I.e. width of a line of 140 Angstroms with 5% scattered light would alter it by 0.7 A. Am I correct?

• What exactly is the scenario you have in mind here? How does the scattering arise? – Thomas Mar 23 at 18:24

Imagine your line as a rectangle of width $$w$$ and depth $$d$$ relative to a normalised continuum.
Without scattered light, the area blocked off by the line is $$wd$$ and if the continuum level is normalised to 1, then the equivalent width is $$wd$$.
Now add 5% scattered light. The height of the continuum is 1.05 (but we're going to renormalise it) and the depth and width of the line are still $$d$$ and $$w$$.