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?

  • $\begingroup$ What exactly is the scenario you have in mind here? How does the scattering arise? $\endgroup$
    – 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 $w$.

Now renormalise by dividing by 1.05. The width of the line is the same, but the depth is smaller by a factor 1.05 and thus so is the equivalent width.

  • $\begingroup$ Thanks for the answer. Ahhh so the width of the line is unaffected and its just the depth that is affected? @ProfRob $\endgroup$
    – Jay D
    Mar 23 at 3:14

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