<p>Imagine your line as a rectangle of width <span class="math-container">$w$</span> and depth <span class="math-container">$d$</span> relative to a normalised continuum.</p>
<p>Without scattered light, the area blocked off by the line is <span class="math-container">$wd$</span> and if the continuum level is normalised to 1, then the equivalent width is <span class="math-container">$wd$</span>.</p>
<p>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 <span class="math-container">$d$</span> and <span class="math-container">$w$</span>.</p>
<p>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.</p>

CC BY-SA 4.0
2021-03-23T03:14:52.813Z
Thanks for the answer. Ahhh so the width of the line is unaffected and its just the depth that is affected? @ProfRob
Jay D
CC BY-SA 4.0
2021-03-22T23:42:58.827Z
2021-03-22T23:42:58.827Z
ProfRob
<p>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 <span class="math-container">$W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$</span>; where <span class="math-container">$F_{\nu}$</span> represents the flux in the line and <span class="math-container">$F_c$</span> represents the flux in the continuum.</p>
<p>The measured equivalent width is <span class="math-container">$W_m = \int_{-\lambda_o}^{\lambda_o} \frac{I(\lambda)*(F_c - F_\nu)}{D_c} \, d\nu$</span> in which <span class="math-container">$\lambda_o$</span> is the spectral range over which the profile can be traced, <span class="math-container">$I(\lambda)$</span> is the instrumental profile, and <span class="math-container">$D_c$</span> is the apparent continuum.</p>
<p>If we choose <span class="math-container">$\lambda$</span> 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 <span class="math-container">$200 A \cdot 0.05$</span> = 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?</p>

CC BY-SA 4.0
2021-03-23T18:24:56.460Z
What exactly is the scenario you have in mind here? How does the scattering arise?
Thomas
<astrophysics><light><spectroscopy><spectra><flux>
How much does the equivalent width of a line change by the introduction of 5% scattered light?
Jay D
# How much does the equivalent width of a line change by the introduction of 5% scattered light?

(Asked by: username_0 on 2021-03-22)

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?

**Tags:** astrophysics, light, spectroscopy, spectra, flux

> **Comments:**
> **username_1**: What exactly is the scenario you have in mind here? How does the scattering arise?  

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## 1 answers

### Answer by username_2 on 2021-03-22. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks for the answer. Ahhh so the width of the line is unaffected and its just the depth that is affected? @username_2  

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