How Does 5% Scattered Light Affect Equivalent Width?

[AlphaCrucis]

Homework Statement

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

Relevant Equations

Below

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
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 depends 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?

 

[Twigg]

Couple of things.

$$W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$$ where $F_{\nu}$ represents the flux in the line and $F_c$ represents the flux in the continuum.

There is an error here, could just be a latex typo. What are the units of the integrand, as written above? Are they consistent?

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.

There is no information given about the instrument, so you are only solving for the equivalent width not the measured equivalent width.

To get you started: How does the addition of 5% scattered light change $F_c$ and $F_\nu$?