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

In summary: 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.
  • #1
AlphaCrucis
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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?
 
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Couple of things.

AlphaCrucis said:
$$W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$$ 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?

AlphaCrucis said:
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##?
 
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1. What is the equivalent width of a line?

The equivalent width of a line is a measurement of the area under a spectral line, which represents the amount of light absorbed or emitted by a particular element or molecule.

2. How is the equivalent width of a line calculated?

The equivalent width of a line is calculated by comparing the observed spectrum to a continuum level, and measuring the area between the two curves.

3. What is the significance of introducing 5% scattered light?

Introducing 5% scattered light can affect the observed equivalent width of a line, as it can contribute to the overall intensity of the spectrum and alter the shape of the spectral line.

4. How does 5% scattered light impact the equivalent width of a line?

The impact of 5% scattered light on the equivalent width of a line depends on the particular spectral line and the amount of scattered light present. In some cases, it may cause a slight increase or decrease in the equivalent width, while in others it may have a more significant effect.

5. Is it possible to accurately measure the equivalent width of a line with 5% scattered light present?

Yes, it is possible to measure the equivalent width of a line with 5% scattered light present. However, it is important to account for the scattered light in the analysis and take it into consideration when interpreting the results.

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