How Does 5% Scattered Light Affect Equivalent Width?

Click For Summary
SUMMARY

The introduction of 5% scattered light affects the equivalent width of a spectral line, defined mathematically as $$W = \int_{-\infty}^{\infty} \bigg(\frac{1-F_{\nu}}{F_c}\bigg) \, d\nu$$. The measured equivalent width is represented by $$W_m = \int_{-\lambda_o}^{\lambda_o} \frac{I(\lambda)*(F_c - F_\nu)}{D_c} \, d\nu$$, where the parameters include the instrumental profile and the apparent continuum. A 200 Angstrom range with 5% scattered light alters the equivalent width by 10 Angstroms, while a 140 Angstrom line width results in a change of 0.7 Angstroms. The discussion highlights the need for clarity regarding the instrument used for measurements.

PREREQUISITES
  • Understanding of equivalent width in spectroscopy
  • Familiarity with spectral line profiles and continuum flux
  • Knowledge of instrumental profiles in spectroscopic measurements
  • Basic grasp of integration in mathematical physics
NEXT STEPS
  • Research the impact of scattered light on spectral measurements
  • Study the mathematical derivation of equivalent width in spectroscopy
  • Explore the effects of instrumental profiles on measured equivalent widths
  • Learn about the calibration of spectroscopic instruments to minimize errors
USEFUL FOR

Astronomers, astrophysicists, and researchers involved in spectral analysis and those interested in the effects of scattered light on equivalent width measurements.

AlphaCrucis
Messages
10
Reaction score
1
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?
 
Last edited by a moderator:
Physics news on Phys.org
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##?
 
Reply
  • Like
Likes   Reactions: AlphaCrucis