How Does 5% Scattered Light Affect Equivalent Width?

AI Thread Summary
The introduction of 5% scattered light affects the equivalent width by altering the continuum flux (F_c) and the line flux (F_ν). The equivalent width is calculated using the integral of the ratio of these fluxes, and the change in equivalent width depends on the spectral range chosen. For a range of 200 Angstroms, the equivalent width could change by 10 Angstroms, while a narrower line width of 140 Angstroms would result in a change of 0.7 Angstroms. Additionally, there are concerns about potential errors in the equations presented, such as unit consistency and the lack of instrument-specific information. Understanding how scattered light impacts the flux measurements is crucial for accurate calculations of equivalent width.
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##?
 
  • Like
Likes AlphaCrucis
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top