How Does FWHM Affect Spectral Resolution Calculation?

[chris_0101]

Homework Statement

FWHM is an expression of the extent of a function, given by the difference between the two extreme values of the independent variable at which the dependent variable is equal to half of its maximum value.

I have this lab that requires me to calculate the limiting Spectral resolution, R, from a measured FWHM value (2.131nm for example) and an assumed natuarl line width, L, of 1GHz. Whenever I solve for the R value it is the same as the FWHM since 1GHz converted to nm is a very small number.

So, Can anyone give me any feedback regarding this.

Thanks

Homework Equations

FWHM = sqrt(R^2 - L^2) - similar to the pythagorean theorem

rearranged for R

R = sqrt(FWHM^2 - L^2)

The Attempt at a Solution

R = sqrt(2.131^2 - 0.3 x 10^-10)

R = sqrt(2.131^2)

R = 2.131

Any help would be greatly appreciated.

 

[anathema]

Thank you for sharing your question and calculations. It is great to see that you are actively working on your lab and seeking feedback. I would be happy to provide some insights and feedback on your approach.

Firstly, your understanding of FWHM and its relation to the independent variable is correct. FWHM is a measure of the width of a function at half of its maximum value, and the difference between the two extreme values of the independent variable at this point is the FWHM value.

Regarding your calculation of the limiting spectral resolution, R, it is important to note that the value of 1GHz converted to nm is indeed a very small number (approximately 0.3 x 10^-10 nm). However, this does not mean that the R value should be the same as the FWHM value. The formula you have used, R = sqrt(FWHM^2 - L^2), is correct, but it is missing a unit conversion factor.

In order to convert the FWHM value from nm to GHz, you need to divide it by the speed of light (c) in nm/s. This will give you the equivalent value in GHz. Then, you can substitute this value into the formula to calculate the R value.

Therefore, the correct calculation would be:

R = sqrt((2.131/3x10^8)^2 - (0.3x10^-10)^2)

R = sqrt(7.1x10^-18 - 9x10^-21)

R = sqrt(7.1x10^-18)

R = 2.66x10^-9 GHz

I hope this helps clarify your calculations. Keep up the good work and don't hesitate to ask for further clarification if needed. Good luck with your lab!