Linewidths - Frequency and Wavelength

  • Context: Graduate 
  • Thread starter Thread starter Naz93
  • Start date Start date
  • Tags Tags
    Frequency Wavelength
Click For Summary
SUMMARY

The discussion centers on the mathematical relationship between linewidths in frequency (Δf) and wavelength (Δλ) using the equation c = fλ, where c is the speed of light. The user expresses confusion regarding the negative correlation derived from the differentiation of the equation, specifically df/dλ = -c/(λ^2), which indicates that an increase in frequency corresponds to a decrease in wavelength. This relationship is confirmed through the mathematical derivation, clarifying that the broader spread of frequencies indeed results in a narrower spread of wavelengths.

PREREQUISITES
  • Understanding of basic physics concepts, particularly the wave equation c = fλ.
  • Knowledge of calculus, specifically differentiation and its application in physics.
  • Familiarity with the concepts of frequency and wavelength in wave mechanics.
  • Basic mathematical manipulation of equations involving variables and constants.
NEXT STEPS
  • Study the implications of the wave equation c = fλ in different contexts, such as optics and acoustics.
  • Learn about the physical significance of linewidths in spectroscopy and their impact on measurement accuracy.
  • Explore advanced calculus techniques for differentiating complex functions in physics.
  • Investigate the relationship between frequency and wavelength in various media, including dispersion effects.
USEFUL FOR

Students and professionals in physics, particularly those studying wave mechanics, optics, and spectroscopy, will benefit from this discussion. It is also relevant for anyone interested in the mathematical relationships governing wave properties.

Naz93
Messages
28
Reaction score
2
Hi,

Probably a really stupid question... but I'm confused about how to relate a line width in frequency to one in wavelength. To me it seems obvious that if there's a broader spread of frequencies, there must be a broader spread of wavelengths, and vice versa - after all, the line is broader.

But... if you relate Δf and Δλ mathematically (f is frequency, since there was no nu on the quick symbols list):

c = fλ

f = c/λ

df/dλ = -c/(λ^2)

so df = dλ * (-c/(λ^2))

I.e. if df is positive, dλ is negative...

What am I doing wrong??
 
Physics news on Phys.org
Where's the problem in that? One slope is positive and the other is negative - that's what you get when you have

λ= c/f and you write down dλ/df

No worries. Have faith in yourself and your maths.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
11K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K