Physics Asked by luvjyotk on November 27, 2020
A Ruby laser emits 1W pulses of light with wavelength $6940 ± 10 overset{lower.5emcirc}{mathrm{A}}$.
What is the coherence length of the ruby laser?
My question is that what would be the value of $∆λ$ in the formula of coherent length i.e.
Coherence length $=(λ^2) /(∆λ)$.
Will it be $10$ or $20$?
You have not given enough information to determine the coherence length of the laser. Some ruby lasers have coherence length up to 100 meters. In 1974, 1 meter of coherence length was a big achievement.
Answered by S. McGrew on November 27, 2020
You have stated that the Coherence length: $L = dfrac{lambda^2}{Delta lambda}$ .
I will assume that you are using the approximation for the formula:
$$L = frac{2 ln 2}{pi} frac{lambda ^2 }{n Delta lambda}$$
Where $n$ is the refractive index of the medium through which the wave is propagating in and is taken as $1$ for air. For this formula, the value of $Delta lambda$ corresponds to the full width half maximum (FWHM) spectral width of the source.
In optics in general the width of various peaks can be given as FWHM, $dfrac{1}{e}$ or $dfrac{1}{e^2}$ for various different applications. You have not made it clear which is being used for the uncertainty you provided. Despite this, I can say that in optical communication the standard for spectral widths is FWHM.
Answered by Omar Essilfie-Quaye on November 27, 2020
The literature gives you generally the width and the wavelength of the light emission. If $Delta lambda << lambda$, you can apply the equation $l_c = frac{lambda^2}{Delta lambda}$.
Typically, the width is given in Hz or GHz. You can transform it with the relation $Delta lambda = frac{lambda^2 Delta nu}{c}$.
For your case, Ruby have emission at 694.3 nm and FWHM~330 GHz. You should found something around 1 mm.
NB: A better explanation can be found in the textbook of W.T. Silfvast (ISBN: 9780521833455), where the author compares the coherence length of mercury lamp and helium-neon laser.
Answered by JohnB on November 27, 2020
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