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Differentiating $nu = dfrac{c}{lambda}$

Physics Asked on June 27, 2021

I am currently studying Laser Systems Engineering by Keith Kasunic. Chapter 1.2.1 Temporal Coherence says the following:

The coherence time $tau_c$ over which the emitted wavelengths are considered to be in phase – that is, are temporally coherent – thus depends inversely on the absolute value of the wavelength difference $vert Delta lambda vert$ or frequency difference $Delta nu = c vert Delta lambda vert / lambda^2$ [from differentiating Eq. (1.1)]:
$$tau_c approx dfrac{1}{2 vert Delta nu vert} = dfrac{1}{2c} dfrac{lambda^2_o}{vert Delta lambda vert} text{[sec]} tag{1.3}$$
clearly showing that a narrow-spectrum laser with small $Delta nu$ has a longer time over which different frequencies propagate before they are no longer considered to be in phase. Typical numbers for a HeNe laser are used for interferometric optical testing are a coherence time $tau_c = 0.33$ nsec and a coherence length $d_c = c tau_c = 100$ mm for a linewidth $Delta lambda = 2$ pm (see Table 1.3). Note that the results shown in Fig. 1.11(b) – where the different wavelengths are out of phase at a time $t approx 10^{-8}$ sec – are not consistent with the estimates from Eq. (1.3), as the equation assumes that $Delta lambda$ is small in comparison with the center wavelength $lambda_o$.

Eq. (1.1) is given as follows:

A laser is a source of both light and heat. Light is an electromagnetic wave with a wavelength $lambda$ and frequency $nu$ with energy propagating at the speed of light $c$ in a vacuum:
$$nu = dfrac{c}{lambda} text{[Hz]} tag{1.1}$$

How does differentiating $nu = dfrac{c}{lambda}$ get us (1.3)? I don’t even understand how this is a function in the first place, since $c$ is just the speed of light in vacuum and $lambda$ is the wavelength.

One Answer

I don't know much about the topic but this can be done as follows :

$$nu=frac{c}{lambda}$$

$$frac{dnu}{dlambda}=-frac{c}{lambda^2}$$ $$frac{Delta nu}{Delta lambda}=-frac{c}{lambda^2}$$ $$|Delta nu|=frac{c}{lambda^2}|Deltalambda|$$

That's the exact thing that is used.

Correct answer by Young Kindaichi on June 27, 2021

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