How to Tune a Helium-Neon Laser's Frequency Over Its Gain Bandwidth?

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To tune a helium-neon (He-Ne) laser's frequency over its gain bandwidth of 1.0 GHz centered at 632.8 nm, the resonator length must be adjusted by ΔL = -1/2λ. The condition for standing waves in the resonator is given by L = mλ/2, where m is the mode number. To determine the longest and shortest wavelengths within the gain bandwidth, one must convert the bandwidth from frequency to wavelength, rather than directly applying a subtraction. This involves calculating the corresponding wavelengths for the frequency limits of the gain bandwidth. Understanding the non-linear relationship between wavelength and frequency is crucial for accurate calculations.
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Homework Statement


A helium-neon (He-Ne) laser has a gain bandwidth (denoted here as the frequency interval
over which the laser gain equals or exceeds the minimum threshold gain) given by ΔG= 1.0 GHz, centred on the λ = 632.8 nm emission wavelength.

Show that in order to tune the frequency of this He-Ne laser over its entire gain
bandwidth, the length of the resonator should be changed by ΔL = -1/2λ.

Homework Equations


L=mλ/2


The Attempt at a Solution


The length and frequency changes are both small compared to the length and fre-
quency, respectively so differentiation of the condition for standing waves in a
resonator must be applied.

The condition for standing waves is :L=mλ/2

but dL/dλ= m/2 , how can i get ΔL = -1/2λ ?
 
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helpcometk said:
The condition for standing waves is :L=mλ/2

You want the fundamental mode, so m=1. The calculation is pretty straightforward. Calculate the longest and shortest wavelength still within the gain bandwidth and calculate the corrsponding resonator lengths needed to get a standing wave for these two wavelengths.
 
thanks for the reply but i can't understand what it means:
longest and shortest wavelength still within the gain bandwidth ,how can you see if wavelength is within the bandwidth?
 
helpcometk said:
how can you see if wavelength is within the bandwidth?

The problem statement says you have a bandwidth of 1 GHz around the central wavelength of 632.8 nm, so your bandwidth goes from 632.8nm-0.5 GHz to 632.8nm+0.5 GHz. The conversion between wavelength and frequency is pretty much the only math involved here. You should be able to do that yourself.
 
this doesn't make sense because 0.5 Ghz corresponds to 0.6 m wavelength, so 632.8*10^-9 -0.6 gives negative wavelength which cannot be acceptable
 
Ok...let me start at the very beginning.

Just do me the favour and calculate the wavelengths corresponding to 0nm +0.5 Ghz (as you did), 100nm+0.5 GHz, 632nm+0.5 GHz and 2000 nm+0.5 GHz by converting wavelength to frequency FIRST and then adding the 0.5 GHz.

And while you are at it please just plot a graph of wavelength vs. corresponding frequency and have a look at whether it is linear or not. Does that help you understand why one cannot just convert 0.5 GHz into a wavelength and subtract it, but have to convert the desired wavelength into a frequency before and then add or subtract the bandwidth from that converted value?
 
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