How Does a Laser's Resonant Cavity Create Multiple Wavelengths?

AI Thread Summary
A laser's resonant cavity can produce multiple closely spaced wavelengths, known as longitudinal modes, due to its ability to sharpen the wavelength of spontaneous emission. In this case, with a cavity length of 60 cm and a central wavelength of 633 nm, the calculation shows that approximately 1,895,735 wavelengths fit the standing wave condition. To achieve a single wavelength, changing the cavity length would be beneficial, as it would allow for constructive interference of wavelengths when n is an integer. The discussion emphasizes the importance of considering the full range of wavelengths, including the specified line width of 0.001 nm, when calculating the modes. Overall, adjusting the cavity length can help isolate a desired wavelength more effectively.
Demroz
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Homework Statement


The fact that a laser's resonant cavity so effectively sharpens the wavelength can lead to the output of several closely spaced laser wavelengths, called longitudinal modes. Here we see how. Suppose the spontaneous emission serving as the seed for stimulated emission is of wavelength 633 nm, but somewhat fuzzy, with a line width of roughly 0.001 nm either side of the central value. The resonant cavity is exactly 60 cm long. (a). How many wavelengths fit the standing wave condition? (b) If only a single wavelength were desired would changing the length of the cavity help? Explain.

Homework Equations



L = n λ / 2

The Attempt at a Solution


n = 2L/λ

n = 1895734.597

For part one I have a feeling that n is not what were solving for. (Plus I'm not using the 0.001 nm width that is given).

part b.

Yes it would help, because if n is an integer number, wavelengths will end up interfering constructively in the cavity.
 
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You calculated n for the central wavelength. What about the lower and upper bound on the wavelength?
Demroz said:
Yes it would help, because if n is an integer number, wavelengths will end up interfering constructively in the cavity.
No, but you'll need (a) to see how (b) is meant.
 
mfb said:
You calculated n for the central wavelength. What about the lower and upper bound on the wavelength?
No, but you'll need (a) to see how (b) is meant.

would I still use the equation above?
 
Sure, it is just a different wavelength value.
 
so would I just add and subtract 0.001 from 633?
 
Sure, what else?
 
I ended up just solving for n, and using only integer values of n, and then adding and subtracting 1 integer value, and solving for lambda, until lambda was out of the limits (633.001 and 629.999)
 
That is way too complicated, but it is possible.
 
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