Why Does Monochromatic Light Wavelength Spread Through a Fast Shutter?

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SUMMARY

The discussion centers on the phenomenon of wavelength spread in monochromatic light when passing through a fast shutter that opens for 10 ns. The energy-time uncertainty relation is identified as the key principle explaining this spread, indicating that a finite shutter opening time introduces uncertainty in the energy of the light. Participants emphasize that the finite duration of 10 ns results in a measurable spread in wavelengths, as demonstrated through calculations involving the speed of light and Fourier transforms of sine waves. The conclusion is that the act of passing through the shutter inherently alters the energy characteristics of the light.

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  • Understanding of the energy-time uncertainty principle
  • Basic knowledge of monochromatic light properties
  • Familiarity with Fourier transforms
  • Concept of wavefunctions in quantum mechanics
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Monochromatic light with wavelength 6000 nm passes through a fast shutter that opens for 10 ns. What will be the spread in wavelengths in the now longer monochromatic light?

I have calculated the spread in wavelength using the energy-time uncertainty relation. But, I don't understand qualitatively why when the monochromatic light goes through the shutter that opens for 10 ns will cause uncertainty to the energy of the light. HOW? Is it because the shutter opening time is finite, and it will lead to a spread in energies? Is 10 ns finite? How do we know it is finite?

If it doesn't pass through the shutter, then it wouldn't spread in energies, right?
 
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Take a sine wave with a wavelength of 6000 nm. Compute how fast light moves in 10ns, convert it into X nm. Now compute the Fourier transform of a sine wave of wavelength 6000 nm that is just X nm long.

You will find that more wavelengths are present than just the original 6000 nm.

Carl
 
CarlB said:
Take a sine wave with a wavelength of 6000 nm. Compute how fast light moves in 10ns, convert it into X nm. Now compute the Fourier transform of a sine wave of wavelength 6000 nm that is just X nm long.

You will find that more wavelengths are present than just the original 6000 nm.

Carl

I have not done Fourier transform of a wavefunction. Is there any other ways that I can comprehend?
 

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