Air Wedge radius of the wire Problem

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SUMMARY

The air wedge problem involves calculating the radius of a wire using the interference of light. When illuminated with 600 nm light, 30 dark fringes indicate a phase difference corresponding to a thickness change of half a wavelength. This results in a maximum thickness of 15 wavelengths, equating to 9000 nm. Dividing by two for the radius yields a final answer of 4500 nm.

PREREQUISITES
  • Understanding of wave interference and dark fringes
  • Knowledge of wavelength and phase difference concepts
  • Familiarity with basic optics and light behavior
  • Ability to perform calculations involving wavelengths and thickness
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  • Study the principles of wave interference in optics
  • Learn about the relationship between wavelength and fringe patterns
  • Explore the concept of air wedges in experimental physics
  • Investigate the calculations involved in determining wire diameters from interference patterns
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Students studying optics, physics educators, and anyone interested in wave interference and its applications in measuring small dimensions.

Zukie91
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Homework Statement
An air wedge is formed between two glass plates separated at one edge by a very fine wire. When the wedge is illuminated from above by 600 nm light, 30 dark fringes are observed. Calculate the radius of the wire.

The attempt at a solution
Wasn't really sure how to go about doing this, also don't have the answer, so i have no way of knowing if i am correct.
I think it has something to do with every time it goes light dark light, is one wavelength. not sure if that is relevant (or even correct) any help is appreciated.
Thanks
 
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For this particular problem, the fringes correspond to integral numbers of wavelengths. As you go along to the next fringe it represents the next integral number wavelength phase difference which corresponds to a change in thickness of the air gap of half a wavelength. Thus there is a relationship between the maximum thickness of the air gap (i.e. the wire) the wavelength and the number of fringes. Can you work out what that must be?
 
well, if there are 30 fringes, and each integral number wavelength phase dfference corresponds to a thickness of half a wavelength, would that mean at its thickest, the width is 15 wavelengths or 9000 nm?
 
Yes that would be correct. But remember that the question asks for the radius of the wire.
 
right, what i have there is the diameter, so i would divide by two and the final answer would be 450 nm. Thanks
 
Zukie91 said:
right, what i have there is the diameter, so i would divide by two and the final answer would be 450 nm. Thanks

Careful! You're on the correct track though. :approve:
 
whoops, typo, meant 4500 nm
 
Zukie91 said:
whoops, typo, meant 4500 nm

very good.
 

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