Simple Newton's Rings problem

  • #1
Jason+Strife
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Homework Statement



If the apparatus for showing Newton's rings is illuminated with light at 6250 A (angstroms), what thickness of film underlies each of the first three light rings?

Homework Equations



2D (the thickness of the film) equals a multiple of 1/2 of a wavelength
1 angstrom = 1 x 10^-10 wavelength


The Attempt at a Solution



I can't see how the angstroms connect to the fact that the thickness must be a multiple of 1/2.
 
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Answers and Replies

  • #2
Doc Al
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Generally Newton's Rings are formed when light is reflected off of a hemispherical lens resting on a sheet of glass. So I don't know what you mean by "film"--there is a space between the lens and the glass sheet.

In any case, the phase difference (for constructive interference) must be an odd multiple of 1/2 wavelength, not just 1/2.
 
  • #3
Jason+Strife
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I apologize for not clarifying wavelength, but I intended to. As for the film, I believe it means some form of glass, but its not really relevant to the solution of my problem.
 
  • #4
Doc Al
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Do you still have a question?
 
  • #5
Jason+Strife
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Homework Statement



If the apparatus for showing Newton's rings is illuminated with light at 6250 A (angstroms), what thickness of film underlies each of the first three light rings?

Homework Equations



2D (the thickness of the film) equals a multiple of 1/2 of a wavelength
1 angstrom = 1 x 10^-10 wavelength


The Attempt at a Solution



I can't see how the angstroms connect to the fact that the thickness must be a multiple of 1/2.
Yes, I don't see a way to connect the 6250 angstroms and the fact that each wavelength must be a multiple of a half. There must be some equation I'm missing, but I just can't figure out the problem.
 
  • #6
Doc Al
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Yes, I don't see a way to connect the 6250 angstroms and the fact that each wavelength must be a multiple of a half.
That's not quite right.

6250 Angstroms is the wavelength; 2D must be an odd multiple of 1/2 wavelength.

An Angstrom is just a unit of length. 1 Angstrom [itex]= 10^{-10}[/itex] meters.
 
  • #7
Jason+Strife
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OK, but how can I use that information (the angstroms) to figure out the thickness for the first three light rings?
 
  • #8
Doc Al
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OK, but how can I use that information (the angstroms) to figure out the thickness for the first three light rings?
By setting 2D equal to the first three odd integer multiples of 1/2 wavelength. (Solve for D.)
 
  • #9
Jason+Strife
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Aha, thank you, I appreciate your help.
 

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