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Interference Maxima and Newton's Rings (Please help with trig!)

  • Thread starter mm2424
  • Start date
  • #1
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Homework Statement


A lens with radius of curvature R sits on a flat glass plate and is illuminated from above by light with wavelength λ (see picture below). Circular interference patterns, Newton's Rings, are seen when viewed from above. They are associated with variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assuming r/R <<1.


Homework Equations



2L = (m + 1/2) λ

The Attempt at a Solution



I understand that we will use 2L = (m + 1/2) λ here. However, I can't figure out how we relate d to R or r. I have my professor's answer key, and he defines θ as the top angle in the picture (formed by R and the normal line to the glass surfaces). He then says that r/R = sinθ which equals θ. Then, he says d = R(1-cosθ) and uses the expansion of cosθ = 1 - θ^2/2 + θ^4/4!, etc.

I can't wrap my head around how he found d = R(1-cosθ). If someone can help me see it, I would be greatly appreciative.

Thanks!
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi mm2424! :smile:
I can't wrap my head around how he found d = R(1-cosθ).
shift that little arrow to the middle of the picture …

it's the gap between the arc (of the circle) and the triangle …

= R - Rcosθ :wink:
 
  • #3
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I'm still missing something, haha. I'm not sure what you mean by shift the arrow to the middle of the picture, and I'm not clear on where Rcosθ comes from. Is it some type of trig relationship involving arcs?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
… I'm not clear on where Rcosθ comes from.
you have a right-angled triangle with hypotenuse R and top angle θ …

so the vertical side has length Rcosθ :smile:
 

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