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Thin film interference (exam tomorrow please help)

  1. Jun 26, 2006 #1
    Newton's rings are observed in monochromatic light reflected from the air gap between a convex spherical piece of glass (radius of curvature R) and a plane slap of glass. The two pieces are in contact in the centre.

    (b)

    Show that the dark fringes have radii given approximately by [itex]r= \sqrt{mR\lambda}[/itex] , lambda is the wavelength of the light and m = 0,1,2...

    My solution:

    Locally it can be regarded as an air gap of thickness t, separated by two parallel sheets of glass. The light reflected from the top layer of the air experiences no phase shift whereas the light reflected from the bottom suffers a pi rad phase shift since it encounters a more dense medium. Hence the condition from destructive interference is

    [itex]2t = m \lambda[/itex] where m = 0,1,2,...

    so

    [itex]t = \frac{m \lambda}{2}[/itex].

    Want to show that

    [itex]r = \sqrt{2tR}[/itex]

    WTS

    [itex]t = \frac{r^2}{2R}[/itex].

    The arc length along the spherical piece of glass is [itex]s = R\theta[/itex] which is approximately the distance along the flat piece of glass (r). So [itex]r \approx R\theta[/itex].

    Then from the right-handed triangle I get [itex]\tan\theta = t/r[/itex] which for small theta is approximately theta. Hence,

    [itex]r \approx R\theta \approx R\frac{t}{r}[/itex] so

    [itex]t \approx \frac{r^2}{R}[/itex]

    which is off by a factor of 1/2. Any ideas on this one?

    Thanks in advance

    James
     
  2. jcsd
  3. Jun 26, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't quite follow your equating the arc length with the radius of the circles. Instead, write the equation of the circular cross section:
    [tex](y - R)^2 + x^2 = R^2[/tex]
    (where the center of the circle/sphere is at (0,R))

    Of course, y = t & x = r, so:
    [tex](t - R)^2 + r^2 = R^2[/tex]
    Solve for r to first order in t (ignore higher order terms) and you'll get your answer.
     
  4. Jun 27, 2006 #3
    Thanks Doc Al. That's a much better method.
     
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