Experiment 17 - Andrew Underwood by 8h7G6kx1

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									Interference in a Thin Film: Newton’s rings

Experiment 17 - Andrew Underwood
                 Interference in a Thin Film: Newton’s rings

                 Purpose

   The purpose of this experiment was to show
    how light interference works in a thin film
   Part 1: Thin wedge
       consecutive lines of fringes
   Part 2: Newton’s Rings
       circles of fringes
                Interference in a Thin Film: Newton’s rings

                Theory

   Index of refraction
       glass   n = 1.5
       air     n = 1.0
   When nA > nB
       rays reflecting from the side
        where n=nA reflect the
        same phase
       rays reflecting from the side
        where n=nB experience a
        phase reversal
                    Interference in a Thin Film: Newton’s rings

                    Theory

   A phase reversal is equivalent to shifting the
    ray by half a wavelength




      Diagram from Physics 294
      Laboratory Manual, page 17-1
             Interference in a Thin Film: Newton’s rings

             Theory

   When waves interfere constructively:
             Interference in a Thin Film: Newton’s rings

             Theory

   When waves interfere destructively:
             Interference in a Thin Film: Newton’s rings

             Theory

   When waves interfere destructively:
              Interference in a Thin Film: Newton’s rings

              Theory Part 1




Diagram from Physics 294
Laboratory Manual, page 17-2
                     Interference in a Thin Film: Newton’s rings

                     Theory Part 1

   Two glass plates at an angle  cause the
    space between them t to increase with x
       t = /2 = x tan 
           Where
               t   is the distance between the two pieces of glass
                   is the wavelength of the light
               x   is the distance between consecutive fringes
                   is the angle between the glass
       tan  = t / x
           Since  is constant, t / x should be constant
           Implies fringes happen at regular intervals
Interference in a Thin Film: Newton’s rings

Theory Part 2
Interference in a Thin Film: Newton’s rings

Theory Part 2
Interference in a Thin Film: Newton’s rings

Theory Part 2
              Interference in a Thin Film: Newton’s rings

              Theory Part 2




Diagrams from Physics 294
Laboratory Manual, page 17-3
                    Interference in a Thin Film: Newton’s rings

                    Theory Part 2

   rm2 = xm2/4 + 2 = mR
       Where
           rm      is the actual radius of the mth fringe
           xm      is the measured diameter
                  is the perpendicular offset from the center of the
            ring
           m       is the order of the ring
               m=0 for the bulls eye
               m=1, 2, 3, … for each consecutive ring
                  is the wavelength of the light
           R       is the curvature radius of the plano-convex lens
Interference in a Thin Film: Newton’s rings

Apparatus
                  Interference in a Thin Film: Newton’s rings

                  Apparatus




Diagram from Physics 294
Laboratory Manual, page 17-2
Interference in a Thin Film: Newton’s rings

Procedure
                 Interference in a Thin Film: Newton’s rings

                 Procedure Part 1

   The distance from an
    arbitrary (m=0) fringe to 20
    consecutive dark fringes was
    measured
    The difference between
    each fringe was calculated
       x = xn - xn-1
   From x we can calculate 
       /2 = x tan 
                 Interference in a Thin Film: Newton’s rings

                 Procedure Part 2

   Position the microscope over the center of
    the bulls eye
   Measure xL and xR to get xm for 10 rings
       xm = |xL - xR|
   Plot xm2 versus m to find R and 2
       xm2/4 + 2 = mR
       xm2 = mR/4 - 2/4
       y = mx       +b
Interference in a Thin Film: Newton’s rings

Analysis
                Interference in a Thin Film: Newton’s rings

                Analysis Part 1

   Error for x                  m
                                  0
                                       Xm (cm)
                                        11.883
                                                 error Xm (cm)
                                                     0.004
                                                                 difference Xm (cm)   error difference (cm)

                                  1     11.941       0.004             0.058                 0.006

       x = (x)2 + (x)2      2
                                  3
                                        11.990
                                        12.040
                                                     0.004
                                                     0.004
                                                                       0.049
                                                                       0.050
                                                                                             0.006
                                                                                             0.006
                                  4     12.090       0.004             0.050                 0.006

       x = 2(x)2             5
                                  6
                                        12.141
                                        12.190
                                                     0.004
                                                     0.004
                                                                       0.051
                                                                       0.049
                                                                                             0.006
                                                                                             0.006
                                  7     12.233       0.004             0.043                 0.006

       x = 2(0.004 cm)2       8
                                  9
                                        12.290
                                        12.332
                                                     0.004
                                                     0.004
                                                                       0.057
                                                                       0.042
                                                                                             0.006
                                                                                             0.006
                                  10    12.380       0.004             0.048                 0.006

       x = 0.006 cm            11
                                  12
                                        12.432
                                        12.486
                                                     0.004
                                                     0.004
                                                                       0.052
                                                                       0.054
                                                                                             0.006
                                                                                             0.006
                                  13    12.530       0.004             0.044                 0.006
                                  14    12.580       0.004             0.050                 0.006
                                  15    12.632       0.004             0.052                 0.006
                                  16    12.684       0.004             0.052                 0.006
                                  17    12.734       0.004             0.050                 0.006
                                  18    12.786       0.004             0.052                 0.006
                                  19    12.838       0.004             0.052                 0.006
                                  20    12.888       0.004             0.050                 0.006
               Interference in a Thin Film: Newton’s rings

               Analysis Part 1

   Average x = 0.050
       Excel “=AVERAGE(x)”
   Standard Deviation x = 0.004
       Excel “=STDEV(x)”
                Interference in a Thin Film: Newton’s rings

                Analysis Part 1

   t = /2 = x tan 
   /2 = x tan 
        for sodium light is 589.3 nm = 5.893*10-7m
       x average was 0.050 cm = 5.0 * 10-4m
   Solve for 
       (5.893*10-7m)/2 = (5.0 * 10-4m ) tan 
        = 3.4 * 10-2 degrees
                Interference in a Thin Film: Newton’s rings

                Analysis Part 1

   For small angles   (tan)
   (tan) = (/2) / x
       (tan) = (5.893*10-7m )/2 / (6*10-5 m)
       (tan) = .004
    = 0.034 +/- .004 degrees
Interference in a Thin Film: Newton’s rings

Analysis Part 2
                           Interference in a Thin Film: Newton’s rings

                           Analysis Part 2

   xm2 = xm2 * (xm/xm)2 + (xm/xm)2
   xm2 = xm2 * 2(xm/xm)2
        x12 = (0.062 cm) * 2(0.006/0.248)2
        x12 = 0.002
        m    xr (cm)   xl (cm)   xm (cm)   error xm (cm)   xm^2 (cm)   error xm^2 (cm)
        0    11.900
        1    11.770    12.018     0.248       0.006          0.062         0.002
        2    11.721    12.064     0.343       0.006          0.118         0.003
        3    11.678    12.092     0.414       0.006          0.171         0.004
        4    11.650    12.124     0.474       0.006          0.225         0.004
        5    11.618    12.152     0.534       0.006          0.285         0.005
        6    11.594    12.180     0.586       0.006          0.343         0.005
        7    11.576    12.202     0.626       0.006          0.392         0.005
        8    11.550    12.226     0.676       0.006          0.457         0.006
        9    11.530    12.246     0.716       0.006          0.513         0.006
        10   11.518    12.260     0.742       0.006          0.551         0.006
                          Interference in a Thin Film: Newton’s rings

                          Analysis Part 2
                                        Xm^2 versus m


              0.600

                                                            y = 0.0555x + 0.0064



              0.500




              0.400
Xm^2 (cm^2)




              0.300




              0.200




              0.100




              0.000
                      0   2         4            6      8                      10   12
                                                 m
             Interference in a Thin Film: Newton’s rings



   Used Excel’s LINEST() function to calculate
    slope and intercept
   Slope = 0.0555 +/- 0.0006                0.0555   0.006
                                             0.0006   0.004

   Intercept = 0.006 +/- 0.004
                Interference in a Thin Film: Newton’s rings

                Analysis Part 2

   xm2 = mR/4 - 2/4
   y = 0.0555x + 0.0064
   y = mx      +b
   To find 2
       - 2/4 = 0.006
       2 = -0.024 cm2
   2/4 = 0.004
       2 = 0.02
   2 = -0.02 +/- 0.02 cm2
                  Interference in a Thin Film: Newton’s rings

                  Analysis Part 2

   xm2 = mR/4 - 2/4
   y = 0.0555x + 0.0064
   y = mx      +b
   To find R
       mR/4 = 0.0555x             where x = m
            = 589.3 nm = 5.893 * 10-5 cm
       (5.893 * 10-5 cm)R/4 = 0.0555
       R=3.767 * 103 cm
                Interference in a Thin Film: Newton’s rings

                Analysis Part 2

   R/4 = 0.0006
   R = 0.0006 * 4 / 
       R = 0.0006 * 4 / 5.893 * 10-5 cm
       R = 40 cm


   R = 3.77 * 103 +/- 0.04 * 103 cm
             Interference in a Thin Film: Newton’s rings

             Conclusion

   We were able to see how rays of light could
    interfere causing fringes
   Dark fringes: destructive interference
   Bring fringes: constructive interference
   Interference explains colours in a thin film
                Interference in a Thin Film: Newton’s rings

                Conclusion

   Measured distance between fringes
   Able to calculate the thin film’s properties
   In Part 1 we were able to calculate
        = 0.034 +/- .004 degrees
   In Part 2 we found
       2 = -0.02 +/- 0.02 cm2
       R = 3.77 * 103 +/- 0.04 * 103 cm
                Interference in a Thin Film: Newton’s rings

                Conclusion

   Errors
       Error in positioning the travelling microscope over
        the same place for each fringe
       Error in taking measurements from the
        micrometer scale
       Error would arise in Part 1 if the fringes were not
        aligned perpendicular to the direction of the
        travelling microscope
                Interference in a Thin Film: Newton’s rings

                Conclusion

   Experiment could be improved
       by using a travelling light sensor instead of a
        microscope
             Interference in a Thin Film: Newton’s rings



   Questions?




   Additional pictures at www.goldenratio.ca

								
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