# 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?