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Experiment 17 - Andrew Underwood

Experiment 17 - Andrew Underwood. 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. Theory. Index of refraction glass n = 1.5 air n = 1.0

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Experiment 17 - Andrew Underwood

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  1. Experiment 17 - Andrew Underwood

  2. 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

  3. 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

  4. Theory • A phase reversal is equivalent to shifting the ray by half a wavelength Diagram from Physics 294 Laboratory Manual, page 17-1

  5. Theory • When waves interfere constructively:

  6. Theory • When waves interfere destructively:

  7. Theory • When waves interfere destructively:

  8. Theory Part 1 Diagram from Physics 294 Laboratory Manual, page 17-2

  9. 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

  10. Theory Part 2

  11. Theory Part 2

  12. Theory Part 2

  13. Theory Part 2 Diagrams from Physics 294 Laboratory Manual, page 17-3

  14. 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

  15. Apparatus

  16. Apparatus Diagram from Physics 294 Laboratory Manual, page 17-2

  17. Procedure

  18. 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 

  19. 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

  20. Analysis

  21. Analysis Part 1 • Error for x • x = (x)2 + (x)2 • x = 2(x)2 • x = 2(0.004 cm)2 • x = 0.006 cm

  22. Analysis Part 1 • Average x = 0.050 • Excel “=AVERAGE(x)” • Standard Deviation x = 0.004 • Excel “=STDEV(x)”

  23. 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

  24. 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

  25. Analysis Part 2

  26. 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

  27. Analysis Part 2

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

  29. 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

  30. 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

  31. 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

  32. 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

  33. 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

  34. 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

  35. Conclusion • Experiment could be improved • by using a travelling light sensor instead of a microscope

  36. Questions? • Additional pictures at www.goldenratio.ca

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