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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 = mR 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 = mR xm2 = mR/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 = mR/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 = mR/4 - 2/4 y = 0.0555x + 0.0064 y = mx +b To find R mR/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? 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