IPhO-37_Expt_A - The 37th International Physics Olympiad...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The 37th International Physics Olympiad Singapore Experimental Competition Wednesday, 12 July, 2006 Sample Solution Part 1 a. A sketch of the experimental setup (not required) Receiver Rotating table G oniometer F ixed arm Goniom eter Mov ab le arm Beam splitter Transmitter Holder Reflector b. Data sheet (not required) Position Meter Position Meter Position Meter Position Meter (cm) reading (cm) reading (cm) reading (cm) reading (mA) 104.0 103.9 103.8 103.7 103.6 103.5 103.4 103.3 103.2 103.1 103.0 102.9 102.8 102.7 102.6 102.5 102.4 102.3 102.2 102.1 102.0 101.9 101.8 101.7 101.6 101.5 101.4 101.3 0.609 0.817 0.933 1.016 1.030 0.977 0.890 0.738 0.548 0.310 0.145 0.076 0.179 0.392 0.623 0.786 0.918 0.988 1.026 1.006 0.945 0.747 0.597 0.363 0.161 0.055 0.139 0.357 (mA) 100.9 100.85 100.8 100.7 100.6 100.4 100.2 100.0 99.8 99.6 99.4 99.3 99.2 99.0 98.8 98.6 98.4 98.2 98..0 97.9 97.8 97.7 97.5 97.2 97.0 96.8 96.6 96.5 (mA) 1.016 1.060 1.090 0.994 0.940 0.673 0.249 0.074 0.457 0.883 1.095 1.111 1.022 0.787 0.359 0.079 0.414 0.864 1.128 1.183 1.132 1.015 0.713 0.090 0.342 0.714 1.007 1.087 96.0 95.8 95.6 95.4 95.3 95.2 95.1 95.0 94.9 94.7 94.5 94.3 94.1 93.9 93.7 93.6 93.5 93.4 93.2 93.0 92.8 92.6 92.4 92.2 92.15 92.1 92.0 91.8 2 0.514 0.098 0.192 0.669 0.870 1.009 1.119 1.138 1.080 0.781 0.403 0.044 0.364 0.860 1.103 1.160 1.159 1.083 0.753 0.331 0.073 0.515 0.968 1.217 1.234 1.230 1.165 0.871 (mA) 91.0 90.9 90.8 90.7 90.6 90.4 90.2 90.0 89.8 89.6 89.5 89.4 89.3 89.2 89.1 89.0 88.8 88.6 88.5 88.4 88.2 88.0 87.9 87.8 87.7 87.6 87.4 87.2 0.925 1.094 1.245 1.291 1.253 0.978 0.462 0.045 0.278 0.809 1.031 1.235 1.277 1.298 1.252 1.133 0.684 0.123 ­0.020 0.123 0.679 1.116 1.265 1.339 1.313 1.190 0.867 0.316 0.589 0.781 0.954 0.609 0.817 0.933 1.016 1.030 0.977 0.890 0.738 0.548 0.310 0.145 0.076 0.179 0.392 0.623 0.786 0.918 0.988 1.026 1.006 0.945 0.747 0.597 0.363 0.161 0.055 0.139 0.357 0.589 0.781 0.954 96.4 96.3 96.2 100.9 100.8 100.8 100.7 100.6 100.4 100.2 100.0 99.8 99.6 99.4 99.3 99.2 99.0 98.8 98.6 98.4 98.2 98.0 97.9 97.8 97.7 97.5 97.2 97.0 96.8 96.6 96.5 96.4 96.3 96.2 1.070 1.018 0.865 1.016 1.060 1.090 0.994 0.940 0.673 0.249 0.074 0.457 0.883 1.095 1.111 1.022 0.787 0.359 0.079 0.414 0.864 1.128 1.183 1.132 1.015 0.713 0.090 0.342 0.714 1.007 1.087 1.070 1.018 0.865 91.6 91.4 91.2 96.0 95.8 95.6 95.4 95.3 95.2 95.1 95.0 94.9 94.7 94.5 94.3 94.1 93.9 93.7 93.6 93.5 93.4 93.2 93.0 92.8 92.6 92.4 92.2 92.15 92.1 92.0 91.8 91.6 91.4 91.2 0.353 0.018 0.394 0.514 0.098 0.192 0.669 0.870 1.009 1.119 1.138 1.080 0.781 0.403 0.044 0.364 0.860 1.103 1.160 1.159 1.083 0.753 0.331 0.073 0.515 0.968 1.217 1.234 1.230 1.165 0.871 0.353 0.018 0.394 87.1 87.0 86.9 91.0 90.9 90.8 90.7 90.6 90.4 90.2 90.0 89.8 89.6 89.5 89.4 89.3 89.2 89.1 89.0 88.8 88.6 88.5 88.4 88.2 88.0 87.9 87.8 87.7 87.6 87.4 87.2 87.1 87.0 86.9 2.0 1.6 87.8 cm Mete r read in g (mA) 101.2 101.1 101.0 104.0 103.9 103.8 103.7 103.6 103.5 103.4 103.3 103.2 103.1 103.0 102.9 102.8 102.7 102.6 102.5 102.4 102.3 102.2 102.1 102.0 101.9 101.8 101.7 101.6 101.5 101.4 101.3 101.2 101.1 101.0 1.2 103.6 cm 0.8 0.4 0.0 ­0.4 88 90 92 94 96 p osition (cm) 3 98 100 102 104 0.034 ­0.018 0.178 0.925 1.094 1.245 1.291 1.253 0.978 0.462 0.045 0.278 0.809 1.031 1.235 1.277 1.298 1.252 1.133 0.684 0.123 ­0.020 0.123 0.679 1.116 1.265 1.339 1.313 1.190 0.867 0.316 0.034 ­0.018 0.178 From the graph (not required) or otherwise, the posit ions of the first maximum po int and th 12 maximum po int are measured at 87.8 cm and 103.6 cm. The wavelength is calculated by l 103. - 87. 6 8 = cm 2 Thus, 1.8 marks 11 l = 2.87 cm. Error analysis l = 2 d , 11 Dl = Dd = 0.05 x2 cm = 0.1 cm. 2 2 Dd = ´ 0 10 = 0 018 cm < 0 02 cm . . . 11 11 4 0.2 marks Part 2 (a) Deduction of interference condit ions A B q1 q1 n t q2 Assume that the thickness of the film is t and refractive index n. Let q be the incident 1 angle and q 2 the refracted angle. The difference of the optical paths DL is: DL = 2 / cos 2 - t tan 2 sin q1 ) (nt q q Law of refract ion: sin q = n sin q 2 1 Thus D = 2t n 2 - sin 2 q 1 L Considering the 180 deg (p) phase shift at the air­ thin film interface for the reflected beam, we have interference condit ions: 2 n 2 - sin 2 q min = m ( = 1 2 3 ...) t l m , , , and for the destructive peak 1 2t n 2 - sin 2 q max = ( ± )l m 2 for the constructive peak If thickness t and wave length l are known, one can determine the refractive index o f the thin film fro m I ­ q 1spectrum (I is the intensit y o f the interfered beam). 5 1 mark (b) A sketch of the experimental setup Receiver Goniomet er Movable arm Pla no­ conv ex cylinder lens θ θ Trans mitter Rotating tabl e Thin film 1 mark Goniometer Fix ed arm Students should use the labeling on Page 2. (c) Data Set X: q1 / degree 40.0 41.0 42.0 43.0 44.0 45.0 46.0 47.0 48.0 49.0 50.0 51.0 52.0 53.0 54.0 55.0 56.0 57.0 Y: Meter reading S/mA 0.309 0.270 0.226 0.196 0.164 0.114 0.063 0.036 0.022 0.039 0.066 0.135 0.215 0.262 0.321 0.391 0.454 0.511 6 58.0 0.566 59.0 0.622 60.0 0.664 61.0 0.691 62.0 0.722 63.0 0.754 64.0 0.796 65.0 0.831 66.0 0.836 67.0 0.860 68.0 0.904 69.0 0.970 70.0 1.022 71.0 1.018 72.0 0.926 73.0 0.800 74.0 0.770 75.0 0.915 Uncertaint y: angle Dq = ±0 5 , current: ±0.001 mA . o 1 0.5 marks 1.2 o 70.5 Meter reading (mA) 1.0 0.8 0.6 0.4 0.2 0.0 o 48 35 40 45 50 55 60 65 70 75 80 o q ( ) o o From the data, q min and q max can be found at 48 and 70.5 respect ively. To calculate the refract ive index, the fo llowing equations are used: 7 0.9 marks 0.6 marks 2 n 2 - sin 2 48 o = m ( = 1 2 3 ...) t l m , , , 1 2t n 2 - sin 2 70. = ( - )l 5 o m 2 and (1) (2) In this experiment, t = 5.28 cm, l = 2.85cm (measured using other method). Solving the simultaneous equations (1) and (2), we get m = o sin 2 70. o - sin 2 48 5 + 0. 5 2 l 2 ( ) 2t m = 4.83 1 mark m = 5 Subst ituting m = 5 in (1), we get n = 1.54 0.5 marks Subst ituting m = 5 in (2), we also get n = 1.54 Error analysis: m l n = sin 2 q + ( ) 2 2 t Dn = = 1 m l sin 2 q + ( ) 2 2t (sin 2q · Dq + m 2 l m 2 l2 Dl Dt ) 2t 2 2t 3 1 m 2 l m 2 l 2 (sin 2q · Dq + 2 Dl Dt ) n 2t 2t 3 o o If we take Dq = ±0.5 = ±0.0087 rad, Dt = ±0.05 cm, Dl = ±0.02 cm, and q = 48 Dn = Thus, 2 1 5 2 ´ 2 85 . 5 ´ 2 85 2 . (0 0087 sin 96 o + . ´ 0 01 + . ´ 0 05 » 0 02 . ) . 2 1 54 . 2 ´ 5 28 . 2 ´ 5 28 . 3 n + Dn = 1.54 ±0.02 0.5 marks 8 Part 3 Sample Solution Task 1 Sketch your final experimental setup and mark all components using the labels given at page 2. In your sketch, write down the distance z (see Figure 3.2), where z is the distance fro m the tip of the pr ism to the central axis o f the transmitter. Transmitter z Lens Prism Prism d Receiver (Students should use labels on page 2.) Task 2 Tabulate your data. Perform the experiment twice. Data Set X: d(cm) DX(cm) 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 Set 1 S1 (mA) 0.78 0.68 0.58 0.50 0.42 0.36 0.31 0.26 0.21 Set 2 S2(mA) 0.78 0.69 0.59 0.51 0.42 0.35 0.31 0.25 0.22 Saverage (mA) 0.780 0.685 0.585 0.505 0.420 0.355 0.310 0.255 0.215 # DS(mA) 2 It (mA) * $ D(It) Y: ln(It (mA) 2 ) DY& 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.6080 0.4690 0.3420 0.2550 0.1760 0.1260 0.0961 0.0650 0.0462 0.016 0.014 0.012 0.010 0.008 0.007 0.006 0.005 0.004 ­0.50 ­0.76 ­1.07 ­1.37 ­1.74 ­2.07 ­2.34 ­2.73 ­3.07 0.03 0.03 0.03 0.04 0.05 0.06 0.06 0.08 0.09 # DS = 0.01 mA (for each set of current measurements) S2 proportional to the intensit y, It $ D(S2 ) = DIt = 2 S ´ DS & DY = D(lnIt) = D(It)/It * 9 Task 3 By plotting appropriate graphs, determine the refractive index, n1, of the prism with error analys is. Write the refract ive index n1, and its uncertaint y ∆n1, of the prism in the answer sheet provided. Least Square Fitting X = d(cm) 0.60 0.70 0.80 0.90 1.00 1.10 1.20 1.30 1.40 SX = 9.00 DX(cm) 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 Y = ln(It) ­0.50 ­0.76 ­1.07 ­1.37 ­1.74 ­2.07 ­2.34 ­2.73 ­3.07 DY 0.03 0.03 0.03 0.04 0.05 0.06 0.06 0.08 0.09 DY2 0.001 0.001 0.001 0.002 0.002 0.003 0.004 0.006 0.009 SY = ­15.648 SDY = 0.469 2 S(DY) = 0.029 10 XY ­0.298 ­0.530 ­0.858 ­1.230 ­1.735 ­2.278 ­2.811 ­3.553 ­4.304 2 X 0.360 0.490 0.640 0.810 1.000 1.210 1.440 1.690 1.960 Y2 0.247 0.573 1.150 1.867 3.010 4.290 5.487 7.469 9.451 SXY = SX2 = SY2 = ­17.596 9.600 33.544 From I t = I 0 exp ( - g d ) , taking natural log on both sides, we obtain: 2 ln( I t ) = -2g d + ln( I 0 ) which is of the form y = mx + c. To calculate the gradient, the fo llowing equat ion was used, where N is the number of data points: m = N å ( XY ) - ( å X ) ( å Y ) 2 N å X 2 - ( å X ) = -3.247 To calculate the standard deviat ion sY o f the individual Y data values, the fo llowing equation was used: 2 s Y = å ( DY ) N - 2 = 0.064 Hence the standard deviat ion in the slope can be calculated: s m = s Y N 2 N å X 2 - ( å X ) = 0.082 From the gradient: 2g = 3.247 ± 0.082 » 3.25 ± 0.08 Using: n1 = k 2 + g 2 2 k2 sin q1 o where q1 = 60 , k2 = 2p/l » 2.20 (using the wavelength determined from earlier part (using l = (2.85 ± 0.02)cm), we obtain: n1 ± Dn1 = 1.434 ± 0.016 » 1.43 ± 0.02 11 Error Analysis for refract ive index of n1 é k 2 + g 2 12 ù ) ú Dk + d d ê( 2 Dn1 = 2 dk2 ê k2 sin q1 ú dg ê ú ë û é k 2 + g 2 1 2 ù ) ú Dg ê ( 2 ê k2 sin 1 ú q ê ú ë û 1 é k 2 + g 2 - 12 é g k 2 + g 2 - 1 2 ù k22 + g 2 ) 2 ù ( 2 ) -( ) ú Dg ú Dk + ê ( 2 Dn1 = ê 2 2 ê sin q1 ê k2 sin 1 ú k2 sin q1 ú q ê ú ê ú ë û ë û = 0.016 » 0.02 where: Dk2 = - 2p l 2 Dl = -0.015 Note: Other methods of error analysis are also accepted. 12 Part 4 Task 1 Top­view of a simple square lattice. a d 0.5 marks Figure 4.1: Schematic diagram of a simple square lattice with lattice constant a and interplaner d of the diagonal planes indicated. Deriving Bragg's Law Condit ions necessary for the observation of diffraction peaks: 1. The angle o f incidence = angle of scattering. 2. The pathlength difference is equal to an integer number of wavelengths. 13 Figure 4.2: Schematic diagram for deriving Bragg's law. h = d sinq (1). The path length difference is given by, 2h = 2d sinq (2). For diffract ion to occur, the path difference must satisfy, 2 d sinq = ml, m = 1, 2, 3... ( 3). a d 0.5 marks Figure 4.3 Illustration of the lattice used in the experiment (this Figure is not required) 14 Fig. 4.4 Actual lattice used for the experiment (this Figure is not required) Fig. 4.4 The actual lattice used in the experiment (not required) Task 2 (a) Lattice Box I on Rotating Table L I Plano­cylindrical Lens on Holder J Digital Multimeter N q A Microwave Transmitter on Holder D z = 180° ­ 2q B Microware Receiver on Holder 1.5 marks Fig. 4.5 Sketch of the experimental set up 15 Task 2(b) & 2(c) Data Set z/° 20.0 21.0 22.0 23.0 24.0 25.0 26.0 27.0 28.0 29.0 30.0 31.0 32.0 33.0 34.0 35.0 36.0 37.0 38.0 39.0 40.0 41.0 42.0 43.0 44.0 45.0 46.0 47.0 48.0 49.0 50.0 140.0 138.0 136.0 134.0 132.0 130.0 128.0 126.0 124.0 122.0 120.0 118.0 116.0 114.0 112.0 110.0 108.0 106.0 104.0 102.0 100.0 98.0 96.0 94.0 92.0 90.0 88.0 86.0 84.0 82.0 80.0 Output current S (mA) 0.023 0.038 0.070 0.109 0.163 0.201 0.233 0.275 0.320 0.350 0.353 0.358 0.354 0.342 0.321 0.303 0.280 0.241 0.200 0.183 0.162 0.139 0.120 0.109 0.086 0.066 0.067 0.066 0.070 0.084 0.080 Intensity 2 I=S 2 (mA) 0.000529 0.001444 0.0049 0.011881 0.026569 0.040401 0.054289 0.075625 0.1024 0.1225 0.124609 0.128164 0.125316 0.116964 0.103041 0.091809 0.0784 0.058081 0.04 0.033489 0.026244 0.019321 0.0144 0.011881 0.007396 0.004356 0.004489 0.004356 0.0049 0.007056 0.0064 0.14 0.12 2 2 Rellattive iIntensiitty (mA) R e a i ve n tens y (mW) q/° 0.1 0.08 0.06 0.04 0.02 0 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 Theta (degree) 2.7 marks Task 2(d) From eq 3 and let m = 1, 2d sin q max = l (4) From Fig. 4.3, 16 (5) a = 2 d Combine eqs (4) and (5), we obtain, a = l 2 sin q max From the symmetry of the data, the peak position is determined to be: qmax = 31° (The theoretical value is qmax = 32°) a = l 2 85 . cm = = 3 913 . cm o 2 sin q max 2 sin 31 (Actual value a = 3.80 cm) [The value 3.55 in the marking scheme is derived from: l 2 83 . cm a = = = 3 58 . cm 2 sin q max 2 sin 34 o where 2.83 cm and 34 deg are the min and max allowed values for wavelength and peak position. Similarly: l 2 87 . cm = = 4 06 . cm 2 sin q max 2 sin 30 o l 2 83 . cm The value 3.55 is derived from: a = = = 3 58 . cm 2 sin q max 2 sin 34 o The value 4.10 is derived from: a = l 2 83 . cm = = 3 49 . cm o 2 sin q max 2 sin 35 l 2 87 . cm The value 4.20 is derived from: a = = = 4 18 ] . cm 2 sin q max 2 sin 29 o The value 3.40 is derived from: a = Error analys is: Known uncertainties: ∆λ = 0.02 cm; ∆q = 0.5 deg = 0.014 rad. (uncertaint y in determining q from graph). 17 From: a = D = a l 2 sin q max D l 2 sin q max Dl = a ( - l d (sin q max ) q D 2 q 2 (sin q max ) d 1 d (sin q max ) q ) D sin q max d q l Dl = a - co t q max Dq ) ( l 0. 2 0 = 3 80( . - co t(32 ) ´ ( 0. 14)) ° - 0 cm 2. 5 8 = 0 112 » 0. . cm 1 Hence: 0.8 marks a ± ∆a = 3.913 ± 0.112 » 3.9 ± 0.1 cm 18 ...
View Full Document

This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online