Solution (The Experimental Question):Task 1 1a.nominal =5´=0.08nominal (degree)0.08If a is the distance between card and the grating and r is the distance between the hole and the light spot so we have,...,21x x f 0,2tan arWe want0 to be zero i.e.r04.0007.0170,10radrad mma mm r 00.4range of visible light (degree)1326中华物理.c om中华物理竞赛网 官方网站 圣才学习网 1c.minR (21.6±0.1) k 05´ = 0.081minRR=(192±1) k0=5´ because= 5´ => R= (21.9±0.1) k =-5´ => R= (21.9±0.1) k1d.Table 1d. The measured parameters(degree)R glass (M )R glass (M )R film (M )R film (M )15.00 3.770.03183315.50 2.580.02132216.00 1.880.0187116.50 1.190.0151.50.517.000.890.0133.40.317.500.680.0119.40.118.000.4860.00510.40.118.500.3650.005 5.400.0319.000.2740.003 2.660.0219.500.2250.002 1.420.0120.000.2000.0020.8800.00520.500.2270.0020.8220.00521.000.3680.003 1.1230.00721.500.6000.005 1.610.0122.000.7750.005 1.850.0122.500.830.01 1.870.0123.000.880.01 1.930.0223.50 1.010.01 2.140.0224.00 1.210.01 2.580.0224.50 1.540.01 3.270.0225.00 1.910.01 4.130.0216.25 1.380.0166.50.516.75 1.000.0140.00.317.250.720.0123.40.217.750.5350.00512.80.118.250.3910.003 6.830.0518.750.2930.003 3.460.0219.250.2350.003 1.760.0119.750.1950.0020.9880.00520.250.2010.0020.7760.00520.750.2730.0030.890.01中华物理竞赛网ww w .100w u li .c om中华物理竞赛网 官方网站 圣才学习网 1e.In =-20 => R glass = (132± 2) k , R film = (518±5) kT filmT film0.25519.250.13419.500.15819.750.19720.000.22720.250.25920.500.27620.750.307GraphicsWe see that: T(= 20.25) = T(= -20)(degree)0.25±0.08T film |=-20°中华理竞赛网ww w .100w li .c omTask 2.2a.sin d d= 2.9 cos() (nm)filmglass filmT TR R T2b.26132.62.8 nmmm 1dand degree0.085where 中华物理竞赛网ww w.u li .c om2c.Table 2c. The calculated parameters using the measured parameters(degree)(nm)I g /C((M )I s /C() (M -1)T film15.04280.2650.005460.0206 3.8815.54420.3880.007580.0195 3.9416.04560.5320.01150.0216 3.8316.254630.7250.01500.0208 3.8816.54700.8400.01940.0231 3.7716.75477 1.000.02500.0250 3.6917.0484 1.120.02990.0266 3.6317.25491 1.390.04270.0308 3.4817.5498 1.470.05150.0351 3.3517.75505 1.870.07810.0418 3.1718.0512 2.060.0960.0467 3.0618.25518 2.560.1460.0572 2.8618.5525 2.740.1850.0676 2.6918.75532 3.410.2890.0847 2.4719.0539 3.650.3760.103 2.2719.25546 4.260.5680.134 2.0119.5553 4.440.7040.158 1.8419.75560 5.13 1.010.197 1.6220.0567 5.00 1.140.227 1.4820.25573 4.98 1.290.259 1.3520.5580 4.41 1.220.276 1.2920.75587 3.66 1.120.307 1.1821.0594 2.720.8900.328 1.1221.5607 1.670.6210.3730.9922.0621 1.290.5410.4190.8722.5634 1.200.5350.4440.8123.0648 1.140.5180.4560.7923.56610.990.4670.4720.7524.06750.8260.3880.4690.7624.56880.6490.3060.4710.7525.07010.5240.2420.4620.77中华物.c om2d.max (I glass )564±5 (nm) max (I film )573±5 (nm)中wwTask 3.3a.Table 3a. The calculated parameters for each measured data point(degree)x (eV) y ( eV 2) 15.002.898126.615.50 2.806121.916.00 2.720108.816.25 2.679107.816.50 2.63998.916.75 2.60092.017.00 2.56386.317.25 2.52777.417.50 2.49169.717.75 2.45760.918.00 2.42455.118.25 2.39246.818.50 2.36040.418.75 2.33033.119.00 2.30027.319.25 2.27120.9119.50 2.24317.0719.75 2.21512.9220.00 2.18810.5120.25 2.1628.5320.50 2.1377.5620.75 2.112 6.2321.00 2.088 5.4321.50 2.0414.0622.00 1.997 3.0222.50 1.954 2.5223.00 1.914 2.2623.50 1.875 1.9824.00 1.838 1.9424.50 1.803 1.8425.001.7691.86中华物理竞赛网ww w .100w u li .c om3b.3c.Am tt A mE x tA y E h tA h t E h A h ggg2222中华物理竞In linear range we have, m=213 (eV), r 2= 0.9986, E g =2.17 (eV)and we have /nm eV 0.071A 1/2 so we find t= 206 (nm)222xm yxymwhere x & y are the mean of error range of x & yyxSo 2eV9.0,eV 014.0yxm 10 (eV) t = tm/(2 m) 5 (nm) 11mmE g02.0gTable 3d. The calculated values of E g and t using Fig. 3E g (eV) E g (eV) t (nm) t (nm)2.170.022065中华物理竞赛网ww w 1w li .c om中华物理竞赛网 官方网站 圣才学习网 This document was created with Win2PDF available at .The unregistered version of Win2PDF is for evaluation or non-commercial use only.中华物理竞赛网ww w .100w u li .c om中华物理竞赛网 官方网站 圣才学习网 。