資料處理：(1)逐差法d/mm0.2990.3020.2920.2910.2960.298=0.293l=7.34cmL=101.80cmD=110.21cm/cm/cm/cm=/cm/g4.794.674.734.34+500/g3.813.503.664.73+1000/g2.782.522.654.69+1500/g1.461.191.564.88+2000/g0.540.240.39+2500/g-0.55-1.58-1.07+3000/g-1.77-2.31-2.04+3500/g-3.32-3.32=4.66cm=0.55cmE/E=D/D+L/L+2d/d+l/l+b/b+F/FD=2mmL=2mmd=0.004mml=0.2mmm=5gE/E0.1710.1600.1610.157(E/E)平均=0.162E=2.078e111.879e111.821e111.821e11=1.900e11E=E/E*E=3.078e10E=(1.9000.3078)e11(N/M•M)（2）作圖法[2006-5-2219:53"/Graph1"(2453877)]LinearRegressionforData1_B:Y=A+B*XParameterValueError------------------------------------------------------------A4.529170.93036B-0.001834.44796E-4------------------------------------------------------------RSDNP-------------------------------------------------------------0.859011.441380.00629------------------------------------------------------------[2006-5-2220:02"/Graph2"(2453877)]LinearRegressionforData1_B:Y=A+B*XParameterValueError------------------------------------------------------------A4.869170.08835B-0.002314.22385E-5------------------------------------------------------------RSDNP-------------------------------------------------------------0.9990.136878<0.0001即M=0.00231E=8DL/(lM)=1.963e11思考题1．利用光杠杆把测微小长度ΔL变成侧b，光杠杆的放大率为2D/L，根据此式能否以增加D减小来提高放大率，这样做有无好处？有无限度？应怎样考虑这个问题？答：有一定的好处，但是如果放大倍数特别大的话，实验过程中微小的误差就可会放大成巨大的误差。所以应当保证放大后观察者能清晰地读值，而不是将放大倍数设的过大。
