12-11由l·sinb=得 sin e 5893×10-7 2n22×152×5×10-3=3.88×10-5 O=3.88×10-5rad=8 ek+1 = -ek 20条明条纹对应平晶厚度差为 △d=19(ek+1 e)1919×6328×10 2×1.5 40×10-(m) 12-13(1)≈gO=a=0048×10-13 0.12 A680×10 (2)ek+1 =3.40×10 2 2×1 680×10-9 (3)l= mm 2n、O2×1×4×10-4 =8.5×10m=0.85(n (4)N=0.12 8.5×10-4=141 .反射光中明条纹的条件为:2ne=k 油膜边缘e=0 油膜中心e=h=12×10°m 2n2e2×1.2×1.2×10 k 4.8 6×10-7 故共可看到五条明条纹(k=0,1,2,3,4) (2)对应各明条纹中心油膜的厚度e 当k=0,1,2,3,4时,对应油膜的厚度分别为:0,2500A,5000A,7500A,10000A132 12-11 由 2 2 sin n l = 得 3.88 10 rad 8 3.88 10 2 1.52 5 10 5.893 10 2 sin 5 5 3 7 2 = = = = = − − − − n l 12-12 ∵ 2 1 2n e e k k + = − = ,∴ 20 条明条纹对应平晶厚度差为 2 1.5 19 6.328 10 2 19 19( ) 7 2 1 = − = = − + n d e e k k 4.0 10 (m) −6 = 12-13 (1) 0.12 0.048 10−13 = = L d tg 4 10 (rad) −4 = (2) 3.40 10 m 2 1 680 10 2 7 9 2 1 − − + = − = = n e e k k (3) 8.5 10 m 0.85(mm) 2 1 4 10 680 10 2 4 4 9 2 = = = = − − − n l (4) 141 8.5 10 0.12 4 = = N − 12-14 (1)∵ n1 n2 n3 ∴ 反射光中明条纹的条件为: 2n2 e = k 油膜边缘 e=0 ∴ k=0 油膜中心 1.2 10 m −6 e = h = ∴ 4.8 6 10 2 2 1.2 1.2 10 7 6 2 = = = − − n e k 故共可看到五条明条纹(k=0,1,2,3,4) (2)对应各明条纹中心油膜的厚度 2n2 k e = 当 k=0,1,2,3,4 时,对应油膜的厚度分别为:0,2500 A ,5000 A ,7500 A ,10000 A