一、选择题(共 33 分) 1. 在真空中波长为 λ 的单色光，在折射率为 n 的透明介质中从 A 沿某路径传播到 B，若 A， B 两点位相差为 3π，则此路径 AB 的光程为( ) (A)1.5λ． (B)1.5nλ． (C)3λ． (D)1.5λ/n．

首先，复习e=Bδlv公式，说明 e正比于Bδ。结合图2.1解释v=2πRn/60（m/s， n(r/min)）; 机械角速度Ω=v/R=2πn/60 ( r/s); 电角速度ω=pΩ=p2πn/60 (rad/s) （记下来）；导体或线圈

光纤(Optical Fiber)即光导纤维，我们讨论通信所用的阶跃光纤。它的简化模型是中心纤芯半径为a，折射率为n1；层半径为b，折射率为n2；外部空气折射率为n0，并满足

Wheat derived from separate ancestors, and which differ in chromosome numbers. One classification of the races, with the probable wild types, chromo￾some numbers (2n) and the cultivated forms is shown in Table

一、寄存器 寄存器是用来寄存数码的逻辑部件，所以必须具备接收和寄存数码的 功能。任何一种触发器都可以构成寄存器，每一个触发器存放一位二进制数 或一个逻辑变量，用n个触发器组成的寄存器就可以存放n位二进制数或n个 逻辑变量

Interpretation over a singleton Let I be , and σ ∈ ΣI. 1. I(A)(σ) = I(∀xA)(σ). 2. I(t)(σ) = a. 3. I(Sx1,···,xn t1,···,tn A)(σ) = I(A)(σ). 4. I0(P), σ(P) ∈ {I(n), Ψ(n)} for every n-ary predicate constant (variable), where

Interpretation An interpretation I of F is , where D is a non-empty set called the domain of individuals. I0 is a mapping defined on the constants of F satisfying 1. If c is an individual constant, then I0(c) ∈ D. 2. If f n is an n-ary function constant, then I0(f n) : Dn → D

Substitutivity of Equivalence Let A,M and N be wffs and let AMN be the result of replacing M by N at zero or more occurrences (henceforth called designate occurrences) of M in A. 1. AMN is a wff. 2. If |= M ≡ N then |= A ≡ AMN