Assignment 6, solutions:

Problem 1:

 For the wave guide switch we have |x₂(d)|² = [|k₁₂|/B₀]²sin²(B₀d)|x₁(0)|².  The maximum transfer of energy occurs from guide 1 to guide 2 occurs when B₀d = np/2, n = odd.  When Db = 0 and B₀d = np/2, n = odd, full transfer occurs.

We need B₀d = np for no transfer to occur.  Let n = 1.

(B₀d )² = (b₁-b₂)²d²/4 + (kd)² = (b₁-b₂)²d²/4 + (p/2)² = p².

(b₁-b₂)²d²  = 3p².  (b₁-b₂)  = (3)^1/2p/d = 2(3)^1/2k.

Problem 2:

(a)  dx₁/dz = ib₀x₁ - ikx₂   and   dx₂/dz = ib₀x₂ - ikx₁.

Let x₁(z) and x₂(z) be proportional to exp(ibz).  Then

.

Setting the determinant of the above matrix equal to zero we have (b-b₀)² - k² = 0.

b₊ = b₀ + k,  b_- = b₀ - k.

The propagation constants of the coupled system are b₊ and b_-.
The mode corresponding to  b₊  has x₁ = -x₂,
x₁(z,t) = Cexp(i(b₊z - wt)) = Cexp(ib₊(z - vt)),
and the mode corresponding to  b_- has x₁ = x₂,
x₁(z,t) = Aexp(i(b_-z - wt)) = Aexp(ib_-(z - vt)).

(b)  The same equations have the same solutions.  For the reflected waves the propagation constants of the coupled system are b₊ and b_- and the mode corresponding to  b₊  has y₁ = -y₂,
y₁(z,t) = C'exp(-i(b₊z + wt)) = Cexp(-ib₊(z + vt)),
and the mode corresponding to  b_- has y₁ = y₂,
y₁(z,t) = A'exp(-i(b_-z + wt)) = A'exp(-ib_-(z + vt)).
(The reflected wave travels into the negative z-direction.)

The boundary conditions of the cavity have to be satisfied.  At z = 0 and z = d we need

x₁ + y₁ = 0,  x₂ + y₂ = 0.

For the mode corresponding to  b₊  we have 
x₁ = Cexp(ib₊(z-vt)),  x₂ = -Cexp(ib₊(z-vt)),  y₁ = C'exp(-ib₊(z+vt)),  y₂ = -C'exp(-ib₊(z+vt)).

z = 0: C + C' = 0.
z = d: Cexp(ib₊d) + C'exp(-ib₊d) = 0.

Therefore exp(ib₊d) - exp(-ib₊d) = 0,  sin(b₊d) = 0,  b₊d = np,  b₀ = np/d - k.
b₊= 2p/l₊, so 2pd/l₊ = np,  d = nl₊/2, we have a standing wave.

Similarly, for the mode corresponding to  b_-  we have b₀ = np/d + k.
We have standing waves.