Assignment 8, solutions:

Problem 1:

tana = A_y/A_z,  cosa = A_z/[A_y² + A_z²]^1/2,  sina = A_y/[A_y² + A_z²]^1/2. 

(a)  Let I' be the transmitted intensity, and I the incident intensity.  The intensity is proportional to the square of the amplitude.  The transmitted amplitude is proportional to cos(q-a).  Therefore

I'/I = cos²(q-a) = [cos(q)cos(a) + sin(q)sin(a)]²

= cos²(q)cos²(a) + sin²(q)sin²(a) + 2cos(q)cos(a)sin²(q)sin(a)

= [(A_ysinq)² + (A_zcosq)² + 2A_yA_zcos(q)sin(q)]/[A_y² + A_z²]

Let q = 90 deg.  Then I'/I = A_y²/[A_y² + A_z²] = tan²a/(tan²a + 1).

Let q = 0 deg.  Then I'/I = A_z²/[A_y² + A_z²] = 1/(tan²a + 1).

(b)  

Problem 2:

Let the optic axis be the x-axis.  Then

E_x = E_x0exp[i(k_ox-wt)] =  E_x0exp[iw((n_ox/c)-t)],

E_y = E_y0exp[i(k_ex-wt)] =  E_y0exp[iw((n_ex/c)-t)],

since k = 2pn/l = 2pfn/(lf) = wn/c.

After a distance L the phase difference is

Df = (wL/c)Dn = (w/c)(l/(2Dn))Dn = wl/(2c) = 2plf/(2c) = p.

incident: a = 45^o, E_x = E_y.

transmitted: E_y = E_xe^±ip = -E_x, a = -45^o.

L = l/(2Dn) = [(633 10^-9)/(2 10^-5)] m = 3.165 cm

Problem 3:

Brewster angle:  tanq_B = n₂/n₁ = 1.5. q_B = 56.31^o.

n₁sinq_B  = n₂sinq_t  ,  q_t = 33.69^o.

q_B + q_t = 90^o.

The Brewster angle for the second interface is tanq_B' = n₁/n₂,  q_B' = 33.69^o.

It satisfies Brewster's condition at the second interface.

Problem 4:

r_12s = sin(q₁ - q₂)/sin(q₁ + q₂).

r_12s is the Fresnel reflection coefficient for s-polarization.

R = |r_12s|² = I_reflected/I_incident.

For the first interface q₁ = q_B and q₂ = q_t. 

r_12s = sin(22.62^o)/sin(90^o) = 0.385

R = 0.148.  85.2% of the s-polarized incident beam is transmitted through the first interface.

For the second interface q₁ = q_t and q₂ = q_B. 

R = 0.148.  72.6% of the s-polarized incident beam is transmitted through the second interface.

100% of the p-polarized incident beam is transmitted through both interfaces.

Problem 5:  (C)

Problem 6:  (B)