3 Quarter-Wave and Half-Wave Stacks

1 Dielectric-Slab as a Fabry-Pérot (FP) Etalon

(a) For the slab shown above, we want to calculate both the transmittance and reflectance via

Figure 1: Slab with refractive index different than the surrounding medium Airy formula. The 1^st interface has the following matrix,

M1=

1 t^∗₁

r1

t1

r^∗₁ t^∗₁

1 t1

!

Then, for the propagation inside the slab, we have a 2^nd matrix, M₂ =

e^iφ 0 0 e^−iφ

whereφ=kon2d. Lastly, we have another matrix representing the transmission betweenn2 to n₁ at the second interface, so

M3=

1 t^∗₂

r2

t2

r^∗₂ t^∗₂

1 t2

!

As is known, the transfer matrix M is the multiplication of the three matrices,

M =M₃M₂M₁=





1 t^∗

r t r^∗ t^∗

1 t





whereas r and t are the overall reflection and overall transmission coefficient, respectively of the Etalon slab.

From the transmission coefficient

t= t₁t₂eⁱφ

|1−r1r2e^2iφ| from t we can calculate the transmittance,

T =|t|²= |t₁t2|²

|1−r₁r₂e^2iφ|²

where for normal incidence

r₁= n₁−n₂ n1+n2

r2 = n2−n1

n₂+n₁ =−r₁ subsititutingr₁ andr₂ into the transmittance equation gives

T = T_max 1 + (²_π^z)²sin²φ .

This is the Airy formula, whereas

Tmax = |t₁t2|²

(1− |r₁r2|)² = (1− |r₁|²)(1− |r₂|²) (1− |r₁r2|)² and

z= πp

|r₁r₂| 1− |r₁r₂| is the Finesse.

Sincer1=−r₂, thenTmax =1 andz= _1−|r^π|r1|

1|². |r₁|=√

R1, whereR1 is the reflectance of the interface. Thus,

z= π√ R₁ 1−R₁

=πn₂−n₁ n2+n1

1 1−(ⁿ_n²⁻ⁿ¹

2+n1)²

=πn₂−n₁ n2+n1

(n₂+n₁)² (n2+n1)²−(n2−n1)²

z=πn₂−n₁ 4n2n1

T = 1

1 +⁽ⁿ_4n²⁻ⁿ¹⁾²

2n1

sin²φ

Given thatφ=k_on₂d, wherek_o= ^ω_c and φ=ωn₂^d_c, where ω_o = cπ

n₂d , soφ= _ω^ω

o. Hence,

T = 1

1 +⁽ⁿ_4n²⁻ⁿ2 ¹⁾²

2n²₁ sin²(^ωπ_ω

o) such that the maxima occurs forω=nω_o.

(b) Now, we specify: n₂=3.5, n₁=1.5, d=200 nm (n2−n1)²

4n²₂n²₁ = 1

n²₂n²₁ '0.036

ωo= cπ

n₂d ' π×(3×10⁻⁸)

3.5×(2×10⁻⁷) '1.34×10¹⁵rad/s

We conclude - from the curve shown above - that the Fabry-Pérot etalon weakly modulates the transmission (by only 4%).

Figure 2: Modulated transmittance vs. angular frequency

2 Air Gap in Glass

(a) Here we have two cases - as depicted in the figure below - one case: the interface between air and glass, and the second case: the interface between glass and air. In either case, the beam is incident at Brewster angle. Recall that the Brewster angle is the angle at which the

Figure 3: Light incidence across air-glass and vice versa

reflectance of P-polarized light is zero, as shown below. The reflectance is calculated from the reflectance coefficient,

rf = n2cscθ1−n1cscθ2

n2cscθ1+n1cscθ2

= 0

where cscθ = _cos¹_θ and n1sinθ1 = n2sinθ2 (from Snell’s law). Thus, the Brewster angle is calculated as

tanθB= n1

n2

wheren1 is the medium of the incident wave.

We have two situations:

Figure 4: Reflectance of P and S polarized light (1)n1=1,n2 =1.5, so tanθB =0.6666, hence θB '33.70

(2)n1=1.5, n2=1, sotanθB =1.5, hence θB2 '56.30 Be careful: from glass to air we have total internal reflection:

n1sinθc=n2sinθ2=n2

sinθc= n2

n₁ = 0.666 θc'41.75^◦ < θB

This means at the Brewster angle, we don’t have zero reflectance but total internal reflection.

(b) For a stack of parallel glass plates filled with air, as depicted below, we want to prove that TM polarized light incident at Brewster angle at the first plate, wouldn’t be reflected while traversing other plates.

Since we hit the first glass plate at Brewster angle '33,70, we shall hit the following plates

Figure 5: P-polarized light traversing a stack at Brewster angle

at smaller angles. Notice that the reason for the light not being reflected while propagating through the plates is because the reflected light interferes destructively. When we replace air with water, theθ_B1 is no longer the same, hence the reflectance is not 0 anymore.

3 Quarter-Wave and Half-Wave Stacks

The stack shown below can be considered as a unit cella=d1+d2 repeated N times, where each of the alternating layer hasn₁ and n₂ and different thicknessd₁ and d₂. In this problem,

Figure 6: Representation of double layer repeated N times the optical path is equal, such thatn₁d₁ =n₂d₂.

We calculate the transfer matrix M for the double layer. So, we should consider two interfaces - as clarified below - matricesM₁ and M₃, as well as two propagation matricesM₂ andM₄.

Figure 7: Alternating interface-propagation across double layer

M_DL =M₄M₃M₂M₁

The four matrices contained in the transfer matrix, could be better grouped into only two matricesM_A=M₄M₃ and M_B=M₂M₁.

MA= 1 2n1n2

(n₁+n₂)e^−iφ2 (n₁−n₂)e^iφ2 (n1−n2)e^−iφ2 (n1+n2)e^iφ2

M_B = 1 2n1n2

(n₁+n₂)e^−iφ1 (n₁−n₂)e^iφ1 (n₁−n₂)e^−iφ1 (n₁+n₂)e^iφ1

MDL =MAMB = e^{−i(φ1+φ2)} 4n²₁n²₂

(n₂+n₁)²−(n₂−n₁)²e^2iφ2 (n²₂−n²₁)e^2iφ1(1−e^2iφ2) (n²₁−n²₂)(1−e^2iφ2) (n1+n2)²e^2iφ1e^2iφ2−(n1−n2)²e^iφ1

r = B

D = (n²₂−n²₁)e^2iφ1(1−e^2iφ2) e^2iφ1[(n₁+n₂)e^2iφ2−(n₂−n₁)²] φ1=φ2 =n2kod2 =n2d2

2π λ

Half-wave:n2d2 = λ 2 φ₂ =π

Quarter-wave:n₂d₂ = λ 4 φ2 = ^π₂

Ifφ₂ =π :r = 0. This means for every two half-wave plates, we have zero reflection, hence full transmission.

Ifφ₂= ^π₂ :e^iπ=−1

r = 2(n²₂−n²₁)

−(n₁+n2)²−(n1−n2)² = n²₁−n²₂ n²₁+n²₂

The reflectance of N-double layers is given byM = (M_DL)^N.

Forumla: The reflectance of N double layers for a Quarter-wave plate stack:

RN =|r_N|²= |C_DL|²

|C_DL|²+ 1 N² M_DL =

A_DL B_DL C_DL D_DL

From that we conclude that, for a sufficient number of double layers of a Quarter-wave plate, the reflectance is drastically reduced to zero, thus we have full transmission.

Now we want to calculateC_DL, so

|C_DL|²=|n²₁−n²₂ 2n²₁n²₂ |²

ForN =1, we should get |r|², but forN −→+∞,RN −→1.

Figure 8: Reflectance as a function of the number of layers N

4 Reflection off a retroreflector

Figure 9: Reflection off a retroreflector

(a) From the figure drawn above, it can be shown using the fact that the reflected beam has the same angle of the incident beam.

(b) The losses are associated with the TIR, namely if TIR condition is not fulfilled according to

sinθc= 1 n,

then you have losses. Here we consider the prism’s refractive index ’n’ and the refractive index outside the prism is 1. Therefore, when the angle of incidence θ is smaller than the critical angleθ_c, part of the light will pass through the prism, thereby causing losses.

Exercises selected fromFundamentals of Photonics, chapter 7, by B.E.A Saleh and M.C. Teich andPhotonic Devices, chapter 1, by Jia-ming Liu.