Ординатура / Офтальмология / Английские материалы / Optics Learning by Computing with Examples using MATLAB_Dieter Moller_2007

FIGURE 6.1 Coordinates for the discussion of the modes of a box with reflecting walls and dimensions a, b, c.

components of Eq. (6.2). For the square of the scalar value one gets

The corresponding wave is a product of three standing waves in the three directions x, y, and z. The standing waves have fixed nodes, and the number of nodes is related to the values of ni as (1 + ni ) for i 1, 2, 3. Therefore, the number of nodes may be used for the characterization of the modes. In FileFig 6.1 the first two graphs show the one-dimension standing waves for the x and y directions for the case of nx 2 and ny 2. The third and fourth graphs are the contour, and surface plots of the amplitude for Mi,k with i 2 and k 2 and the fifth and sixth graph, show the intensity. In the third through sixth graphs we see six node lines.

FileFig 6.1 (N1RECBOX)

Graphs of sine functions in one and two dimensions with nodes depending on the integers n1, n2. The modes are numbered by Mik, where i gives the number of nodes in the x direction and k in the y direction. Note that the node lines are the same for the graphs of amplitude and intensity.

N1RECBOX

Modes of the Rectangular Box in Two Dimensions

Standing sine waves in x and y directions. Mode number constants. x direction n1 and a; y direction n2 and b. The wave in each direction is shown as well as contour and surface plots. The square is also shown as surface plot.

1. One dimension

2. Amplitude, 2 D

M11i,j : y1(xi ) · y2(yj ) n1 ≡ 2 a ≡ 40 n2 ≡ 2 b ≡ 40 N ≡ 20.

3. Intensity, 2D

MM11i,j : (y1(xi ) · y2(yj ))2.

Application 6.1.

1.Print out M00, M01, M10, M11, and M22.

2.Make graphs of M00, M01, M10, and M11, using a cosine function for x. The corresponding boundary condition would be an “open end" of the box.

3.Make graphs of M00, M01, M10, and M11, using cosine functions for x and y.

6.2STRATIFIED MEDIA

In the chapter on geometrical optics we discussed image formation with spherical mirrors. We also discussed astronomical telescopes composed of an objective lens and magnifiers. Lenses always reflect some of the incident light, according to Fresnel’s formulas. To reduce the reflection, an antireflection coating was developed. A thin film of a specific material was vacuum-deposited on the lens surface, preventing most of the incident light of a specific wavelength from being reflected. In the same chapter we also discussed laser cavities. The mirrors of a laser cavity need to have high reflectivity only for a limited wavelength range, but the reflectivity should be close to one in order to obtain a high gain.

In this section we show that stacks of dielectric layers may have very high reflectance or transmittance. Applications are antireflection coatings of lenses and mirrors for laser cavities with extremely high reflectivity. We study the

FIGURE 6.2 Three media of index of refraction n1 1, n2, and n3. In each medium we consider a forward and a backward traveling wave.

reflection and transmission of light incident on stacks of dielectric layers. All layers have the thickness d (1/4)(λ/n) n/4λ, but some may have different refractive indices from others.

6.2.1 Two Interfaces at Distance d

When calculating Fresnel’s formulas, we equated the fields on both sides of a boundary and obtained a set of linear equations for the amplitudes of the reflection and the transmission coefficients. This method is called the boundary value method and is now applied to two interfaces at distance d. In the chapter on interference we studied the plane parallel plate and summed up all reflected and transmitted waves. This method is called the summation method and is less rigorous than the boundary value method. We consider three media and assume normal incident light. In each medium we assume forward and backward traveling waves (Figure 6.2). We do not show the time dependence of the waves, giving us these equations for incident, reflected, and transmitted waves at the two interfaces:

1. before the first interface

We take the electrical field vector perpendicular to the plane of incidence, which is the case of perpendicular polarization discussed in Chapter 5. We first

take into account the electrical field vectors at boundary 1, where Y 0. The sum of Eqs. (6.4) and (6.5) is equal to the sum of Eqs. (6.6) and (6.7). At the boundary for Y d the sum of Eqs. (6.6) and (6.7) is equal to the sum of Eqs. (6.8) and (6.9). We obtain the following equations

where k 2πn2/λ, k 2πn3/λ, and d is the distance between the two interfaces. In a similar manner we obtain a second set of two equations for the B-fields. This is similar to the calculations of Fresnel’s formulas in Chapter 5, applying the corresponding boundary conditions for the B-field at Y 0 and

Y d.

In order to derive a vector-matrix formulation, we abbreviate the sum of the amplitudes before the first boundary (Eq. (6.10)), by E1,d 0. Since Eq. (6.11) was derived using the magnetic field vector we call the amplitude B1,d 0. In a similar way we use E2,d 0 and B2,d 0 for the sum of the amplitudes after the first boundary. For the amplitudes before and after the second boundary, we use E2,d d and B2,d d , and E3,d d and B3,d d , respectively. We may now write

We want to express the vector Field3,d d as the product of matrices times the vector Field1,d 0, and have

The manipulation of the matrices is shown in FileFig 6.2.

FileFig 6.2 (N2SYMATR)

Demonstration of the matrix manipulations and multiplication of M2,d M2−,01 and the resulting matrix M2.

Using this matrix approach, we discuss some applications.

6.2.2 Plate of Thickness d (λ/2n2)

We apply Eq. (6.22) to a plate of thickness d (λ/2n2)q, where q is an integer. The thickness of the plate is a multiple of half a wavelength divided by the refractive index n2 of the material of the plate. Using k 2πn2/λ and d (λ/2n2)q we have for the product kd qπ. At the boundary we have for the exponentials e−ik d e−ikd . Both are 1 for even q. Equation (6.22) is now

The matrix M is the unit matrix and the result is that the medium of thickness d has no effect on the transmitted fields. The same result is obtained when q is odd. A similar result has been obtained in Chapter 2 for the plane parallel plate. All incident power of the particular wavelength λ will be transmitted when the plate has the thickness d (λ/2n2)q.

1.Special case of zero thickness for demonstration of reflected and transmitted intensities.

2.Antireflection coating. Graph of reflected intensity depending on the refractive index of the coating.

N4SYMULANTa and N4SYMULANTb are only on the CD.

Application 6.5.

1.Using the correct expression for T (see chapter 5), derive formulas for the transmitted and reflected intensities for one interface with refractive indices n1 and n2 1. Show that T + R 1.

2.Find the refractive index for an antireflection coating material for silicon

(n 3.4).

3.Use polyethylene (n 1.5) as a coating of silicon and calculate the percentage of reflected amplitude and intensity.

6.2.5Multiple Layer Filters with Alternating High and Low Refractive Index

High-reflecting dielectric mirrors are composed of a large number of dielectric layers with alternating high and low refractive indices. We extend Eq. (6.22) to f − 1 layers of equal thickness and obtain

For a rigorous derivation see Born and Wolf (1964; p.66).

We apply Eq. (6.31) to a sequence of double layers, consisting of a high and a low refractive index, assuming that the refractive index outside is 1. Assuming N double layers the sequence of the refractive indices is then

The thickness of one layer is assumed to be one-quarter of a wavelength and we have the product kd [(2πn2/λ)(λ/4n2)], resulting in sin(kd) 1 and cos(kd) 0. For one double layer we get for the product of MH times ML

and for N double layers