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

Application 5.6.

1.Make graphs of rp2, rs2, tp2 and ts2 for the case where n1 < n2, compare to the graphs of the intensities, and find out which quantities, just squared, are

useful for calculation of corresponding intensities and which are not.

2. Make a graph of the factor α (n2 cos θ )/(n1 cos θ) for n1 < n2 and n1 > n2, compare to FileFig 5.5.

5.6.7 Total Reflection and Evanescent Wave

We look at the transmitted amplitude for the case where the angle of incidence is larger than the critical angle. We have from (5.40)

where Eto may be calculated from Fresnel’s formulas. Using the law of refraction as k1 sin θi k2 sin θt we may rewrite Et depending on the angle of incidence:

This wave is called the evanescent wave, traveling in the medium with the lower index of refraction in the −Y direction. It is composed of a traveling wave and an attenuation factor. The attenuation factor is

where k2 (2π/λ)n2.

In FileFig 5.7 we show graphs of the attenuation factor depending on different angles of incidence. One observes the rapid decrease of the magnitude depending on penetration depth −Yt .

FileFig 5.7 (M7FREVA)

Graph of the attenuation factor of the amplitude of the evanescent wave for

(a) n1 1.5, n2 1, and critical angle θc 41.81, and θ1 θc + 2;

(b) nn1 3.4, nn2 1, and critical angle θc 17.105 and θ2 θc + 2 and λ 0.0005mm, depending on coordinate −Y .

M7FREVA

Penetration into the Less Dense Medium at Total Reflection

Exponential factor for decrease of amplitude into the less dense medium with −Y for two different refractive indices n1 and nn1 and n2 nn2. θc is the critical angle. The value a is used to “be off” the critical angle. First we set

To study different angles, make refractive indices the same for both and change a to values larger than 2.

Application 5.7.

1.Study the attenuation factor for a fixed angle of incidence. Make a graph for three indices of refraction.

2.Study the attenuation factor depending on the angle of incidence. Make a graph for three angles of incidence for a large and a small difference of the refractive indices n1 and n2.

5.7POLARIZED LIGHT

5.7.1 Introduction

In contrast to solutions of the scalar wave equation the solutions of Maxwell’s equations are vector waves. In the discussion of Fresnel’s formulas, we considered the two components of the electrical field vector, parallel and perpendicular to the plane of incidence. The two electrical field vectors vibrate in directions perpendicular to each other and each vector presents linear polarized light.Linearly polarized light may be produced by reflection under the Brewster angle at the surface of a dielectric material or when light is reflected on wire gratings with a wavelength larger than the periodicity constant.

The superposition of two linear polarized light vectors will result in linearly polarized light, but only if there is no phase difference between the two vibrations. A phase difference may be produced by using total internal reflection and will result in elliptically or circularly polarized light. The incident light is reflected at a denser medium and the two components, the p-component and the s-component, have a fixed phase angle between them, which is assumed to be zero. After reflection, each of the two reflected components “picks up" a different phase angle and the superposition results in a phase angle between the two components.

There are dielectric materials with different refractive indices in different directions of the material. Plastic films, produced by stretching, may transmit

FIGURE 5.7 Waves with electrical vectors EY and EZ propagating in the X direction.

For two waves, traveling in the X direction and having the same refractive index in the Y and Z directions (Figure 5.7) we have

taking equal amplitudes of the electrical field vectors.

Assuming that the material has a refractive index n1 for the wave vibrating in the Y direction and index n2 for the wave vibrating in the Z direction with corresponding wave vectors k1 and k2, we write

In Figure 5.8 we show an example of two waves with a phase difference of φ.

5.7.4 Half-Wave Plate, Phase Shift of π

We consider the case where φx π and have from Eq. (5.88) for the corresponding distance X π/(k2 − k1). At this distance there is a phase difference of π between the EY and the EZ components, compared to the EY and the EZ components at X 0 (Figure 5.9). We apply this to the case of quartz using ne for the Z component and n0 for the Y component and k1 2πn0/λ, k2 2πne/λ and have

This distance X is very small, but any odd integer of X will have the same effect. Therefore one may use a plate of thickness

where m is an integer. Since ne is larger than n0 we have a positive value for Lh, and therefore quartz has been marked above as a positive crystal.

The case of calcite is reversed. We have ne smaller than n0 so Lh is a negative value. Consequently calcite is called a negative crystal.

In FileFigs.8 we first look at the plane X 0. In the first graph the Y and Z components are plotted as functions of X and in the second graph the Z component is plotted against the Y component. The third and fourth graphs show what happens in the plane X Lh, after a phase shift of π. In the third graph the Y and Z components are plotted as functions of X and the phase change of the Z component is shown. In the fourth graph the Z component is plotted against the Y component. The direction of the resulting vibration of the

two waves is shifted by 90◦ from the second to the fourth graph (and not by π (or 180◦).

FileFig 5.8 (M8POLIN)

Graphs of the superposition of the EY and EZ components before entering the plate, where the phase angle φX 0, and at the plate X L with phase angle

φX π.

M8POLIN is only on the CD.

Application 5.8. Make graphs for φX −π(−180) and compare with φX 0 and φX (180) and with Figure 5.9.

5.7.5 Quarter Wave Plate, Phase Shift π/2

We now consider the case where φX π/2 and have the distance X (π/2)/(k2 − k1). There is a phase difference of π/2 between the EY and EZ components at this distance, compared to the EY and EZ components at X 0 (Figure 5.10). We apply this to the case of quartz using ne for the Z component, and n0 for the Y component, and k1 2πn0/λ, k2 2πne/λ, and have

This distance is very small, but any odd integer of X will have the same effect. Therefore one may use

FIGURE 5.10 Phase relation of the two components of polarized light: (a) before entering; and

(b) emerging from a quarter-wave plate.

where m is an integer. Since ne is larger than n0 we have a positive value for Lq , and therefore quartz has been marked above as a positive crystal, and calcite is called a negative crystal.

For the quarter-wave plate we have, with Eqs. (5.89) and (5.90),

To make a graph of the superposition of the EY and EZ components, we take the real parts of the fields as the values at the Y - and Z- axis of Eqs. (5.95) and (5.96)

We may write Eqs. (5.99) and (5.100), for a certain time interval, as EY cos(−2πx1/360) and EZ cos(−2πx1/360 + π/2). The time interval corresponds to a certain distance in the direction of propagation and to a certain angle interval x1/360. In FileFig 5.9 we show four graphs, corresponding to intervals of angles from 1◦ to 90◦, 1◦ to 160◦, 1◦ to 235◦, and 1◦ to 315◦. Looking onto the paper, in the direction of the source, we see that the resulting vibration describes a circle. The circle develops for positive φX +π/2 in a counterclockwise direction and the light is called left polarized. Considering EY cos(−2πx1/360) and EZ cos(−2πx1/360−π/2), for negative φX −π/2, the circle develops in the clockwise direction and the light is called right polarized.

FileFig 5.9 (M9POELIP)

Graphs of the superposition of the EY and EZ components with positive and negative phase angle φX. Four graphs for four different time spans are shown.

M9POELIP

Circular and Elliptically Polarized Light

Graphs for circular and elliptically polarized light turning “left or right.” Four graphs are shown, extending from 0 to 90, 0 to 160, 0 to 235, and 0 to 315 degrees. The angle ranges (x) correspond to chosen time ranges. Left and right polarized light is described by positive or negative π/2 in one component: Positive

: we have y Ey A cos(−x), yy Ez A cos(−x + ) −A sin(−x); negative : we have y Ey A cos(−x), yy Ez A cos(−x − ) A sin(−x). We write for Ez bA sin(x). When looking in the direction of the incoming light, b −1 is for “left” polarized light (counterclockwise), b 1 for “right” polarized light (clockwise).