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

Application 5.2.

1.Study, for a large angle, the changes of the Brewster angle and the slope of the reflection coefficient. Choose n1 from a value close to n2 to values much smaller than n2. Save the case n1 smaller than n2 for FileFig 5.4.

2.Make graphs for n1 1 and n2 for two different refractive indices. Study the changes of the argument, the Brewster angle, and the slope of the reflec-

tion coefficient at large angles, that is, at grazing angle incidence. Make the difference of n2 and n2 larger and smaller.

3.Look at the transmission coefficients for the p- and s-cases and plot them on one graph. Study the difference of the two for different refractive indices n2.

5.6.5Light Incident on a Less Dense Medium, n1 > n2,

Brewster and Critical Angle

5.6.5.1 Brewster Angle

In FileFig 5.3 we have plotted the reflection and transmission coefficients for light incident on a less dense medium. One sees that the general shape of the reflection and transmission coefficients are the same as found for the case n1 < n2 discussed in Section 5.6.4, and the Brewster angle is present for the parallel case.

5.6.5.2 Critical Angle and Phase Shifts

At a larger angle than the Brewster angle, called the critical angle θc, the curves for the absolute value of rp and rs approach 1, while the transmission coefficients decrease to 0. All arguments of rp and rs , and tp and ts decrease after the critical angle. We apply the law of refraction to this case and find that the refraction angle becomes imaginary when the angle of incidence exceeds the critical angle given by

where n1 is the medium with the larger index of refraction, from which the light is incident on the interface. The question of real and imaginary refraction angles is studied in FileFig 5.4.

FileFig 5.3 (M3FRN2S)

Fresnel’s formulas for the case n1 > n2. Graphs are shown of the absolute value and the argument of rp, rs , and tp, ts , depending on the angle of incidence θ. The choice of n1 1 and n2 1.5 presents the case for light incident on a glasslike material. The Brewster angle appears again for rp and for both rp and

Application 5.3.

1.Study reflection and transmission coefficients by changing from n1 > n2 to n1 ≈ n2 and n1 < n2 and compare with FileFig 5.2. Observe the different

shape of ts and tp as they appear in FileFigs.3 and 4. This is an indication that ts2 and tp2 are not to be confused with the transmitted intensity.

2.Plot on the same graph, for n 1.5 and two different indices of refraction, the reflection coefficients and look at the differences of the curves for larger and smaller differences of the two indices.

3.Change the indices of refraction and observe how the critical angle is changing.

FileFig 5.4 (M4SNELL)

(1) a graph of θ2(θ1) for n1 < n2; (2) a graph of θ2(θ1) for n1 > n2. The graph ends at the critical angle; and (3) real and imaginary parts, separately for n1 > n2. One extends from zero to the critical angle, the other from the critical angle to 90◦.

M4SNELL is only on the CD.

5.6.6 Reflected and Transmitted Intensities

In contrast to what was done in the application of the boundary conditions, we now calculate the energy flow. At the boundary, we equated the sum of the fields of incident and reflected amplitude on one side with the transmitted amplitude on the other side. Now we want to find out how much of the incident energy is reflected and transmitted. From Eq. (5.19) we have for the Poynting vector S for a vacuum

FIGURE 5.4 Areas q, q , and q are cross-sections of the power flow of incident, reflected, and refracted light. Q is the common area at the interface.

For a medium with refractive index n we have with Eq. (5.22) the absolute value of S,

Let us call the incident energy per area Si , the reflected Sr , and the refracted St . In Figure 5.4 we have indicated the areas through which the energy flows. The area q corresponds to the incident energy and q and q to the reflected and refracted energy, respectively. At the interface we have to use the same area Q for incident, reflected, and refracted energy and have to multiply Si , Sr , and St by a cosine factor corresponding to the projections of q, q , q onto the X–Z plane. We then have

(1/2)(n1/µ0c)Eio2 cos θ

5.6.6.1 The Case n1 < n2

We call the reflectance R Ero2 /Eio2 r2, where r stands for both the parallel and perpendicular cases. Using conservation of energy,

one can avoid using the factor α. One can calculate R and R from Fresnel’s formulas, using the amplitude reflection coefficients, and then use 1−R to obtain T . In FileFig 5.5 we have plotted Rp, Rs , Tp 1 − Rp and Ts 1 − Rs for the case where n1 < n2, from 0 to 90◦.

FileFig 5.5 (M5FRINTN2L)

Graphs of Rp, Rs , Tp 1 − Rp, and Ts 1 − Rs depending on the angle of incidence for the case where n1 < n2.

M5FRINTN2L

Intensities

Fresnel’s formulas as function of angle of incidence for n1 < n2 for Rp rp2,

Rs rs2, and Tp 1 − Rp, T s 1 − Rs.

1. Amplitude reflection coefficients

2. Reflection: intensities

Rp(θ) : rp(θ)2 Rs(θ) : rs(θ)2.