Reflection and Refraction

Abstract

In the previous chapter, a plane wave propagating in a homogeneous nonconducting isotropic medium was considered. In this chapter, we will examine what happens to an electromagnetic wave at a plane boundary between two non-conducting media with different electromagnetic properties. The laws of refraction and reflection will be applied to determine the directions of propagation of the reflected and transmitted waves. Their amplitudes will be obtained from Fresnel’s coefficients. We will also evaluate what fraction of the energy in a plane wave incident on a dielectric boundary is reflected, and what fraction is transmitted. As electromagnetic theory of light states that light is an electromagnetic wave, all included in this chapter is applicable in optics, provided that the size of the object is large compared with the wavelength.

Appendices

Solved Problems

Problems A

13.1 :

Calculate the angle between the refracted ray and the normal to the surface between the two media in the cases shown in Fig. 13.5.

Fig. 13.5

Ray of light incident at 45${}^\circ$ on: a air-water interface and b water-air interface

Solution

(a) From Snell’s law, $n_1\,\sin \, \theta _\mathrm{i}=n_2\,\sin \, \theta _\mathrm{t}$, we obtain

$$\sin \, \theta _\mathrm{t}=\frac{n_1}{n_2}\,\sin \, \theta _\mathrm{i}. $$

Fig. 13.6

As refractive index for water is greater than that for air, in a bending is toward the normal while in b bending is away from the normal

Note that the angle of incidence $\theta _\mathrm{i}$ and that of refraction $\theta _\mathrm{t}$ are measured with respect to the surface normal. The incident ray, normal, and refracted ray, all lie in the same plane (plane of incidence). $n_1$ represents the refractive index of the medium in which the incident wave travels, while $n_2$ is the refractive index of the medium in which the refracted wave moves. In the case under study, we have: $n_1=1.00$, $\theta _\mathrm{i}=45^\circ$, and $n_2=1.33$. Thus, it follows that

$$ \sin \, \theta _\mathrm{t}=\frac{n_1}{n_2}\,\sin \, \theta _\mathrm{i}=\frac{1.00}{1.33}\, \sin 45^\circ =0.532 \,\Rightarrow \, \theta _\mathrm{t}=\sin ^{-1}0.532=32.1^\circ . $$

The angle of refraction is $32.1^\circ$, less than $45^\circ$, as shown in Fig. 13.6a. In this case that light passes from air to water, i.e. from a medium of lower index to a higher index, the light ray is bent toward the normal.

(b) In the second case, light passes from water, $n_1=1.33$, to air, $n_2=1.00$. Snell’s law yields

$$ \sin \, \theta _\mathrm{t}=\frac{n_1}{n_2}\,\sin \, \theta _\mathrm{i}=\frac{1.33}{1.00}\, \sin 45^\circ =0.940 \,\Rightarrow \, \theta _\mathrm{t}=\sin ^{-1}0.940=70.1^\circ . $$

As light passes from a higher index to lower index, the ray is bent away from the normal, as shown in Fig. 13.6b.

13.2 :

Figure 13.7 shows a medium (glass) with refractive index $n=3$, limited by two spherical surfaces whose radii are R and 2R, respectively. At the center of the sphere of radius R, there is a point light source. Determine the values of $\alpha$ that delimit the external surface through which light does not emerge.

Fig. 13.7

Sphere of radius 2R and refractive index $n=3$. A spherical hole of radius R has been cut in the large sphere. A point light source is located at the center O

Solution

For any ray emerging from the point source, the angle of incidence at the air-to-glass boundary is zero since the rays propagate in the radial direction. Therefore, the refracted ray does not change its direction at the first interface (air-glass). Figure 13.7 shows a ray emerging from the light source that makes an angle $\alpha$ with the horizontal, which does not suffer a change in direction at point A on the air-glass interface. At the interface between the material and air, the angle between the incident ray and the normal to the interface, the radial direction CB, is denoted by $\theta _i$. Applying Snell’s law to the glass-air interface, the critical angle obtained is

$$ n \,\mathrm{sin}\,\theta _\mathrm{c}=1\times \mathrm{sin}\,\frac{\pi }{2} \Rightarrow \mathrm{sin}\,\theta _\mathrm{c}=\frac{1}{n}, $$

where it is assumed that the refractive index of air is approximately 1. Therefore, the incident ray on the second interface will be totally reflected, if the following condition is satisfied

$$ \mathrm{sin}\,\theta _\mathrm{i}\ge \frac{1}{n}. $$

From the figure and the sine theorem, it follows that

$$ \frac{\mathrm{sin}\,\theta _\mathrm{i}}{R}=\frac{\mathrm{sin}\,\alpha }{2R}\Rightarrow \mathrm{sin}\,\theta _\mathrm{i}=\frac{\mathrm{sin}\,\alpha }{2}.$$

Thus, the values of $\alpha$ for which rays do not emerge from the spherical surface are given by

$$ \frac{\mathrm{sin}\,\alpha }{2}\ge \frac{1}{n} \Rightarrow \mathrm{sin}\,\alpha \ge \frac{2}{n}=\frac{2}{3} \Rightarrow 41.8^\circ \le \alpha \le 138.2^\circ . $$

Due to the symmetry of the problem, this condition is also satisfied in the lower hemisphere.

13.3 :

A light ray is incident from air on a plane-parallel glass plate at an angle $\theta _\mathrm{i}$, as shown in Fig. 13.8. The plate has a refractive index n and a thickness t. Derive the expression for the lateral displacement $\ell$ of the emerging ray.

Fig. 13.8

Ray impinging obliquely on a glass plate of refractive index n

Solution

Figure 13.9 shows the path followed by a ray through the plate. Snell’s law at the point of incidence A, on the air-glass interface, gives

$$\begin{aligned} 1\times \mathrm{sin}\,\theta _\mathrm{i}=n\, \mathrm{sin}\,\theta _\mathrm{t}\Rightarrow \mathrm{sin}\,\theta _\mathrm{t}=\frac{ \mathrm{sin}\,\theta _\mathrm{i} }{n}\,\,\,\,\,\Rightarrow \,\,\,\,\,\cos \theta _\mathrm{t}=\sqrt{1-\frac{\mathrm{sin^2}\theta _\mathrm{i}}{n^2}}.\end{aligned}$$

(13.14)

In the same way, at the interface between air and glass, we obtain

$$ n\, \mathrm{sin}\,\theta '_\mathrm{i}=1\times \, \mathrm{sin}\,\theta '_\mathrm{t}.$$

From the geometry of the ray path shown in Fig. 13.9, we have $\theta '_\mathrm{i}=\theta _\mathrm{t}$. Then, the above two equations yield the following relationship

$$ n\sin \, \theta '_\mathrm{i}=n \sin \,\theta _\mathrm{t}=\sin \, \theta _\mathrm{i}=\sin \, \theta '_\mathrm{t}, $$

which implies that $\theta '_\mathrm{t}= \theta _\mathrm{i}$, i.e. the direction of the ray emerging from the plate is the same as that of the incident ray impinging on the plate. However, there is a lateral displacement $\ell$ between the two rays.

Fig. 13.9

Lateral displacement of a ray after passing through a glass plate

From the geometry shown in Fig. 13.9, the lateral displacement $\ell$ can be calculated as follows

$$\left. \begin{array}{l} CD=AC \, {\sin }\,(\theta _\mathrm{i}-\theta _\mathrm{t})\\ AC=\frac{AB}{\cos \theta _\mathrm{t}}=\frac{t}{\cos \theta _\mathrm{t}}\end{array}\right\} \Rightarrow \ell =CD=\frac{t}{\cos \theta _\mathrm{t}}\,{\sin }\,(\theta _\mathrm{i}-\theta _\mathrm{t})\Rightarrow $$

$$ \ell =\frac{t}{\cos \, \theta _\mathrm{t}}\,\left( {\sin }\,\theta _\mathrm{i}\cos \,\theta _\mathrm{t}-\cos \,\theta _\mathrm{i}{\sin }\,\theta _\mathrm{t} \right) =t\,\left( {\sin }\,\theta _\mathrm{i}-\cos \,\theta _\mathrm{i}{\tan }\,\theta _\mathrm{t} \right) .$$

The value of ${\tan }\,\theta _\mathrm{t}$ can be obtained from (13.14), $\tan \,\theta _\mathrm{t}={\sin }\,\theta _\mathrm{i}/\sqrt{n^2-{\sin ^2}\,\theta _\mathrm{i}}$. Substituting ${\tan }\,\theta _\mathrm{t}$ into the equation obtained for $\ell$, it follows that

$$\ell =t \,{\sin }\,\theta _\mathrm{i}\left[ 1-\frac{\cos \theta _\mathrm{i}}{\sqrt{n^2-{\sin ^2}\theta _\mathrm{i}}} \right] . $$

This equation gives the lateral displacement $\ell$ in terms of the angle of incidence $\theta _\mathrm{i}$, the refractive index n, and the thickness of the plate t.

13.4 :

Consider a simple lens formed by a glass half sphere, with radius R and refractive index $n_1$, surrounded by air, whose refractive index is denoted by $n_2$. A ray of light, parallel to the axis of rotational symmetry, and at a distance H from the axis, is incident on the flat surface, as shown in Fig. 13.10. If $H\ll R$, find the location of point F where the emerging ray from the glass intersects the axis.

Solution

Figure 13.10 shows a ray of light propagating in a direction parallel to the axis, which impinges from air on the flat surface of a lens at normal incidence. Therefore, the direction of the ray remains unchanged according to Snell’s law. Then, the ray passes through the glass and hits the glass-air interface making an angle $\theta _\mathrm{i}$ with the normal (coincident with the radial line). Finally, the ray emerges from the lens at angle $\theta _\mathrm{t}$ (the angle of refraction). From the geometry (see Fig. 13.10) and the condition that $H\ll R$, it is found that

$$ \sin \, \theta _\mathrm{i}= \frac{H}{R} \approx \tan \, \theta _\mathrm{i}\approx \theta _\mathrm{i},$$

where the small-angle approximation (angle $\theta$ approaches 0) has been used, i.e. $\sin \, \theta \approx \tan \, \theta \approx \theta$ and $\cos \, \theta \approx 1$.

By applying Snell’s law to the glass-air interface, it follows that the angle of refraction is equal to

$$ \sin \, \theta _\mathrm{t}=\frac{n_1}{n_2}\,\sin \, \theta _\mathrm{i}\Rightarrow \theta _\mathrm{t}\approx \frac{n_1}{n_2}\,\frac{H}{R}.$$

Then, for small angles, the following relation can be obtained

$$ \tan \,(\theta _\mathrm{t}-\theta _\mathrm{i})\approx \theta _\mathrm{t}-\theta _\mathrm{i} \approx \frac{n_1}{n_2}\,\frac{H}{R}-\frac{H}{R}=\frac{n_1-n_2}{n_2}\,\frac{H}{R}.$$

Fig. 13.10

Path followed by a single ray through the lens. Rays travelling parallel to the axis and close to it converge to point F, the focal point of the lens

The distance from F, the intersection point of the ray emerging from the glass with the axis, to the flat surface is denoted by L in Fig. 13.10. According to the geometry, point F is located a distance

$$ L=R\, \cos \, \theta _\mathrm{i}+\frac{H}{\tan (\theta _\mathrm{t}-\theta _\mathrm{i})}\approx R+H \, \frac{n_2}{n_1-n_2}\, \frac{R}{H} \approx \frac{n_1}{n_1-n_2}\, R\,,$$

from the front of the lens. Therefore, all parallel rays close to the axis converge to point F after passing through the lens. This point is called the focal point of the lens.

13.5 :

A point light source at an unknown distance H under water yields an illuminated circular area with diameter 12 m, seen from the air side of the interface. Find H. Assume that the refractive index of water is 1.33.

Solution

Figure 13.11 shows a light source at a distance H under water. Rays emerging from O in all directions are incident on the water-air interface. On the water surface, directly above the source, the area illuminated is a circle of diameter D. The maximum radius of this circle is determined by the critical angle for refraction. Refraction can only occur if the ray incident on the interface makes an angle with the normal line smaller than the critical angle.

For ray $OO'$, at normal incidence, there is no change in the direction that the wave is travelling, whereas for ray OP, the incident angle being $\theta$, the transmitted ray into air changes its direction relative to the normal to the surface. As $n_1=1.33>n_2=1$, it follows that $\theta _\mathrm{t }>\theta$. It should be noted in Fig. 13.11 that the greater the angle that the ray emitted by the source makes with $OO'$, which is equal to the angle of incidence at the water-air interface, the greater the angle of refraction. At angles of incidence greater or equal to the critical angle, light rays will experience total reflection and, hence, no light is emerging from the water to the air. The critical angle at the water-air interface is given by

$$ n_1\,\sin \,\theta _\mathrm{c}=n_2\,\sin \,90^\circ \Rightarrow \theta _\mathrm{c}=\sin ^{-1}\left( \frac{1}{n_1} \right) =\sin ^{-1}\left( \frac{1}{1.33} \right) =48.75^\circ .$$

Note in Fig. 13.11 that ray OC hits the interface at the critical angle. Then, the point light source is submerged below surface at a distance,

$$ H=\frac{O'C}{\tan \,\theta _\mathrm{c}}=\frac{6}{\tan \,48.75^\circ }\,\mathrm{m }=5.26 \,\, \mathrm{m}, $$

where $O'C$ is the radius of the illuminated circle.

13.6 :

A prism is made of a glass whose refractive index varies with wavelength: $n_\mathrm{p}=1.60-0.10 \, \lambda ,$ where $\lambda$ represents the wavelength in vacuum in $\upmu$m. A beam of white light is incident on a face of the prism at an angle of 45${}^\circ$, as shown in Fig. 13.12. If the wavelength of the red light is 0.750 $\upmu$m and that of violet light 0.390 $\upmu$m, find the angular dispersion.

Fig. 13.11

A point light source under water emitting rays that impinge on the water-air interface

Solution

A ray of white light incident obliquely on a prism is twice refracted as it passes through it. As the refractive index is dependent on the wavelength, the angle of refraction varies with the wavelength and light is hence dispersed into all the colors of the visible spectrum. The visible spectrum ranges in wavelength from approximately 0.390 $\upmu$m, for violet light, to 0.750 $\upmu$m, for red light. Longer wavelengths have smaller refractive indexes and are refracted less than shorter wavelengths.

We have for violet light, the lower end of the visible spectrum, a refractive index, $n_\mathrm{v}$, and an angle of refraction, $\theta _\mathrm{tv}$:

$$ n_\mathrm{v}=1.60-0.10 \times 0.390=1.561, \quad \mathrm{sin}\, \theta _\mathrm{i}=n_\mathrm{v}\,\mathrm{sin}\,\theta _\mathrm{tv}\Rightarrow \mathrm{sin}\,\theta _\mathrm{tv}=\frac{\mathrm{sin}\,45^\circ }{1.561}\Rightarrow \theta _\mathrm{tv}=26.94^\circ . $$

From the triangle ABC (see Fig. 13.12) the angle of incidence at B, $\theta '_\mathrm{iv}$, can be obtained,

$$ 180^\circ =(90^\circ -\theta _\mathrm{tv})+60^\circ +(90^\circ -\theta '_\mathrm{iv}) \Rightarrow \theta '_\mathrm{iv}=33.06^\circ .$$

Therefore, for violet light, the angle of refraction for the refracted ray emerging from the second face of the prism, $\theta '_\mathrm{tv}$, will be

$$ n_\mathrm{v}\,\mathrm{sin}\,\theta '_\mathrm{iv}=\sin \,\theta '_\mathrm{tv}\Rightarrow 1.561 \,\mathrm{sin}\,33.06^\circ =\mathrm{sin}\,\theta '_\mathrm{t_\mathrm{v}} \Rightarrow \theta '_\mathrm{tv}=58.38^\circ . $$

Fig. 13.12

Dispersion of light by a prism. $\theta _\mathrm{i}=45^\circ$

In the same way, for red light, the longer wavelength of the visible spectrum, the refractive index being $n_\mathrm{r}$, we have

$$ n_\mathrm{r}=1.60-0.10 \times 0.750=1.525, \quad \mathrm{sin}\, \theta _\mathrm{tr}=\frac{\mathrm{sin}\, 45^\circ }{1.525}\Rightarrow \theta _\mathrm{tr}=27.62^\circ . $$

The angle of incidence at the second face will be

$$ \theta '_\mathrm{ir}=60^\circ -27.62^\circ =32.38^\circ , $$

and the angle of refraction at the second face

$$ 1.525\,\mathrm{sin}\,32.38^\circ =\mathrm{sin}\,\theta '_\mathrm{tr} \Rightarrow \theta '_\mathrm{tr}=54.75^\circ .$$

Therefore, the angular dispersion is given by

$$ \theta '_\mathrm{tv}-\theta '_\mathrm{tr} =58.38^\circ -54.75^\circ =3.63^\circ . $$

13.7 :

(a) Determine the Fresnel coefficients for normal incidence. (b) A monochromatic plane wave with an amplitude of 10 V/m is incident normally on the plane surface of an air-glass interface. Find the amplitude and phase for the reflected and transmitted waves in the following two cases: (1) When the wave is incident from the air side; (2) when the wave is incident from the glass side. The index of refraction for glass is 1.5. Take the index of air to be one.

Solution

(a) At normal incidence $\theta _\mathrm{i}=0$, Snell’s law (13.2) gives for the angle of refraction: $\sin \theta _\mathrm{t}=0$, $\theta _\mathrm{t}=0$, since $\theta _\mathrm{t}\le \pi /2$. Therefore, the direction of propagation of the incident wave remains unchanged in the process of reflection and refraction.

For $\theta _\mathrm{i}=0$, the meaning of plane of incidence is lost, since the two vectors defining this plane are parallel, and therefore, the distinction between the parallel and perpendicular components is of no interest. The Fresnel coefficients (13.4) and (13.5) for $\theta _\mathrm{i}=\theta _\mathrm{t}=0$ reduce to

$$\begin{aligned} r_{\Vert }\equiv & {} \frac{E_\mathrm{r \Vert }}{E_\mathrm{i \Vert }}=\frac{n_2-n_1}{n_1+n_2}\quad ;\quad r_{\bot }\equiv \frac{E_\mathrm{r \bot }}{E_\mathrm{i \bot }}=\frac{n_1-n_2}{n_1+n_2},\\t_{\Vert }\equiv & {} \frac{E_\mathrm{t \Vert }}{E_\mathrm{i \Vert }}=\frac{2n_1}{n_1+n_2} \quad ;\quad t_{\bot }\equiv \frac{E_\mathrm{t \bot }}{E_\mathrm{i \bot }}=\frac{2n_1}{n_1+n_2}. \end{aligned}$$

The $r_{\Vert }$ and $r_{\bot }$ coefficients lead to the same direction of oscillation for the electric field of the reflected wave and, therefore, at normal incidence, the reflection coefficient of a plane wave is independent on the wave’s polarization. We can conclude that for normal incidence the ratio of amplitudes of the reflected and incident waves and the ratio of the amplitude of the transmitted wave to that of the incident wave can be expressed, respectively, as,

$$\begin{aligned} r=\frac{E_\mathrm{r}}{E_\mathrm{i}}=\frac{n_1-n_2}{n_1+n_2}\quad \mathrm{and} \quad t=\frac{E_\mathrm{t}}{E_\mathrm{i}}=\frac{2n_1}{n_1+n_2}, \end{aligned}$$

(13.15)

where a positive r means that reflected wave oscillates in phase with the incident wave. On the other hand, a negative coefficient means that the reflected wave oscillates $\pi$ radians out of phase with the incident wave.

Fig. 13.13

The amplitude and phase relations for harmonic plane waves incident normally on an interface: a air-glass and b glass-air. The wave is assumed to be linearly polarized

It follows from (13.15) that if $n_1<n_2$, the electric vector in the reflected wave is in the opposite direction to that of the incident electric field. As coefficient t always has a positive sign, the transmitted wave oscillates in phase with the incident wave. In this case, the amplitude of the transmitted wave is smaller than that of the incident wave. On the other hand, for $n_1>n_2$, the reflected wave oscillates in phase with the incident wave, and the amplitude of the transmitted wave is larger than that of the incident wave.

(b) For the air-glass interface ($n_1=1$ and $n_2=1.5$), (13.15) gives: $r=-0.2$ and $t=0.8$. Then, for an incident wave with amplitude 10 V/m, the amplitudes of the reflected and transmitted waves will be:

$$ E_\mathrm{0r}=r \times E_\mathrm{0i}=-0.2 \times 10=-2\,\,\,\mathrm{V /m}\quad ;\quad E_\mathrm{0t}=t\times E_\mathrm{0i}=0.8 \times 10=8 \,\,\,\mathrm{V/m}. $$

Figure 13.13a shows the amplitude and phase relations for this case; the phase of the reflected wave is shifted by $\pi$ with respect to the incident wave.

For the glass-air interface ($n_1=1.5$ and $n_2=1$), the results are shown in Fig. 13.13b, which corresponds to $r=0.2$ and $t=1.2$, the resulting amplitudes being::

$$ E_\mathrm{0r}=r \times E_\mathrm{0i}=0.2 \times 10=2\,\,\,\mathrm{V/m}\quad ;\quad E_\mathrm{0t}=t\times E_\mathrm{0i}=1.2 \times 10=12 \,\,\,\mathrm{V/m}. $$

The reflected wave does not change phase, as expected. Figure 13.13 shows the continuity of the tangential component of the electric field across the interface.

13.8 :

(a) Use the equations for reflectance and transmittance for normal incidence to prove that energy is conserved. (b) For the cases considered in the previous problem for an air-glass interface, determine what fraction of the incident energy is reflected and what fraction is transmitted.

Solution

(a) At $\theta _\mathrm{i}=0$, (13.11) and (13.15) give for the reflectance:

$$\begin{aligned} R=R_{\Vert }=R_{\bot }=r^2=\left( \frac{n_1-n_2}{n_1+n_2} \right) ^2.\end{aligned}$$

(13.16)

Analogously, for the transmittance, (13.12) and (13.15) lead to:

$$\begin{aligned} T=T_{\Vert }=T_{\bot }=\frac{n_2}{n_1}\,t^2=\frac{4n_1n_2}{(n_1+n_2)^2}.\end{aligned}$$

(13.17)

By adding R and T,

$$ R+T=\left( \frac{n_1-n_2}{n_1+n_2} \right) ^2+\frac{4n_1n_2}{(n_1+n_2)^2}=\frac{n_1^2+n_2^2-2n_1n_2+4n_1n_2}{(n_1+n_2)^2} =\frac{n_1^2+n_2^2+2n_1n_2}{(n_1+n_2)^2}=1. $$

As R and T represent, respectively, the reflected and transmitted energy divided by the incident energy (across unit area per unit time), this result means that the reflected energy plus the transmitted energy is equal to the energy incident on the interface, and, therefore, energy (per unit surface area per unit time) is conserved.

(b) For the air-to-glass interface, $n_1=1$ and $n_2=1.5$, the reflection and transmission coefficients, calculated from (13.16) and (13.17), are:

$$ R=0.04 \quad \mathrm{and }\quad T=0.96, $$

which means that 4$\%$ of the incident energy (per unit area and unit time) is reflected and 96$\%$ is transmitted.

For the glass-to-air interface, $n_1=1.5$ and $n_2=1$, we find that $R=0.04$ and $T=0.96$. Then, through a flat-glass window panel, due to the multiple reflection and refraction in the two interfaces, the total power reflected is approximately 8$\%$.

13.9 :

Plot R and T for normal incidence in terms of the quotient $n_1/n_2$.

Solution

Let the quotient $n_1/n_2$ be denoted by x. Then, (13.16) and (13.17) can be written as

$$ R=\left( \frac{x-1}{x+1} \right) ^2\quad ;\quad T=\frac{4x}{(x+1)^2}\,. $$

Fig. 13.14

Reflectance (dashed line) and transmittance (continuous line) in terms of $x=n_1/n_2$ for normal incidence

Figure 13.14 shows reflectance R and transmittance T in terms of $x=n_1/n_2$. For $x\approx 1$, i.e. $n_1 \approx n_2$, all the energy is transmitted. The smaller the difference of the indices of the two media, the less energy is carried by the reflected wave. When R equals T, we have

$$ R=T \Rightarrow \left( \frac{x-1}{x+1} \right) ^2=\frac{4x}{(x+1)^2}\Rightarrow x^2-6x+1=0 \Rightarrow x=0.17\,\,\mathrm{and}\,\,5.83. $$

For $x=0.17$ and 5.83, it follows that $R\approx T \approx 0.5$. Figure 13.14 shows that for $x>1$, R and T increases and decreases, respectively, with x. Analogously, for $x<1$, R increases and T decreases with the same proportion ($T=1-R$) as x becomes smaller. Therefore, the greater the difference between the properties of the two media, the greater the energy reflected and the smaller the energy transmitted. Note that R yields the same result for a given value of x and its reciprocal, 1 / x. The same holds for T.

13.10 :

For an air-to-water interface, plot the reflectance, $R_{\Vert }$ and $R_{\bot }$, as a function of incident angle $\theta _\mathrm{i}$. Also plot the water-to-air reflectance. Assume that water has an index of refraction of 1.33. Take the index of air to be one.

Solution

It follows from (13.11) and (13.4) that reflectance for the parallel and perpendicular polarization is given, respectively, by

$$ R_{\Vert }=\left( \frac{n_2\cos \,\theta _\mathrm{i}-n_1\cos \,\theta _\mathrm{t}}{n_1\cos \,\theta _\mathrm{t}+n_2\cos \,\theta _\mathrm{i} }\right) ^2\quad ;\quad R_{\bot }=\left( \frac{n_1\cos \,\theta _\mathrm{i}-n_2\cos \,\theta _\mathrm{t}}{n_1\cos \,\theta _\mathrm{i}+n_2\cos \,\theta _\mathrm{t} }\right) ^2. $$

We have to represent $R_{\Vert }$ and $R_{\bot }$ in terms of $\theta _\mathrm{i}$ for the air-water interface first and then for the water-air interface.

(a) The plot of the reflectance for the case of $n_1=1$ and $n_2=1.33$ is show in Fig. 13.15. We find that for normal incidence the fraction of power reflected is 2$\%$ ($R_{\Vert }$ = $R_{\bot }=0.02$). For the perpendicular polarization, the fraction of energy reflected increases with the angle of incidence and when $\theta _\mathrm{i}$ approaches 90$^\circ$, 100$\%$ of the power is reflected. For the component in the plane of incidence, the reflectance decreases to zero at Brewster’s angle, $\theta _\mathrm{B}=\tan ^{-1}=n_2/n_1=1.33=53.06^\circ$, and then exhibits the same behaviour as the reflectance for the perpendicular polarization.

Fig. 13.15

Reflectance $R_{\Vert }$ (continuous line) and $R_{\bot }$ (dashed line) plotted versus $\theta _\mathrm{i}$ for the air-water interface, $n_1=1$ and $n_2=1.33$. Brewster’s angle is $\theta _\mathrm{B}=53.06^\circ$

(b) For the interface water-air, $n_1>n_2$, we can find a critical angle $\theta _\mathrm{c}$ at which there is total reflection and then $R_{\Vert }=1$ and $R_{\bot }=1$. From (13.3), the critical angle $\theta _\mathrm{c}=\sin ^{-1}(1/1.33)=48.75^\circ$. Beyond the critical angle, the wave is said to undergo total reflection. Figure 13.16 shows the plot of the reflectance versus $\theta _\mathrm{i}$ for both components. For angles of incidence greater than the critical angle, the reflectance (modulus) is 1, hence the incident energy is reflected and no energy is transmitted to the second medium. At normal incidence, we find the same values for the reflectance as those for the air-water interface. The reflectance for the perpendicular polarization increases with $\theta _\mathrm{i}$ until the critical angle is reached. For the component in the plane of incidence, the reflectance becomes zero at Brewster angle $\theta _\mathrm{B}=36.94^\circ$, as expected. For $\theta _\mathrm{i}>\theta _\mathrm{B}$, $R_{\Vert }$ increases sharply with $\theta _\mathrm{i}$ until the incident angle equals the critical angle.

Fig. 13.16

Reflectance $R_{\Vert }$ (continuous line) and $R_{\bot }$ (dashed line) plotted versus $\theta _\mathrm{i}$ for a water-air interface $n_1=1.33$ and $n_2=1$. Brewster’s angle is $\theta _\mathrm{B}=36.94^\circ$ and critical angle $\theta _\mathrm{c}=48.75^\circ$

Problems B

13.11 :

Figure 13.17 shows an optical fiber surrounded by a material of lower refractive index, known as cladding. Find the maximum angle of incidence $\theta _\mathrm{i}$ for rays incident on the core’s end face to be trapped inside the core. Consider that air has a refraction index of 1.

Solution

Figure 13.17 shows a ray meeting the air-core boundary at an angle $\theta _\mathrm{i}$, measured relative to a line normal to the boundary. The angle of refraction at $P_1$ is $\theta _\mathrm{t}$. At $P_2$ on the interface between the core and the cladding, Snell’s law gives

$$ n_1\, \sin \,\theta _\mathrm{i}'=n_2 \, \sin \,\theta _\mathrm{t}', $$

where $n_1$ and $n_2$ are the indices of refraction of the core and cladding ($n_1>n_2$), respectively. Note that $\theta _\mathrm{t}$ and $\theta '_\mathrm{i}$ are complementary angles, i.e. $\theta _\mathrm{t}=90^\circ -\theta _\mathrm{i}'$. Then, we can infer the following relationship, between the angles of incidence at $P_1$ and $P_2$, by applying Snell’s law at $P_1$ on the interface between the air and the core:

$$\begin{aligned} n_\mathrm{a}\,\sin \,\theta _\mathrm{i}=n_1\,\sin \,\theta _\mathrm{t} =n_1\,\sin (90^\circ -\theta _\mathrm{i}')=n_1\, \cos \,\theta _\mathrm{i}'\Rightarrow \cos \,\theta _\mathrm{i}'=\frac{n_\mathrm{a}\,\sin \, \theta _\mathrm{i}}{n_1}, \end{aligned}$$

(13.18)

where $n_\mathrm{a}$ is the refractive index of air ($n_\mathrm{a}\approx 1$). The critical angle at $P_2$ is determined by

$$ \sin \theta _\mathrm{c}'=\frac{n_2}{n_1}. $$

For internal reflection to take place, the angle of incidence at $P_2$ must satisfy

Fig. 13.17

Sketch of an optical fiber showing the core, the cladding, and the path followed by a ray incident at the core’s end face with an angle $\theta _\mathrm{i}$

$$\begin{aligned} \theta _\mathrm{i}' \ge \theta _\mathrm{c}' \Rightarrow \cos \, \theta _\mathrm{i}' \le \cos \, \theta _\mathrm{c}'=\sqrt{1-\sin ^2\,\theta _\mathrm{c}'}=\sqrt{1-\left( \frac{n_2}{n_1}\right) ^2 }. \end{aligned}$$

(13.19)

Rays that meet the core-cladding boundary at an angle greater than the critical angle are completely reflected. For this condition to be satisfied, it follows from (13.18) and (13.19) that the following equation must hold

$$ \cos \, \theta _\mathrm{i}'=\frac{n_\mathrm{a}\,\sin \, \theta _\mathrm{i}}{n_1}\le \sqrt{1-\left( \frac{n_2}{n_1}\right) ^2 } \Rightarrow \sin \theta _{ \mathrm i}\le \frac{1}{n_\mathrm{a}} \,\sqrt{n_1^2-{n_2}^2}. $$

Then, the maximum angle of incidence is given by

$$\begin{aligned} \sin \theta _{ \mathrm i,max}= \frac{1}{n_\mathrm{a}} \,\sqrt{n_1^2-{n_2}^2}. \end{aligned}$$

(13.20)

Using this equation, we can calculate the maximum angle of incidence (acceptance angle) at which the rays incident on the core’s end face are trapped inside the core by total internal reflection.

13.12 :

The refractive index of mammalian tissues can be measured by using a fiber optic cladding method [105] based on substituting the usual cladding by the tissue under study and utilizing the principle of total internal reflection. If a He-Ne laser, with wavelength 632 nm, is used as a light source, the core made from fused quartz with refractive index $n_\mathrm{q}=1.457$ at 632 nm, and the half-angle of the emergent cone of light from the output of the optical fiber is $23.8^\circ$, find the refractive index of the tissue.

Solution

In reference [105], a method for measuring the refractive index of mammalian tissues is described, which is based on the principle of internal reflection at the core-cladding interface. If the refractive indices of air and quartz are known, and the angle of the emergent cone of light from the output of the fiber is measured, the refractive index of the tissue can be calculated from (13.20).

Figure 13.18 shows a sketch of a typical optical fiber. The core is made from fused quartz with refractive index $n_\mathrm{q}=1.457$, at the wavelength of He-Ne laser light, and the cladding is a tissue with a refractive index $n_\mathrm{t}<n_\mathrm{q}$. The incident beam comes from air and enters the fiber at the acceptance angle, $\theta _\mathrm{a}$, which is the maximum angle of a ray hitting the fiber core that is kept within the core. Then, total reflection takes place at the quartz-tissue interface. It should be noted that the half-angle of the cone at the exit of the fiber is equal to $\theta _\mathrm{a}$. It follows from (13.20) that the refractive index of the tissue, $n_\mathrm{t}$, can be expressed in terms of the refractive index of the core, $n_\mathrm{q}$, the refractive index of air, $n_\mathrm{a}$, and the aperture angle, $\theta _\mathrm{a}$,

$$ \mathrm{sin}\,\theta _\mathrm{a}=\frac{1}{n_\mathrm{a}}\, \sqrt{n_\mathrm{q}^2-n_\mathrm{t}^2}\Rightarrow n_\mathrm{t}=\sqrt{n_\mathrm{q}^2-(n_\mathrm{a}\,\mathrm{sin}\,\theta _\mathrm{a})^2}.$$

By substituting the numerical values into the above equation, the refractive index of the tissue is

$$ n_\mathrm{t}=\sqrt{1.457^2-(1\times \mathrm{sin}\,23.8^\circ )^2}=1.40.$$

13.13 :

(a) Determine the phase velocity for a harmonic plane wave of frequency f propagating in a homogeneous ionized gas with N electrons per unit volume. (b) Calculate the lowest frequency of the wave that can propagate through the ionized gas. (c) For a wave perpendicularly incident on the interface between vacuum and a layer of an ionized gas with $N=10^{10}$ m$^{-3}$, determine the frequencies of the waves that can penetrate into the layer. What are such frequencies for $N=10^{12}$ m$^{-3}$? (d) For oblique incidence on an interface between vacuum and a layer of an ionized gas with $N=10^{12}$ m$^{-3}$, calculate the lowest frequency of the wave that can penetrate into the layer if the angle of incidence is $\theta _\mathrm{i}=30^\circ$. Mass of electron $m_\mathrm{e}=9.1 \times 10^{-31}$ kg and electronic charge $-e=-1.6 \times 10^{-19}$ C.

Fig. 13.18

Sketch showing an optical fiber. An incident light ray is first refracted and then undergoes total internal reflection at the core-cladding interface. Cladding is substituted by mammalian tissue for which refractive index is to be measured. Light acceptance cone is shown

Solution

(a) Electrons are much lighter than positive ions and, therefore, they are accelerated more by the electric field of electromagnetic waves passing through the ionized gas. At a given point in the medium, the electric field associated with a plane electromagnetic wave of angular frequency $\omega =2 \pi f$ can be expressed as $\mathbf E=\mathbf E_{_0}\, \mathrm{sin}\,(\omega t)$. If there are N electrons per unit volume, with mass $m_\mathrm{e}$ and charge e, the equation of motion of each electron and the velocity $\mathbf v$ due to the electric field of the wave are given by

$$ m_\mathrm{e}\, \frac{d \mathbf v}{dt}=-e\, \mathbf E_{_{0}}\, \mathrm{sin}\,(\omega t) \Rightarrow \mathbf v=\frac{e}{m_\mathrm{e }\omega }\,\mathbf E_{_{0}}\, \cos (\omega t).$$

Note the integration constant can be disregarded by choosing an appropriate origin in time. The current density $\mathbf j$ is

$$ \mathbf j=-Ne \, \mathbf v=-\frac{Ne^2}{m_\mathrm{e}\omega }\,\mathbf E_{_0}\, \cos (\omega t).$$

Maxwell’s equation for $\mathbf B$ becomes

$$\begin{aligned} \nabla \times \mathbf B=\mu _{_0}\mathbf j+\mu _{_0}\varepsilon _{_0}\,\frac{\partial \mathbf E}{\partial t}= & {} -\frac{\mu _0 Ne^2 }{m_\mathrm{e}\omega }\,\mathbf E_{_0}\, \cos (\omega t)+\mu _{_0} \varepsilon _{_0} \omega \mathbf E_{_0}\, \cos (\omega t)\\= & {} \mu _{_0} \varepsilon _{_0} \left[ 1-\frac{Ne^2}{m_\mathrm{e }\varepsilon _{_{0}}\omega ^2}\right] \omega \mathbf E_{_0}\, \cos (\omega t)\\ {}= & {} \mu _{_0} \varepsilon _{_0} \left[ 1-\frac{Ne^2}{m_\mathrm{e}\varepsilon _{_{0}}4 \pi ^2 f^2}\right] 2\pi f \mathbf E_{_0}\, \cos (2 \pi f t). \end{aligned}$$

The result obtained shows that the propagation of electromagnetic waves in an ionized gas with N electrons per unit volume can be analyzed as if the wave propagates in a dielectric with “an effective permittivity” $\varepsilon$,

$$ \varepsilon =\varepsilon _{_{0}}\left( 1- \frac{Ne^2}{m_\mathrm{e }\varepsilon _{_{0}}4 \pi ^2 f^2}\right) =\varepsilon _{_{0}}\left( 1- \frac{f_\mathrm{c}^2}{f^2}\right) , $$

where $f^2_\mathrm{c}=Ne^2/ m_\mathrm{e }\varepsilon _{_{0}}4\pi ^2$. Hence, in the ionized gas, the phase velocity and the refractive index n can be expressed as

$$\begin{aligned} v=\frac{1}{\sqrt{\mu _{_{0}}\varepsilon } }=\frac{1}{\sqrt{\mu _{_{0}}\varepsilon _{_{0}}(1-f^2_\mathrm{c}/f^2)} }=\frac{c}{\sqrt{1-f^2_\mathrm{c}/f^2} }\Rightarrow n=\frac{c}{v}=\sqrt{1-\frac{f^2_\mathrm{c}}{f^2}}.\end{aligned}$$

(13.21)

(b) When $f<f_\mathrm{c}$, (13.21) shows that the phase velocity is imaginary, which does not make physical sense. Waves with frequencies less than $f_\mathrm{c}$ cannot propagate through the ionized gas. On the other hand, if $f>f_\mathrm{c}$, electromagnetic waves will propagate in the ionized gas. The frequency $f_\mathrm{c}$ is referred to as the critical frequency (or cutoff frequency). For electrons, with $e=1.6\times 10^{-19}$ C, $m_\mathrm{e}=9.1 \times 10^{-31}$ kg,

$$\begin{aligned} f_\mathrm{c}=\sqrt{\frac{Ne^2}{m_\mathrm{e}\varepsilon _{_{0}}4 \pi ^2} }=\sqrt{\frac{N\times (1.6\times 10^{-19})^2}{9.1 \times 10^{-31}\times 8.854\times 10^{-12}\times 4 \pi ^2} }=\sqrt{80.5N}\approx 9 \sqrt{N}. \end{aligned}$$

(13.22)

(c) For perpendicular incidence, $\theta _\mathrm{i }=\theta _\mathrm{t }=0$, i.e. the direction of the refracted ray is the same as that of the incident. The velocity of propagation in the ionized layer is given by (13.21). For $N=10^{10}$ m$^{-3}$, (13.22) gives $f_\mathrm{c}=0.9$ MHz, while for $N=10^{12}$ m$^{-3}$ we have $f_\mathrm{c}=9$ MHz. Therefore, for $N=10^{10}$ m$^{-3}$, if a wave with frequency $f<0.9$ MHz impinges perpendicularly on the ionized gas layer, the wave will be totally reflected. On the other hand, the wave will penetrate through the gas if $f>0.9$ MHz. For perpendicular incidence and $N=10^{12}$ m$^{-3}$, waves with frequencies greater than 9 MHz can penetrate through the gas.

(d) For critical incidence, $\theta _\mathrm{t}=\pi /2$, with $n_1=1$ and the index of refraction of the the ionized gas given by (13.21), Snell’s law gives

$$ n_1\mathrm{sin}\, \theta _\mathrm{i}= n_2\mathrm{sin}\, \theta _\mathrm{t} \Rightarrow 1\times \mathrm{sin}\, \theta _\mathrm{i}= \sqrt{1-\frac{f_\mathrm{c}^2}{f^2}}\,\mathrm{sin}\, \left( \frac{\pi }{2}\right) $$

$$ \Rightarrow \frac{f_\mathrm{c}^2}{f^2}=1-\mathrm{sin}^2\, \theta _\mathrm{i}=\cos ^2\theta _\mathrm{i} \Rightarrow f=\frac{f_\mathrm{c}}{\cos \theta _\mathrm{i}}.$$

Then, for $\theta _\mathrm{i}=30^\circ$, the lowest frequency of the waves that can propagate through the gas is

$$ f=\frac{f_\mathrm{c}}{\cos \theta _\mathrm{i}}=\frac{9}{\cos 30^\circ }\,\mathrm{MHz}\approx 10.4\,\mathrm{MHz},$$

where $f_\mathrm{c}=9$ MHz for $N=10^{12}$ m$^{-3}$, as seen in (c). Frequencies lower than 10.4 MHz make $\mathrm{sin}\,\theta _\mathrm{t}$ in Snell’s law be greater than 1 and, hence, such frequencies cannot penetrate into the gas. On the other hand, waves with $f>10.4$ MHz will propagate through the gas.

13.14 :

The ionosphere is a layer of ionized gas around the earth. If the refractive index of the ionosphere can be expressed as $n_\mathrm{i}=\sqrt{1-\frac{\lambda ^2}{C}}$, where $\lambda$ is the wavelength of the wave and C is a constant, (a) find the wavelength of the shortest radio wave that can be totally reflected by the ionosphere. It is assumed that the ionosphere has a sharp boundary at an altitude H above the surface of the Earth. The wave is emitted, at a given angle $\theta _\mathrm{e}$, by an emitter E on the surface of the earth, as shown in Fig. 13.19. (b) For $\theta _\mathrm{e}=0$ and $\pi /2$, find the wavelength of the shortest totally reflected wave. (c) For $H\approx 300$ km, $R\approx 6371$ km, and $C\approx 10^{3}$ m$^2$, find the wavelength of the shortest radio wave that can be totally reflected from the ionosphere.

Fig. 13.19

Emitter E on the surface of Earth emits a plane wave with angle $\theta _\mathrm{e}$ that is reflected when entering the ionosphere. $\theta _\mathrm{i}$ and $\theta _\mathrm{t}$ denote the angles of incidence and that of refraction, respectively, at the boundary “atmosphere-ionosphere”. R represents the radius of Earth

Solution

In a region extending from a height of about 50 km to over 500 km, molecules of the atmosphere are ionized by radiation from the Sun. This region is called the ionosphere. The altitude and character of the ionized layers depend on the nature of the solar radiation and on the composition of the atmosphere. Ionization of the ionosphere varies greatly with the time of day, the season, and other factors. An important feature of the ionosphere is that it makes it possible for the reflection of radio waves. However, only those waves within a certain wavelength range (or frequency range) will be reflected. Critical wavelengths (or frequencies) change with time of day, atmospheric conditions, and the emission angle. The greater the density of electrons, the shorter the wavelength (or the higher the frequencies) that can be totally reflected. The electron density of the ionosphere ranges from about $10^{10}$ m$^{-3}$, in the lowest layer, to $10^{12}$ m$^{-3}$ in the highest layer. At night, the lower regions become very much depleted of free electrons, and only radio waves with the longest wavelengths can be totally reflected.

In this problem, a simple model of the ionosphere is considered, consisting of a layer at an altitude H above the surface of the Earth. It is assumed that ionization varies sharply at the boundary between the atmosphere and the ionosphere.

(a) Figure 13.19 shows a wave leaving the Earth at an angle $\theta _\mathrm{e}$. From Fig. 13.19, it follows,

$$ \frac{\mathrm{sin}\,\theta _\mathrm{i}}{R}=\frac{\mathrm{sin}\,(\pi -\theta _\mathrm{e})}{H+R}=\frac{\mathrm{sin}\,\theta _\mathrm{e}}{H+R}\Rightarrow \mathrm{sin}\,\theta _\mathrm{i}=\frac{R}{R+H}\,\mathrm{sin}\,\theta _\mathrm{e}=\frac{\mathrm{sin}\,\theta _\mathrm{e}}{1+H/R}. $$

Snell’s law at the interface “atmosphere-ionosphere” gives

$$\begin{aligned} n_\mathrm{a}\, \mathrm{sin}\,\theta _\mathrm{i}= n_\mathrm{i}\, \mathrm{sin}\,\theta _\mathrm{t}\Rightarrow 1 \times \frac{\mathrm{sin}\,\theta _\mathrm{e}}{1+H/R}=\sqrt{1-\frac{\lambda ^2}{C}}\times \mathrm{sin}\,\theta _\mathrm{t}, \end{aligned}$$

(13.23)

where the refractive index for the atmosphere $n_\mathrm{a}$ is assumed to be approximately equal to 1.

Total reflection begins for $\theta _\mathrm{t}=\pi /2$, which gives the smallest wavelength in (13.23) for total reflection to occur. Then, the smallest wavelength, denoted by $\lambda _\mathrm{c}$, which can be totally reflected for a given $\theta _\mathrm{e}$ is given by

$$\begin{aligned} 1 \times \frac{\mathrm{sin}\,\theta _\mathrm{e}}{1+H/R}=\sqrt{1-\frac{\lambda _\mathrm{c}^2}{C}}\times \mathrm{sin}\,\left( \frac{\pi }{2}\right) \Rightarrow \lambda _\mathrm{c}=\left\{ C\left[ 1-\left( \frac{\mathrm{sin}\,\theta _\mathrm{e}}{1+{H}/{R}} \right) ^2 \right] \right\} ^{1/2}. \end{aligned}$$

(13.24)

Thus, waves with $\lambda < \lambda _\mathrm{c}$ are transmitted into the ionosphere, whereas waves with $\lambda >\lambda _\mathrm{c}$ are totally reflected.

(b) At vertical incidence, $\theta _\mathrm{e}=0$, (13.24) gives

$$\begin{aligned} \lambda _\mathrm{c}=\sqrt{C}, \end{aligned}$$

(13.25)

which corresponds to the largest $\lambda _\mathrm{c}$. For $\theta _\mathrm{e}=\pi /2$, then the wave is sent off in the direction of the horizon, thus it follows from (13.24) that

$$\begin{aligned} \lambda _\mathrm{c}=\left\{ C\left[ 1-\left( \frac{1}{1+{H}/{R}} \right) ^2 \right] \right\} ^{1/2}\approx \left\{ C\left[ 1-\left( 1-\frac{2H}{R} \right) \right] \right\} ^{1/2}=\left[ C\times \frac{2H}{R}\right] ^{1/2}, \end{aligned}$$

(13.26)

which corresponds to the smallest wavelength obtained with the model proposed that can be totally reflected from the ionosphere.

Fig. 13.20

Communications between stations on the Earth using the reflection of waves from the ionosphere

(c) Equation (13.26) provides the smallest $\lambda _\mathrm{c}$,

$$ \lambda _\mathrm{c}\approx \left[ C\times \frac{2H}{R}\right] ^{1/2}=\left[ 10^3\times \frac{2\times 300 }{6371}\right] ^{1/2}=9.7\,\,(\mathrm{m}).$$

Therefore, if we wish to use the ionosphere as a reflector of radio waves for communicating between stations on the Earth, as shown in Fig. 13.20, waves shorter than approximately 10 m cannot be used. On the other hand, if we wish to communicate with a satellite beyond the ionosphere, we must use shorter wavelengths to ensure wave penetration through the ionosphere.

13.15 :

Linearly polarized light is incident along the normal of face AB of a glass prism of refractive index $n=1.5$, as shown in Fig. 13.21. Calculate the percentage of the intensity of incident light reflected back by the prism when light emerges from the glass into the air in the opposite direction of the incident beam.

Solution

When the incident light impinges on face AB at normal incidence, from Snell’s law we have: $\theta _\mathrm{i}=\theta _\mathrm{t}=0$. At the air-glass interface, (13.5) gives for the transmission coefficient, denoted by $t_1$,

$$ t_1=t_{\parallel }=t_{\perp }=\frac{2n_\mathrm{a}}{n_\mathrm{a}+n},$$

where $n_\mathrm{a}$ represents the refractive index of air, $n_\mathrm{a}\approx 1$. Then, the amplitude of the transmitted wave at face AB, $E_{0t1}$, in terms of the amplitude of the incident wave, $E_{0i}$, can be expressed as $E_{0t1}=t_1E_{0i}$.

Intensity $I_\mathrm{w}$ (the average Poynting vector) can be expressed in terms of n as

$$\begin{aligned} I_\mathrm{w}=\frac{1}{2}\,\varepsilon v E_0^2=\frac{1}{2}\,\varepsilon _0 n c E_0^2, \end{aligned}$$

(13.27)

where it has been taken into account that $n=c/v=\sqrt{\varepsilon _r \mu _r}$, and that for materials with negligible magnetic properties $\mu _r\simeq 1$ and, hence, $n\simeq \sqrt{\varepsilon _r}$. Therefore, at face AB, the intensities of the incident beam $I_\mathrm{i}$ and that of transmitted beam $I_\mathrm{t1}$ can be written, respectively, as

$$ I_\mathrm{i}=\frac{1}{2}\varepsilon _0n_\mathrm{a}cE_{0i}^2 \quad \mathrm{and}\quad I_\mathrm{t1}=\frac{1}{2}\varepsilon _0ncE_{0t1}^2=\frac{1}{2}\varepsilon _0nct_1^2E_{0i}^2. $$

Then, the ratio of the intensity of the transmitted wave to that of the incident wave, results in

$$\begin{aligned} \frac{I_\mathrm{t1}}{I_\mathrm{i}}=\frac{n}{n_\mathrm{a}}\,t_{t1}^2=\frac{n}{n_\mathrm{a}}\left( \frac{2n_\mathrm{a}}{n_\mathrm{a}+n}\right) ^2=\frac{4nn_\mathrm{a}}{(n+n_\mathrm{a})^2}. \end{aligned}$$

(13.28)

Fig. 13.21

Light enters the prism along the normal of face AB, undergoes total internal reflection twice from the sloped faces, and exits again through face AB

The light continues straight on until it hits the back face AC. Total internal reflection occurs at AC when

$$ n\, \mathrm{sin }\,\theta _\mathrm{c}=n_\mathrm{a}\,\mathrm{sin }\,\frac{\pi }{2}\Rightarrow \mathrm{sin }\,\theta _\mathrm{c}=\frac{n_\mathrm{a}}{n}=\frac{1}{1.5} \Rightarrow \theta _\mathrm{c}=\mathrm{sin }^{-1}\frac{1}{1.5}=41.8^\circ . $$

The light strikes the surface AC at 45$^\circ$, which is greater than the critical angle. After that, the totally reflected ray falls on face CB at 45$^\circ$ and it is again totally reflected. Note that no light is refracted out of the prism at faces AC and BC. Following the ray path shown in Fig. 13.21, the ray then hits surface AB along its normal and exits again through the glass-air interface. At this interface, (13.5) gives for the transmission coefficient

$$ t_2=\frac{2n}{n+n_\mathrm{a}}, $$

where it has been taken into account that $\theta _\mathrm{i}=\theta _\mathrm{t}=0$. Denoting the intensity of the beam transmitted to air by $I_\mathrm{t2}$, at glass-air interface AB, the quotient of the intensities of the transmitted and incident beams can be expressed as

$$ \frac{I_\mathrm{t2}}{I_\mathrm{t1}}=\frac{n_\mathrm{a}}{n}\,t_2^2=\frac{n_\mathrm{a}}{n}\left( \frac{2n}{n_\mathrm{a}+n}\right) ^2=\frac{4n_\mathrm{a} n}{(n+n_\mathrm{a})^2},$$

where $I_\mathrm{t1}$ corresponds to the incident intensity. As the light undergoes total reflection twice, at BC and AC interfaces, the intensity incident on the glass-air interface (for the exiting beam) is the same as that transmitted through the air-glass interface (for the entering beam).

Substituting $I_\mathrm{t1}$ given by (13.28) into the above equation, it follows that

$$ \frac{I_\mathrm{t2}}{I_\mathrm{i}}=\frac{4n n_\mathrm{a}}{(n+n_\mathrm{a})^2}\times \frac{4n_\mathrm{a} n}{(n+n_\mathrm{a})^2}=\frac{(4n n_\mathrm{a})^2}{(n+n_\mathrm{a})^4}=\frac{(4\times 1.5 \times 1)^2 }{(1.5+1)^4}=0.92. $$

Then, the intensity of the beam exiting from face AB is $92\%$ of the intensity of the entering beam.

13.16 :

A plane harmonic wave linearly polarized is incident at the Brewster angle on an interface between two dielectric media with $n_1=1.2$ and $n_2=1.5$ (Fig. 13.22). The electric field of the incident wave makes an angle of 60$^\circ$ with the normal to the plane of incidence. If the intensity of the incident wave is 2 Wm$^{-2}$, determine the intensities for the reflected and transmitted waves.

Solution

Equation (13.7) for the Brewster angle gives an angle of incidence of

$$ \tan \theta _\mathrm{B}=\frac{n_2}{n_1}=\frac{1.5}{1.2}=1.25 \Rightarrow \theta _\mathrm{B}=51.34^{\circ }=\theta _\mathrm{i}. $$

Then, according to (13.6) the angle of refraction is given by (Fig. 13.22)

$$ \theta _\mathrm{t}+\theta _\mathrm{B}=90^{\circ }\Rightarrow \theta _\mathrm{t }=90^{\circ }-51.34^{\circ }=38.66^{\circ }. $$

Fig. 13.22

Plane wave incident obliquely on a plane boundary at the Brewster angle. The reflected wave is linearly polarized along a direction normal to the plane of incidence

The incident linearly polarized wave can be expressed as

$$ \mathbf E_\mathrm{i}=\mathbf E_\mathrm{0i}\cos (\mathbf k_\mathrm{i}\cdot \mathbf r -\omega t)=[E_\mathrm{0i\parallel }\mathbf u_{\parallel }+E_\mathrm{0i\perp }\mathbf u_{\perp }]\,\cos (\mathbf k_\mathrm{i}\cdot \mathbf r -\omega t), $$

$$ \left. \begin{array}{l} E_\mathrm{0i\parallel }=E_\mathrm{0i}\mathrm{sin}\,60{^{\circ }},\\ E_\mathrm{0i\perp }=E_\mathrm{0i}\cos 60^{\circ }, \end{array}\right\} $$

where $E_\mathrm{0i}=|\mathbf E_\mathrm{0i}|$ is the amplitude of the incident wave and $E_\mathrm{0i\parallel }$ and $E_\mathrm{0i\perp }$ are the amplitudes of the components parallel and normal to the plane of incidence, respectively. $\mathbf u_{\parallel }$ and $\mathbf u_{\perp }$ represent the corresponding unit vectors in these directions. The components of the incident wave, parallel and perpendicular to the incidence plane, can then be expressed as

$$ \left. \begin{array}{l} E_\mathrm{i\parallel }=E_\mathrm{0i\parallel }\cos (\mathbf k_\mathrm{i}\cdot \mathbf r -\omega t),\\ E_\mathrm{i\perp }=E_\mathrm{0i\perp }\cos (\mathbf k_\mathrm{i}\cdot \mathbf r -\omega t).\end{array}\right\} $$

From the intensity of the incident wave $I_\mathrm{i}$, it is found from (13.27) that the amplitude of the incident wave is given by

$$ I_\mathrm{i}=\frac{1}{2}\,\varepsilon _0 n_1 c E_\mathrm{0i}^2 \Rightarrow 2=\frac{1}{2}\times 8.85\times 10^{-12}\times 1.2 \times 3\times 10^8 \,E_\mathrm{0i}^2 \Rightarrow E_\mathrm{0i}=35.43 \,\,\,\mathrm{N/C}.$$

Then, the amplitudes of the parallel and perpendicular components are, respectively,

$$ \left. \begin{array}{l} E_\mathrm{0i\parallel }=E_\mathrm{0i}\,\mathrm{sin} 60^{\circ }=35.43\,\mathrm{sin} \,60{^{\circ }}\, \mathrm{N/C}=30.68 \,\,\mathrm{N/C} ,\\ E_\mathrm{0i\perp }=E_\mathrm{0i}\cos 60{^{\circ }} =35.43\cos 60{^{\circ }}\,\mathrm{N/C}=17.72 \,\,\mathrm{N/C}. \end{array}\right\} $$

From (13.4), and $n_1=1.2$, $n_2=1.5$, $\theta _\mathrm{i}=51.34^\circ$, and $\theta _\mathrm{t}=38.66^\circ$, the coefficients for the reflected wave are: $r_{\perp }=-0.22$ and $r_{\parallel }=0$, as expected for Brewster incidence. Then, the amplitudes of the components of the reflected wave are, respectively,

$$ \left. \begin{array}{l} E_\mathrm{0r\parallel }=0,\\ E_{0r\perp }=r_{\perp }E_\mathrm{0i\perp }=-0.22\times 17.72\simeq -3.90 \,\,\,\mathrm{N/C}. \end{array}\right\} $$

Finally, the intensity of the reflected wave will be

$$ I_\mathrm{r}=\frac{1}{2}\,\varepsilon _0 n_1 c E^2_\mathrm{0r}=\frac{1}{2}\times 8.85\times 10^{-12}\times 1.2 \times 3\times 10^8 \times (0+3.90^2)\, \mathrm{W/m^2}=0.024 \,\, \mathrm{W/m^2}.$$

In the same way, for $n_1=1.2$, $n_2=1.5$, $\theta _\mathrm{i}=51.34^\circ$, and $\theta _\mathrm{t }=38.66^\circ$, (13.5) yields: $t_{\parallel }=0.80$ and $t_{\perp }=0.78$. Then, the amplitudes of the components of the transmitted wave are found to be

$$ \left. \begin{array}{l} E_\mathrm{0t\parallel }=t_{\parallel }\, E_\mathrm{0i\parallel }=0.80\times 30.68\,\mathrm{N/C}=24.54 \,\,\mathrm{N/C},\\ E_\mathrm{0t\perp }=t_{\perp }E_\mathrm{0i\perp }=0.78\times 17.72\,\mathrm{N/C}=13.82 \,\,\,\mathrm{N/C}. \end{array}\right\} $$

Thus, the intensity of the transmitted wave will be

$$\begin{aligned} I_\mathrm{t}= & {} \frac{1}{2}\,\varepsilon _0 n_2 c E^2_\mathrm{0t}=\frac{1}{2}\,\varepsilon _0 n_2 c \left( E^2_\mathrm{0 t \parallel }+E^2_\mathrm{0 t \perp }\right) \\= & {} \frac{1}{2}\times 8.85\times 10^{-12}\times 1.5 \times 3\times 10^8 \left( 24.54^2+13.82^2\right) \, \mathrm{W/m^2}=1.58\,\, \mathrm{W/m^2}. \end{aligned}$$

Problems C

13.17 :

A linearly polarized plane wave of frequency 100 MHz is incident on an air-glass interface at an angle of 30${}^\circ$. The refractive index of the glass is $n=1.60$. The electric vector of the incident beam makes an angle of 45$^\circ$ with the plane of incidence and has an amplitude of 25 V/m. (a) Find the reflection and transmission coefficients. (b) Write the expressions for the reflected and transmitted electric fields. (c) Describe the polarization state of the reflected and transmitted beams.

Solution

(a) Application of Snell’s law to the air-glass interface with $\theta _\mathrm{i}=30^\circ$, $n_1\approx 1$, $n_2 =1.6$ gives $\theta _\mathrm{t}=18.2^\circ$. From (13.4) and (13.5), we have for the Fresnel coefficients

$$ r_{\parallel }=\frac{1.60 \times \cos \,30^\circ -1 \times \cos \,18.2^\circ }{1 \times \cos \,18.2^\circ +1.60 \times \cos \,30^\circ }=0.19 \quad t_{\Vert }=\frac{2\times 1 \times \cos \,30^\circ }{1 \times \cos \,18.2^\circ +1.60 \times \cos \,30^\circ }=0.74 $$

$$ r_{\perp }=\frac{1 \times \cos \,30^\circ -1.60 \times \cos \,18.2^\circ }{1 \times \cos \,30^\circ +1.60 \times \cos \,18.2^\circ }=-0.27\quad t_{\perp }=\frac{2\times 1 \times \cos \,30^\circ }{1 \times \cos \,30^\circ +1.60 \times \cos \,18.2^\circ }=0.73 $$

(b) Let the XY plane be the plane of incidence, as shown in Fig. 13.23. The propagation vector for the incident wave can be then expressed as

$$\begin{aligned} \mathbf k_\mathrm{i}=\frac{2 \pi }{\lambda }\, \left( \mathrm{sin}\,30^\circ \, \mathbf u_x- \cos \,30^\circ \, \mathbf u_y\right) =\frac{2 \pi }{3}\, \left( \frac{1}{2}\, \mathbf u_x-\frac{\sqrt{3}}{2}\, \mathbf u_y\right) \,\,\,\mathrm{m^{-1}}, \end{aligned}$$

(13.29)

where $\lambda$ denotes the wavelength in air, equal to $\lambda \approx c/f=3\times 10^{8}/100\times 10^{6}\,\mathrm{m}=3$ m.

Fig. 13.23

Linearly polarized plane wave incident obliquely on an air-glass boundary

The incident wave is linearly polarized along a direction that makes 45$^\circ$ with the plane of incidence. Then, the amplitude of the components parallel and perpendicular to the plane of incidence are $E_\mathrm{0i\parallel }=E_\mathrm{0i}\, \cos \, 45^\circ$ and $E_\mathrm{0i\perp }=E_\mathrm{0i}\,\mathrm{sin}\, 45^\circ$, respectively, where $E_\mathrm{0i}$ is the amplitude of the incident wave. The components parallel and perpendicular of the incident electric field are then given by

$$\begin{aligned} E_\mathrm{i \parallel }= & {} E_\mathrm{0i\parallel }\, \cos \left( \mathbf k_\mathrm{i}\cdot \mathbf r -\omega t \right) =E_\mathrm{0i}\, \cos \, 45^\circ \, \cos \left( \mathbf k_\mathrm{i}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2} \, \cos \left[ \frac{2\pi }{3}\left( \frac{1}{2}\,x -\frac{\sqrt{3}}{2}\,y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m}\\= & {} 17.68 \, \cos \left( 1.05\,x -1.81\,y-6.28\times 10^{8} t \right) \,\,\mathrm{V/m},\\ E_\mathrm{i \perp }= & {} E_\mathrm{0i\perp } \, \cos \left( \mathbf k_\mathrm{i}\cdot \mathbf r -\omega t \right) =E_\mathrm{0i}\,\mathrm{sin}\, 45^\circ \, \cos \left( \mathbf k_\mathrm{i}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2} \, \cos \left[ \frac{2\pi }{3}\left( \frac{1}{2}\,x -\frac{\sqrt{3}}{2}\,y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m}\\= & {} 17.68 \, \cos \left( 1.05\,x -1.81\,y-6.28\times 10^{8} t \right) \,\,\mathrm{V/m}, \end{aligned}$$

where it has been taken into account that the position vector is $\mathbf r=(x,y,z)$, $\mathbf k_\mathrm{i}$ is given by (13.29), and $\omega =2\pi f=2\pi \times 10^{8}$ s$^{-1}$.

The propagation vectors for the reflected and transmitted waves are, respectively,

$$\begin{aligned} \mathbf k_\mathrm{r}= & {} \frac{2 \pi }{\lambda }\, \left( \mathrm{sin}\,30^\circ \, \mathbf u_x+ \cos \,30^\circ \, \mathbf u_y\right) =\frac{2 \pi }{3}\, \left( \frac{1}{2}\, \mathbf u_x+ \frac{\sqrt{3}}{2}\, \mathbf u_y\right) \,\,\,\mathrm{m^{-1}},\\ \mathbf k_\mathrm{t}= & {} \frac{2 \pi }{\lambda }\,n\, \left( \mathrm{sin}\,18.2^\circ \, \mathbf u_x- \cos \,18.2^\circ \, \mathbf u_y\right) =\frac{2 \pi }{3}\times 1.60 \left( 0.31\, \mathbf u_x- 0.95\, \mathbf u_y\right) \,\,\,\mathrm{m^{-1}}. \end{aligned}$$

Then, the reflected wave has the following components

$$\begin{aligned} E_\mathrm{r \parallel }= & {} E_\mathrm{0i\Vert } \,r_{\parallel }\, \cos \left( \mathbf k_\mathrm{r}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2} \times 0.19\,\cos \left[ \frac{2 \pi }{3}\, \left( \frac{1}{2}\, x+ \frac{\sqrt{3}}{2}\, y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m},\\ {}= & {} 3.36\,\cos \left( \,1.05 x+ 1.81\, y-6.28\times 10^{8} t \right) \,\,\mathrm{V/m},\\ E_\mathrm{r \perp }= & {} E_\mathrm{0i\perp } \, r_{\perp }\,\cos \left( \mathbf k_\mathrm{r}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2}\times (-0.27) \, \cos \left[ \frac{2 \pi }{3}\, \left( \frac{1}{2}\, x+ \frac{\sqrt{3}}{2}\, y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m},\\ {}= & {} -4.77 \, \cos \left( 1.05\, x+ 1.81\, y-6.28\times 10^{8} t \right) \,\mathrm{V/m}\\ {}= & {} 4.77 \, \cos \left( 1.05\, x+ 1.81\, y-6.28\times 10^{8} t +\pi \right) \,\,\mathrm{V/m}, \end{aligned}$$

whereas for the components of the transmitted wave, we have

$$\begin{aligned} E_\mathrm{t \parallel }= & {} E_\mathrm{0i_{\Vert } } \,t_{\parallel }\, \cos \left( \mathbf k_\mathrm{t}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2} \times 0.74\,\cos \left[ \frac{2 \pi }{3}\times 1.60 \left( 0.31\, x- 0.95\, y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m}\\ {}= & {} 13.08\,\cos \left( 1.04\, x- 3.18\, y-6.28\times 10^{8} t \right) \,\,\mathrm{V/m},\\ E_\mathrm{t \perp }= & {} E_\mathrm{0i_{\perp } } \, t_{\perp }\,\cos \left( \mathbf k_\mathrm{t}\cdot \mathbf r -\omega t \right) \\ {}= & {} 25\times \frac{\sqrt{2}}{2}\times 0.73 \, \cos \left[ \frac{2 \pi }{3}\times 1.60 \left( 0.31\, x- 0.95\, y\right) -2\pi \times 10^{8} t \right] \,\mathrm{V/m}\\ {}= & {} 12.90 \, \cos \left( 1.04\, x- 3.18, y-6.28\times 10^{8} t \right) \,\,\mathrm{V/m}. \end{aligned}$$

(c) The phase difference between the perpendicular and parallel components of the reflected wave is $\left( \varphi _{0\perp }-\varphi _{0\parallel }\right) _\mathrm{r}=\pi$. Therefore, the wave is linearly polarized, the electric field of the reflected wave makes an angle with the plane of incidence of

$$ \tan \,\theta _\mathrm{r} =\frac{E_\mathrm{r\perp }}{E_\mathrm{r\parallel }}=\frac{-4.77}{3.36}\Rightarrow \theta _\mathrm{r}=-54.84^\circ .$$

For the transmitted wave, $\left( \varphi _{0\perp }-\varphi _{0\parallel }\right) _\mathrm{t}=0$, and, hence, the transmitted wave is linearly polarized, its electric field making an angle with the plane of incidence of

$$ \tan \,\theta _\mathrm{t} =\frac{E_\mathrm{t\perp }}{E_\mathrm{t\parallel }}=\frac{12.90}{13.08}\Rightarrow \theta _\mathrm{t}=44.60^\circ .$$

13.18 :

A right-handed circularly polarized electromagnetic wave is incident on a glass-air interface, as shown in Fig. 13.24. The incident wave has an intensity of $20\times 10^{-4}$ Wm$^{-2}$. When the reflected wave is linearly polarized along a direction perpendicular to the plane of incidence, the angle of incidence is $33.69^\circ$. (a) Find the intensity of the reflected wave. (b) Describe the state of polarization of the transmitted wave. (c) Determine the intensity of the transmitted wave.

Fig. 13.24

Circularly polarized plane wave incident on a plane glass-air interface

Solution

Since the incident wave is circularly polarized and the reflected wave is linearly polarized, the wave is incident at the Brewster angle, $\theta _\mathrm{i}=\theta _\mathrm{B}=33.69^\circ$. The angle of refraction can be obtained from (13.6), $\theta _\mathrm{t}=90^\circ -33.69^\circ =56.31^\circ$. At the Brewster incidence, (13.7) gives for the refractive index of the glass through which the electromagnetic wave is propagating

$$ \tan \,\theta _\mathrm{B}=\frac{n_2}{n_1}\Rightarrow \tan \,33.69^\circ =\frac{n_2}{n_1} \Rightarrow n_1= \frac{n_2}{\tan \,33.69^\circ }=\frac{1}{\tan \, 33.69^\circ }=1.50,$$

where the refractive index of air is assumed to be $n_2 \approx 1$.

The amplitude of the components of the incident wave can be calculated from the intensity of the incident wave $I_\mathrm{i}$, which for a circularly polarized wave is given by

$$ I_\mathrm{i}=\varepsilon _0 n_1 c E_\mathrm{0i}^2 \Rightarrow 20\times 10^{-4} = 8.85\times 10^{-12}\times 1.5 \times 3\times 10^8 \,E_\mathrm{0i}^2 \Rightarrow E_\mathrm{0i}=0.71 \,\,\,\mathrm{N/C}.$$

Then, the parallel and perpendicular components can be expressed as

$$ \left. \begin{array}{l} E_\mathrm{i\parallel }=E_\mathrm{0i}\,\mathrm{sin}(\mathbf k_\mathrm{i}\cdot \mathbf r-\omega t+\pi /2)=E_\mathrm{0i}\, \cos (\mathbf k_\mathrm{i}\cdot \mathbf r-\omega t) =0.71\, \cos (\mathbf k_\mathrm{i}\cdot \mathbf r-\omega t)\,\,\,\mathrm{N/C}, \\ E_\mathrm{i\perp }=E_\mathrm{0i}\, \mathrm{sin} (\mathbf k_\mathrm{i}\cdot \mathbf r-\omega t)=0.71\, \mathrm{sin} (\mathbf k_\mathrm{i}\cdot \mathbf r-\omega t)\,\,\,\mathrm{N/C}, \end{array}\right\} $$

where it has been taken into account that for a right-handed circularly polarized wave $(\varphi _{0\parallel }-\varphi _{0\perp })_\mathrm{i}=\pi /2$.

(a) For $n_1=1.5$, $n_2=1$, $\theta _\mathrm{i}=33.69^\circ$, and $\theta _\mathrm{t }=56.31^\circ$, (13.4) for the reflection coefficients gives $r_{\parallel }=0$, as expected, and $r_{\perp }=0.38$. Then, the amplitudes of the components of the reflected wave are, respectively,

$$ \left. \begin{array}{l} E_\mathrm{0r\parallel }=0,\\ E_{0r\perp }=r_{\perp }E_\mathrm{0i}=0.38\times 0.71=0.27 \,\,\,\mathrm{N/C}. \end{array}\right\} $$

Thus, the intensity of the reflected linearly polarized wave is equal to

$$ I_\mathrm{r}=\frac{1}{2}\,\varepsilon _0 n_1 c E^2_\mathrm{0r}=\frac{1}{2}\times 8.85\times 10^{-12}\times 1.5\times 3\times 10^8 \times 0.27^2\, \mathrm{W/m^2}=1.45 \times 10^{-4}\,\, \mathrm{W/m^2}.$$

(b) Equation (13.5) yields for the transmission coefficients: $t_{\parallel }=1.50$ and $t_{\perp }=1.38$. The amplitudes of the components of the transmitted wave will be

$$ \left. \begin{array}{l} E_\mathrm{0t\parallel }=t_{\parallel }\, E_\mathrm{0i}=1.50\times 0.71=1.07 \,\,\mathrm{N/C},\\ E_\mathrm{0t\perp }=t_{\perp }E_\mathrm{0i}=1.38 \times 0.71=0.98 \,\,\,\mathrm{N/C}, \end{array}\right\} $$

and the components of the transmitted wave are then given by

$$ \left. \begin{array}{l} E_\mathrm{t\parallel }=1.07\, \cos (\mathbf k_\mathrm{t}\cdot \mathbf r-\omega t) =1.07\,\mathrm{sin} (\mathbf k_\mathrm{t}\cdot \mathbf r-\omega t+\pi /2) \,\,\mathrm{N/C}, \\ E_\mathrm{t\perp }=0.98\, \mathrm{sin} (\mathbf k_\mathrm{t}\cdot \mathbf r-\omega t) \,\,\mathrm{N/C}, \end{array}\right\} $$

which corresponds to a right-handed elliptically polarized wave since $(\varphi _{0\parallel }-\varphi _{0\perp })_\mathrm{t}=\pi /2$ and $E_{0\mathrm t\parallel } \ne E_{0\mathrm t\perp }$.

(c) Finally, the intensity of the transmitted wave can be expressed as

$$\begin{aligned} I_\mathrm{t}= & {} \frac{1}{2}\,\varepsilon _0 n_2 c E^2_\mathrm{0t}=\frac{1}{2}\,\varepsilon _0 n_2 c \left( E^2_\mathrm{0 t \parallel }+E^2_\mathrm{0 t \perp }\right) \\= & {} \frac{1}{2}\times 8.85\times 10^{-12}\times 1 \times 3\times 10^8 \times \left( 1.07^2+0.98^2\right) \, \mathrm{W/m^2}=2.79 \times 10^{-3} \,\, \mathrm{W/m^2}. \end{aligned}$$

13.19 :

A linearly polarized plane harmonic wave is incident normally on a plane boundary between two dielectric media, denoted 1 and 2, with permeabilities $\mu _1=\mu _2=\mu _0$ and refractive indices $n_1$ and $n_2$, respectively, see Fig. 13.25a. In order to eliminate the reflected wave, a plane parallel dielectric layer is inserted between the two media, as shown in Fig. 13.25b. Determine the thickness of the layer L and its refractive index $n'$ so that the condition of no reflection is fulfilled.

Fig. 13.25

a A plane wave is incident normally on the interface between two dielectrics. b A dielectric layer is inserted between the two media in order to eliminate the reflected wave

Solution

Let OY direction be along the polarization direction of the incident electromagnetic plane wave, as shown in Fig. 13.26. Then, for simplicity, using complex notation, the electric field vector of the incident wave can be written as

$$ \mathbf E_{_\mathrm{i}}=\left( E_\mathrm{0i}\, \mathbf u_{y}\right) e^{j\omega \left( t-\frac{x}{v_1}\right) }=\left( E_\mathrm{0i}\, \mathbf u_{y}\right) e^{j \left( \omega t-k_1 x\right) },$$

where $v_1$ is the velocity in medium 1 and the wavenumber $k_1=\omega /v_1$. The corresponding magnetic field $\mathbf H$ is given by

$$ \mathbf H_{_\mathrm{i}}=\left( \frac{E_\mathrm{0i}}{\mu _0 v_1} \, \mathbf u_{z}\right) e^{j \left( \omega t-k_1 x\right) }=\left( \frac{n_1E_\mathrm{0i}}{\mu _0 c} \, \mathbf u_{z}\right) e^{j \left( \omega t-k_1 x\right) }. $$

Transmitted and reflected waves are generated by the incident wave impinging on the dielectric layer at the interface $x=0$. In the dielectric layer, the transmitted wave in turn will give rise to reflected and refracted waves at $x=L$. The wave returning back into the layer will be reflected and transmitted, in turn, at $x=0$. As a result of the multiple reflections at the two boundaries, two set of waves will be propagating in opposite directions through the dielectric layer, denoted by $\mathbf E'_\mathrm{t}$ and $\mathbf E'_\mathrm{r}$ in Fig. 13.26. Note that these waves can be considered to be the result of the interference of the multiple reflected and transmitted waves at the two boundaries. The resulting wave from the superposition of the transmitted waves into medium 2 is denoted by $\mathbf E_\mathrm{t}$ and the resulting reflected wave travelling in medium 1 is represented by $\mathbf E_\mathrm{r}$. These waves can be expressed as

$$ \begin{array}{ll} \mathbf E_\mathrm{r}= \left( E_\mathrm{0r}\, \mathbf u_{y}\right) e^{j\omega \left( t+\frac{x}{v_1}\right) }=\left( E_\mathrm{0r}\, \mathbf u_{y}\right) e^{j \left( \omega t+k_1 x\right) },\quad &{}x<0,\\ \mathbf E'_\mathrm{t}= \left( E'_\mathrm{0t}\, \mathbf u_{y}\right) e^{j\omega \left( t-\frac{x}{v'}\right) }=\left( E'_\mathrm{0t}\, \mathbf u_{y}\right) e^{j \left( \omega t-k' x\right) },\quad &{}0<x<L,\\ \mathbf E'_\mathrm{r}= \left( E'_\mathrm{0r}\, \mathbf u_{y}\right) e^{j\omega \left( t+\frac{x}{v'}\right) }=\left( E'_\mathrm{0r}\, \mathbf u_{y}\right) e^{j \left( \omega t+k' x\right) },\quad &{}0<x<L,\\ \mathbf E_\mathrm{t}= \left( E_\mathrm{0t}\, \mathbf u_{y}\right) e^{j\omega \left[ t-\frac{(x-L)}{v_2}\right] }=\left( E_\mathrm{0t}\, \mathbf u_{y}\right) e^{j \left[ \omega t-k_2( x-L)\right] },\quad &{}x>L, \end{array} $$

where $E_\mathrm{0r}$, $E'_\mathrm{0t}$, $E'_\mathrm{0r}$, and $E_\mathrm{0t}$ represent complex amplitudes. The boundary conditions must be satisfied simultaneously at the two interfaces.

The field $\mathbf H$ associated with the above expressions for the electric fields can be written as

$$ \begin{array}{ll} \mathbf H_\mathrm{r}=\frac{n_1E_\mathrm{0r}}{\mu _0 c} \,(- \mathbf u_{z})e^{j \left( \omega t+k_1 x\right) },\quad &{}x<0,\\[4pt] \mathbf H'_\mathrm{t}=\frac{n'E'_\mathrm{0t}}{\mu _0 c} \,( \mathbf u_{z})e^{j \left( \omega t-k' x\right) },\quad &{}0<x<L,\\[4pt] \mathbf H'_\mathrm{r}=\frac{n'E'_\mathrm{0r}}{\mu _0 c} \,( -\mathbf u_{z})e^{j \left( \omega t+k' x\right) },\quad &{}0<x<L,\\[4pt] \mathbf H_\mathrm{t}=\frac{n_2E_\mathrm{0t}}{\mu _0 c} \,(\mathbf u_{z})e^{j \left[ \omega t-k_2( x-L)\right] },\quad &{}x>L. \end{array}$$

The continuity of $E_y$ requires that at $x=0$

$$ E_\mathrm{0i}+E_\mathrm{0r}=E'_\mathrm{0t}+E'_\mathrm{0r},$$

where the common factor $e^{j\omega t}$ is omitted. At $x=L$, the boundary conditions gives

$$ E'_\mathrm{0t}e^{-jk'L}+E'_\mathrm{0r}e^{jk'L}=E_\mathrm{0t}. $$

The continuity of $H_z$ requires that at $x=0$

$$ n_1 E_\mathrm{0i}-n_1 E_\mathrm{0r}=n' E'_\mathrm{0t}-n' E'_\mathrm{0r}.$$

For $x=L$, the continuity of $H_z$ gives

$$ n'E'_\mathrm{0t}e^{-jk'L}-n'E'_\mathrm{0r}e^{jk'L}=n_2E_\mathrm{0t}.$$

Fig. 13.26

Coordinate system in a plane parallel dielectric layer. $\mathbf E_\mathrm{i}$ is the incident wave, $\mathbf E_\mathrm{r}$ is the reflected wave, and $\mathbf E_\mathrm{t}$ represents the resulting wave transmitted out of the layer. As a result of the multiple reflections at the boundaries, two waves, $\mathbf E'_\mathrm{r}$ and $\mathbf E'_\mathrm{t}$, are propagating in the layer in opposite directions

Rearranging the above set of equations, the following system of linear equations is obtained

$$\begin{aligned} -E_\mathrm{0r}+E'_\mathrm{0t}+E'_\mathrm{0r}+0= & {} E_\mathrm{0i},\\ n_1E_\mathrm{0r}+n'E'_\mathrm{0t}-n'E'_\mathrm{0r}+0= & {} n_1E_\mathrm{0i},\\ 0+E'_\mathrm{0t}e^{-jk'L}+E'_\mathrm{0r}e^{jk'L}-E_\mathrm{0t}= & {} 0,\\ 0+n'E'_\mathrm{0t}e^{-jk'L}-n'E'_\mathrm{0r}e^{jk'L}-n_2E_\mathrm{0t}= & {} 0,\ \end{aligned}$$

where $E_\mathrm{0i}$ is assumed to be known. From the condition of no reflection, $E_{0r}$ must be zero, and, therefore, by applying Crammer’s rule, it follows that

$$ \left| \begin{array}{cccc} E_\mathrm{0i}&{} 1&{}1&{}0\\ n_1E_\mathrm{0i}&{} n'&{}-n'&{}0\\ 0&{}e^{-jk'L}&{}e^{jk'L}&{}-1\\ 0&{}n'e^{-jk'L}&{}-n'e^{jk'L}&{}-n_2\end{array} \right| =0. $$

Manipulation of the determinant, dividing the first column by $E_\mathrm{0i}$, and subtracting the first row multiplied by $n_1$ from the second row, gives

$$ \left| \begin{array}{cccc} 0&{} n_1-n'&{}n_1+n'&{}0\\ n_1&{} n'&{}-n'&{}0\\ 0&{}e^{-jk'L}&{}e^{jk'L}&{}-1\\ 0&{}n'e^{-jk'L}&{}-n'e^{jk'L}&{}-n_2\end{array} \right| =0. $$

Multiplying the third row by $n'$ and addition and substraction of the third and fourth rows leads to

$$ \left| \begin{array}{cccc} 0&{} n_1-n'&{}n_1+n'&{}0\\ n_1&{} n'&{}-n'&{}0\\ 0&{}2n'e^{-jk'L}&{}0&{}-n'-n_2\\ 0&{}0&{}2n'e^{jk'L}&{}-n'+n_2\end{array} \right| =0. $$

Evaluating this determinant, we have

$$ (n_1-n')\left( 2n'e^{jk'L}\right) (n'+n_2)-(n_1+n')\left( 2n'e^{-jk'L}\right) (-n'+n_2)=0\Rightarrow $$

$$(2n_1n'-2n'n_2)\mathrm{cos}(k'L)+j\sin (k'L) (-2n'^2+2n_1n_2)=0.$$

The real part is zero if

$$ (2n_1n'-2n'n_2)\mathrm{cos}(k'L)=0 \Rightarrow n_1=n_2 \,\,\,\mathrm{or}\,\,\, \mathrm{cos}(k'L)=0 \Rightarrow k'L=\frac{\pi }{2}+N \pi .$$

Only the second solution is valid since the first one would imply that there is no change in medium properties. For the imaginary part to be zero

$$\sin (k'L) (-2n'^2+2n_1n_2)=0 \Rightarrow \sin (k'L)=0 \,\,\,\,\,\,\mathrm{or}\,\,\,\,\,\, n'=\sqrt{n_1n_2},$$

where only the second solution is valid. The first condition cannot be satisfied because the solution chosen for the real part to be zero is $\mathrm{cos}(k'L)=0$ and, therefore, ${\sin }(k'L)=\pm 1$.

Therefore, the conditions for no reflection are

$$ k'L=\frac{\pi }{2}+N \pi \,\,\, \,\,\,\mathrm{and}\,\,\, \,\,\,n'=\sqrt{n_1n_2}.$$

Note that the first condition determines the thickness of the layer, whereas the second one the value of the refractive index. The first condition can be expressed as

$$ k'L=\frac{\pi }{2}+N \pi \Rightarrow L=\frac{\lambda '}{4}(1+2N)\Rightarrow n'L=\frac{\lambda _0}{4}(1+2N),$$

where $\lambda '$ is the wavelength in the dielectric layer and $\lambda _0$ the wavelength in vacuum. The dielectric layer thickness must be an odd multiple of $\lambda '/4$ for no light to be reflected from the surface and the dielectric layer will then be an antireflection coating.