Trabajo_grupal_203058_60-1.docx

TEORIA ELECTROMAGNETICA Y DE ONDAS UNIDAD 2: PASO 3: MODOS DE PROPAGACIÓN Y POLARIZACIÓN EN LAS ONDAS ELECTROMAGNÉTICAS

PRESENTADO POR: JOSE ANTONIO SERRANO LUGO JESUS DAVID LOZANO JUAN JOSE PERDOMO CARLOS EDUARDO DIAZ DANIA CAROLINA GONZALEZ

PROGRAMA: TEORIA ELECTROMAGNETICA Y DE ONDAS GRUPO: 203058_60

TUTOR DEL CURSO: WILLIAM MAURICIO SAENZ

ESCUELA DE CIENCIAS BASICA E INGENIERIA UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA OCTUBRE 2018

INTRODUCTION

With the passage of time, the human being can be easier than it is, when feeding, to go to the doctor when something does not work well in his organism, he has the need to communicate and be attentive to what is happening in the world and around you. For this reason the method of propagation of the waves is studied, which propagates in any dielectric material, including air, but does not propagate very well in conductors with lost, for example, seawater. There are several propagation methods that are most known in the wave of the radio that propagates through the terrestrial waves with the energy transmitted from the source, later the energy is received from the side of the receiving antenna. other waves are light, infrared rays, ultraviolet rays, X-rays, gamma rays. These waves are not visible to the human being and must be analyzed with indirect methods such as schemes.

The transmission medium constitutes the physical medium through which the sender and receiver can communicate in a data transmission system. We distinguish two types of media: guided and unguided This guide Will deal with topics related to modes of propagation and polarization in electromagnetic waves and the behavior of waves in different transmission media used in telecommunications.

Each student in the group has to answer the following questions using academic references to support the research: . Explain how refraction and reflection phenomenon could be used in transmission systems. Reflection is the change of direction of a wave, which, when coming into contact with the surface of separation between two changing media.(wikipedia) In itself, the reflection wave can be produced in two common media, which are the atmosphere and the terrestrial one. In the reflection corresponds to the fact that the ground plane acts as an equipotential, that is to say a reflecting surface. This propagation system is called propagation by terrestrial reflection and is very common inRadio frequency (RF) applications. (guias unad pag 249) Ilustración 1 Propagación por reflexión en el plano de tierra

Fuentes: guías UNAD

Refraction is the change of direction and speed that a wave experiences when passing from one medium to another with different refractive index. It only occurs if the wave strikes obliquely on the separation surface of the two media and if they have different indices of refraction. The refraction originates in the change of velocity of propagation of the indicated wave. (wikipedia) Atmospheric reflection, however, has the disadvantage that optimal reception of the signals changes according to the time of day and weather conditions, due to that the atmosphere, due to the effect of thermal exchange, tends to expand or contract, which causes change in the heights of reflection and increase of the attenuation by having different distances to get to the receiver.

Ilustración 2 Propagación por reflexión atmosférica

Fuentes: guías UNAD

Atmospheric reflection, however, has the disadvantage that optimal reception of the signals changes according to the time of day and weather conditions, due to that the atmosphere, due to the effect of thermal exchange, tends to expand or contract, which causes change in the heights of reflection and increase of the attenuation by having different distances to get to the receiver. (unad guias pag. 250) . Explain the Snell law, the critical angle and practical applications where it could be used. Ley de Snell These angles coincide with those formed by the Poynting vector with the normal vector a Each side of the border, and the relationship between them is determined by the so-called "Snell's Law". (unad guias pag. 284) Snell's law (also called Snell-Descartes law) is a formula used to calculate the refractive angle of light by traversing the separation surface between two means of propagating light (or any electromagnetic wave) with a refractive index. Different. (wikipedia)

Unlike the critical angle of total reflection, the Brewster angle is presented for any combination of the refractive indices n1 and n2. If it is observed for example in the Figure 121, you can see how the reflection coefficient becomes zero just at the Brewster angle, which in this case is close to 60º. (guias unad pag, 297)

Ilustración 3 Gráfico del coeficiente de Fresnel de reflexión en magnitud y ángulo

Fuentes: guías UNAD

Some of the applications: Refraction allows us to have lenses, magnifying glasses, prisms and rainbows. Even our eyes depend on this curvature of light. Without refraction, we would not be able to focus the light on our retina. Optical fiber is a transmission medium that is used mainly in data networks. It consists of a very thin thread of plastic or glass material by means of which pulses of light or data to be transmitted are sent. The beam of light propagates through the fiber by means of a reflection angle that passes over the angle that is the limit of the total reflection. This light source that is used can be a laser or also a led light. Otic fibers are widely used in telecommunications because they have the advantage of being able to carry information over long distances at very fast speeds. (www.euston96.com) .Explain the physics behind lenses for optical diseases. In physics, geometric optics is part of Snell's phenomenological laws of reflection and refraction. From them, it is enough to make geometry with the light rays to obtain the formulas that correspond to the mirrors, diopter and lenses, thus obtaining the laws that govern the optical instruments to which we are accustomed. Geometric optics uses the notion of luminous ray; is an approximation of the behavior that corresponds to electromagnetic waves (light) when the objects involved are of a size much greater than the wavelength used; this allows to disregard the effects derived from diffraction, behavior linked to the wave nature of light. (wikipedia)

. What is the importance of the index of refraction? Refractive Index: The ratio of the speed of light in vacuum and the speed of light in the medium whose index is calculated is called refractive index. The refractive index of a medium is a measure of how much the speed of light (or other waves such as sound waves) is reduced within the medium. The refractive property of a material is the most important property of any optical system that uses refraction. It is a reverse index that indicates the thickness of the lenses according to a given power, and the dispersive power of the prisms. It is also used in chemistry to determine the purity of chemical reagents and for the rendering of refractive materials in 3D computer graphics. (wikipedia) From the Law of Snell, the appearance of an interesting phenomenon is deduced, which occurs when the refractive index of the substance in which the wave originates (n1) is greater than that of the substance to which it tries to pass. When this happens, the angle of refraction becomes equal to and there is no refracted ray that is, all the energy of the incident ray returns to the medium from which it comes. This condition is called internal total reflection and can be presented in optics, but also with all kinds of electromagnetic waves. To find the angle of critical incidence from which total reflection is presented, we start from Snell's Law by making the sine of the refraction angle equal to unity. (guias unad pag. 286). . Is it possible that a single light beam be refracted into several beans? Explain. The speed of light depends on the medium through which it travels, so it is slower the more dense the material and vice versa. Therefore, when the light passes from a less dense medium (air) to a denser one (glass), the ray of light is refracted approaching the normal and therefore, the angle of refraction will be smaller than the angle of incidence. In the same way, if the ray of light passes from a denser medium to a less dense one, it will be refracted away from the normal one and, therefore, the angle of incidence will be less than that of refraction. So we can say that refraction is the change of direction of the propagation that light experiences when passing from one medium to another. on the other hand, the speed of the penetration of light in a medium other than a vacuum is related to the wavelength and, when a beam of white light passes from one medium to another, each color undergoes a slight deviation. This phenomenon is known as light scattering. For example, when reaching a denser medium, shorter waves lose speed over longer ones (eg, when white light passes through a prism). The short wavelengths are up to four times more dispersed than the long ones, which explains why the sky looks bluish, since for that range of colors the refractive index is higher and disperses more. (wikipedia, s.f.)

Taking into account this we can say that the refraction angle in the beans is greater, which refraction is very minimal, unlike a premium the refraction angle is closer to the normal, for this reason the refraction is higher. Ilustración 4 refracción rayos de luz prima

Fuentes: imagen tomada de la pagina https://es.123rf.com/photo_60985795_3d-ilustraci%C3%B3n-de-prisma-dual-yrefracci%C3%B3n-rayos-de-luz-con-haces-de-luz-.html

. How is it possible to use the Brewster angle in practical applications? The reflection of light from surfaces is part of our daily life. However, a wonderful property of optics is the possibility of having zero reflection. That is, reflection does not occur from a clean and perfect interface illuminated under a single angle of incidence of p-polarized light. This phenomenon is described by Brewster's law and provides the so-called Brewster angle for the media involved, depending on the refractive indices of the media in the system. (nanocienciainforma.wordpress.com, s.f.) One of the important applications was the Brewster microscope which allows the visualization of molecular monolayers in real time water - air. Also other applications such as cameras, lenses among others. Ilustración 5 microscopio ángulo de brewster

Fuentes: imagen tomada de la pagina https://nanocienciainforma.wordpress.com/microscopio-angulo-de-brewsterbam/

1. When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is 35𝑜 What are (a) the index of refraction of the glass and (b) the Brewster angle?

180°= 𝜃𝑝 + 90° + 𝜃𝑟 180° = 𝜃𝑝 90° + 35° 𝜃𝑝 = 180° + 90 + 35 𝜽𝒑 = 𝟓𝟓°

el ángulo de Brewster

tan 𝜃𝑝 =

𝑛2 𝑛1

𝑛1 = 𝑣𝑎𝑐𝑖𝑜

𝑛1 = 1

𝑛2 = 𝑣𝑖𝑑𝑟𝑖𝑜

𝑛2 =?

𝑛2 = 𝑛1 tan 𝜃𝑝 𝑛2 = 1 tan(55°) 𝒏𝟐 = 𝟏. 𝟒𝟐m el índice de refracción del vidrio

2. In the following figure 𝒏𝟏 = 𝟏. 𝟑, 𝒏𝟏 = 𝟏. 𝟏, and 𝒏𝟑 = 𝟏. 𝟖, light refracts from material 1 into material 2. If it is incident at point A at the critical angle for the interface between materials 2 and 3, what are (a) the angle of refraction at point B and (b) the initial angle 𝜽? If, instead, light is incident at B at the critical angle for the interface between materials 2 and 3, what are (c) the angle of refraction at point A and (d) the initial angle 𝜽

a) 𝑛1 = 1.8, 𝑛2 = 1.3, and 𝑛3 = 1.1 1.1

𝑠𝑖𝑛𝜃2 = 1.3 𝜃2 = 57.79° 𝜆

Angle of incidence = 2 − 57.79° = 32.21° So the angle of refraction is 57.79° b) 𝑛1 𝑠𝑖𝑛𝜃 = 𝑛2 𝑠𝑖𝑛𝜃2 𝑠𝑖𝑛𝜃 = 1.3𝑠𝑖𝑛37.79° 𝑠𝑖𝑛𝜃 =

1.3𝑠𝑖𝑛37.79° 1.8

𝑎𝑟𝑐𝑠𝑖𝑛 = (

1.3𝑠𝑖𝑛37.79° 1.8

)

𝜃 = 37.66° c) For critical angle at B 𝜆

𝑛2 𝑠𝑖𝑛𝜃 = 𝑛3 sin(2) 1.1

𝑠𝑖𝑛𝜃2 = 1.3 𝜃2 = 57.79° 𝜆

Angle of incidence at A=2 − 57.79° = 32.21°

Angle of refraction at A= 1.3𝑠𝑖𝑛32.21° = 1.1𝑠𝑖𝑛𝜃 𝜃=

1.3sin(32.21°) 1.1

=

𝜃 = arcsin(0.62994) 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑎 𝑖𝑠: 𝜃 = 39.04° d) Initial angle 𝜃 𝑛1 𝑠𝑖𝑛𝜃 = 𝑛2 𝑠𝑖𝑛𝜃2 1.8𝑠𝑖𝑛𝜃 = 1.3sin(57.79°) 𝜃 = arcsin 1.3sin(57.79°) 𝜃 = arcsin(0.61107) 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙: 𝜃 = 37.66°

3. Choose one of the following problems, taken from “Cutnell, J. D., & Johnson, K. W. (2009), John Wiley & Sons Ltd., New York & Paz, A. (2013)” solve it and share the solution in the forum. Perform a critical analysis on the group members’ contributions and reply this in the forum. . A laser that is inside a tank of water, shoots a ray of light upwards. If the laser is located h1=12cm under water. At what minimum horizontal distance d1 from the laser pointer should you shoot so that the beam does not leave the water?

Solution 𝒊𝒏𝒅𝒊𝒄𝒆 𝒅𝒆 𝒓𝒆𝒇𝒓𝒂𝒄𝒄𝒊𝒐𝒏 𝒅𝒆𝒍 𝒂𝒈𝒖𝒂 = 𝒏 = 𝟏. 𝟓

𝒊𝒏𝒅𝒊𝒄𝒆 𝒅𝒆 𝒓𝒆𝒇𝒓𝒂𝒄𝒄𝒊𝒐𝒏 𝒅𝒆𝒍 𝒂𝒊𝒓𝒆 = 𝒏 = 𝟏 The critical angle for the water is calculated, taking as a reference the total reflection to generate in this way that the light emitted by the laser does not come out of the water

𝑠𝑒𝑛(𝜃𝑐 ) =

𝑛2 1 = = 0.666 𝑛1 1.5

𝜃𝑐 = 𝑠𝑒𝑛−1 (0.666) = 41.8° 𝑎𝑛𝑔𝑢𝑙𝑜 𝑐𝑟𝑖𝑡𝑖𝑐𝑜

With this Angle you get a trigonometric relationship between depth and distance

𝑑1 = 𝑡𝑎𝑛𝜃1 ℎ1

Despejando

𝑑1 = ℎ1 𝑡𝑎𝑛𝜃1 𝑑1 = 12𝑐𝑚 ∗ tan(41.8° ) 𝑑1 = 10.72𝑐𝑚

This is the minimum distance from the horizontal that is needed so that the beam emitted by the laser does not come out of the water

4. In the following figure a 2 m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. Sunlight is incident at angle 𝜃 = 60𝑜. What is the length of the shadow of the pole on the level bottom of the pool? 2.5 m-long 30 cm 𝜃 = 50𝑜

𝛳1 𝛳

AIR

0.3 m

WATER 𝛳2

POLE 2.2 m

𝛳1

𝛳2

SHADOW

Solution

𝛳1 = 90 − 𝛳 Where 𝛳 = 50° 𝛳1 = 90 − 50 𝛳1 =40° Having the pole height out of the water = 0.3 m and the angle with respect to the normal 𝛳1 = 40°, we can find 𝑥2

𝑇𝑎𝑔40° =

𝑋2 0.3

Tang 40° ∗ (0.3) = 𝑋2

𝑋2 = (0.84) ∗ 0.3 𝑋2 = 0.252𝑚

Now by Snell's law and knowing the refractive indices of air and water

𝛳1 air = 1 𝛳2 water = 1.33

1 = 𝑠𝑖𝑛40° 1.33 𝑠𝑖𝑛𝜃2 = (0.75)(0.64)

= 𝑠𝑖𝑛𝜃2

𝑠𝑖𝑛𝜃2 = 0.48 𝜃2 = 28.8°

𝑠𝑖𝑛50° =

0.3𝑚 ℎ𝑖𝑝

0.3𝑚 = 0.39 𝑚 𝑠𝑖𝑛50° 𝑎𝑑𝑦 𝑐𝑜𝑠50° = 0.39 𝑎𝑑𝑦 = 0.39 ∗ 𝑐𝑜𝑠50° = 0.25 𝑚 ℎ𝑖𝑝 =

𝑜𝑝𝑢 2.2𝑚 𝑜𝑝𝑢 = 2.2 ∗ 𝑇𝑎𝑛(28.8°) = 1.21 𝑚 𝑇𝑎𝑛(28.8°) =

Length of the shadow of the pole on the level bottom of the pool 1.21m

CONCLUSIONS

* Know and apply Snell's law the critical angle in the field of transmission. * Solve exercises related to refractive angles. * Apply the knowledge studied in real exercises that can be presented in the professional field.

REFERENCIAS

guias unad pag 249. (s.f.). Obtenido de www.glossary.oilfield.slb.com (s.f.). guias unad pag, 297. Obtenido de http://espectroelectromagntico.blogspot.com/2011/09/aplicaciones-del-espectro.html (s.f.). guias unad pag. 286. Obtenido de lasondas.webnode.es/ondas-planas/ nanocienciainforma.wordpress.com. (s.f.). Obtenido de nanocienciainforma.wordpress.com: https://nanocienciainforma.wordpress.com/microscopio-angulo-de-brewster-bam/ (s.f.). unad guias pag. 250. (s.f.). unad guias pag. 284. Obtenido de http://espectroelectromagntico.blogspot.com/2011/09/aplicaciones-del-espectro.html wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/Reflexi%C3%B3n_(f%C3%ADsica) wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/Refracci%C3%B3n wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/Ley_de_Snell wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/%C3%93ptica_geom%C3%A9trica wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/%C3%8Dndice_de_refracci%C3%B3n wikipedia. (s.f.). Obtenido de wikipedia: https://es.wikipedia.org/wiki/Refracci%C3%B3n www.euston96.com. (s.f.). Obtenido de https://www.euston96.com/ley-de-snell/