COLLABORATIVE WORK
Step 3 – 3 – Understand Understand the wave’s behavior in open and enclosure médiums
Estudents Jonh Alejandro Escobar Cód.: 1.115.079.532 Leidy Carolina López Delgado Cód.: 1096237125 Yeisson Alexander Enríquez Cód.: 12747933 José Daniel Ochoa Cód.: 1096211141 Franklin Silva Cód.: 1096223805 Grupo del curso
203058_48 Presentado a Omar Leonardo Leyton
UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA ELECTROMAGNETIC THEORY AND WAVES 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 v ery well in conductors with lost, for example, seawater.
There are several propagation methods that are m ost 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 ra ys. 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:
1. Explain how refraction and reflection phenomenon could be used in transmission systems. The Refraction, works as a transmission method, for example in c ommunications, when it is possible to transmit telephonics, television, etc., at great speed and distance, without the need for electronic signals, the principle of optical fiber which is composed of a set of filaments, which consist of a central plastic or glass core that has a high refractive index, surrounded by a layer of similar material that has a lower refractive index, hence when the light reaches the surface It is limited by a lower re fractive index, which is why it is largely reflected. Inside the light cable they are reflected against the walls in the angles that are very open, which makes the light signals are transported without losses by long distances. In the transmission of signals when an adaptation is not generated, transmission and reflection take place. When a signal transmission occurs in a transmission medium, the signal that we transmit from the source (emitter) to the load. When the impedances of the system are adapted ZS = Z0 = ZL, the whole incident wave is transmitted to the load, but when there is no adaptation, a part of the incident wave is transmitted and another part is reflected and returns in the opposite direction to this , taking the name of reflected wave. The sum of the waves twice a year, the transmission of radio waves, the standing waves in the transmission lines, are extremely dangerous for the physical integrity of the components.
2. Explain the Snell law, the critical angle and practical applications where it could be used.
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. The same states that the multiplication of the refractive index by the sine of the angle of incidence with respect to the normal is constant for any ray of light striking the separating surface of two media. That is, the component of the refractive index parallel to the surface is con stant.
Although Snell's law was formulated to explain the phenomena of refraction of light it can be applied to all types of waves crossing a separation surface between two media in which the velocity of propagation of the wave varies.
1sin 1 = 2sin 2 The light rays passing through the two media are refracted, that is, they change their propagation direction depending on the quotient between the refractive indices n1 and n2. The angles θ are the angles that are formed with the normal line, with θ1 the angle of the incident wave and θ2 the angle of the refracted wave *Critical angle
A critical angle or angle limit is defined as the angle from which there is no refraction and all the incident light is reflected to the same medium from which it comes. Only total reflection can occur if the index of the medium in which we find ourselves is higher than the index of the medium to which we are going
One of the applications of this phenomenon in optics
3.
Explain the physics behind lenses for optical diseases.
A utensil of daily life for many people but which, however, most do not know exactly how they work, many say that they "extend" or "close" objects, what happens is that they increase the angle of vision and we appreciate the size or distance of an object is by the angle of vision, that is why very large things look small at great distances. a normal person, the rays penetrate the eye and form their focus on the retina in this way:
Being d, the focal distance, theta the angle of vision, and next to it is how the retina acts to concentrate the rays, in many cases the lens of the eye does not make a good focus and it is when people distort the images. To correct this, we sought to change the angle in which light enters the lens of the eye as a small help, this is where the physics of refraction of light is applied a little. Refraction is the deviation of the path of light waves when they pass from one medium to another with different optical density (), that is, when a ray of light passes through an object of a certain optical density, it changes its trajectory when it leaves. Observe in the following image:
4. What is the importance of the index of refraction?
The use of the refractive index is important in various production processes, it has become increasingly necessary due to the requirements of current quality regulations, which include the entire production chain from the cultivation of raw materials, its reception and the elaboration of final products in the industries of the chemical, agroalimentary and pharmaceutical sector, among others. The determination of the refractive index is used, for example, to know the composition or purity of a sample, through an instrument called a refractometer. The operating principle of a refractometer is based on the speed of light that depends on the medium in which it travels, if a ray of light crosses slanted from one medium to another of different density, it changes its direction when it crosses the surface.
5. Is it possible that a single light beam be refracted into several beans? Explain.: When the light passes from one medium to another, a change in its direction occurs due to the different velocity of propagation that light has in different material media. When we divide the speed of light in the vacuum between one in a transparent medium, we obtain a value, it is the refractive index of that medium
6. How is it possible to use the Brewster angle in practical applications? Light has a property that human eyes do not see: polarization. Light is a vibration and polarization is the direction in which vibration occurs. Normally there is no direction of privileged vibration so the light chooses to vibrate in all: depolarized light or "natural" light is said. The thing changes when the light is reflected. When reflected, the light can acquire a predominant direction of vibration dep ending on the angle with which it falls on the reflecting surface. It is polarized at all if the angle of incidence is the so-called Brewster angle. Therefore, if we have a source of light and see it through reflection, there will be an area in which the light will be polarized, that in which the light has incised at Brewster
angle, while the rest is more or less depolarized. The polarization can be detected by looking through a polarizer, which is a sheet that allows more or less light to pass depending on the polarization that it has and that is what the polarized glasses are made.
Choose one of the following problems, taken from “Cutnell, J. D., & Johnson, K. W. (2009). Physics. John Wiley & Sons Ltd., New York. doi, 10, S0142-9612”
JOHN:
43
1. When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is what are (a) the index of refraction of the glass and (b) the Brewster angle?
The Angle that forms the X axis with the ray refracted is the same that forms the normal with the reflected Angle that would be equal to the incident ray and in this case it would be the Brewster Angle (b) Therefore the Brewster Angle is the same 47° We use Snell's law to find the index of refraction
1sin=1 = sin47
1sin 1 = 2sin 2
Index of refraction in the vacuum = 1
2sin=2 = sin43 2 sisinn4743 = 1.072
Index of refraction in the glass =?
We cleared
Index of refraction in the glass =1.072
= 1.7 = 1.5 = 1.3
2. JOSE OCHOA: In the following figure , , and , 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 ? 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 ? If, instead of all that, light is incident at point A at Brewster’s angle for the interface between materials 2 and 3, what are (e) the angle of refraction at point B and (f) the initial angle ?
3. LEIDY LOPEZ: In about A.D. 150, Claudius Ptolemy gave the following measured values for the angle of incidence and the angle of refraction for a light beam passing from air to water:
Assuming these data are consistent with the law of refraction, use them to find the index of refraction of water. These data are interesting as perhaps the oldest recorded physical measurements.
4. YEISSON: 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 . What is the length of the shadow of the pole on the level bottom of the pool?
= 60
0.5 mts
2 mts
60
Distancia de la sombra reflejada
Solution To meet the length of the shadow of the pole part pole distance and angles of incidence and refraction. The incidence angle is 60 ° Obtain the angle of refraction
= =1.=51 = 1.15 (60°) = 0.577 = −(0.577) = 35°
The index of refraction of water is of
The index of refraction of the aires is of Thus
The angle of refraction is
Now we apply basic trigonometry in a triangle
= 35° D
We calculate the angle unknown from other angles 1
= 180° = 180° 60° 90° = 30° = = 1 = . = 0.5(∗(60°3)0° )= 0.29
We use the sine theorem
We calculate the angle unknown from other angles 2
= 180° = 180° 35° 90° = 55° = = 2 = . = 1.5(∗(55°3)5° )= 1.05 = 1+ 2 = 0.29 + 1.05 = 1.34
We use the sine theorem
Therefore, the length of the pole will be
FRANKLIN
= 1.8 = 1. 3
= 1.6
5. In the following figure, light from ray A refracts from material 1 ( ) into a thin layer of material 2 ( ), crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3 ( ). (a) What is the value of incident angle ? (b) If is decreased, does part of the light refract into material 3? Light from ray B refracts from material 1 into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3. (c) What is the value of incident angle ? (d) If is decreased, does part of the light refract into material 3?
== 4.2.85 = 80 4.8 1 10− 8 = 4.8 2(2 ∗5 ∗10) (364 ∗ ) (8) = 0.432 − 10 √ µ£ = 1.80∗002∗49∗ 1364 ∗ 10−
= 42.4.81∗42.910115121 = 2 = 101.258 (=) =8.368() = 878.92 = 3 3 = 3.41∗10−
Graphics
Zircon 30
Incidence Angle
Precious stones Index of refraction Refraction Angle Title Zircon
1,8
Incidence A Refraction Angle
16,12762021 Zircon [Incid θi=30°
θr= 16,13°
Zircon [Incidence Angle = 30°, Refraction Angle = 16,13°] Zircon -2,5
4,330127019
0
0
1,388888889
-4,803226796 θ i=30°
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.
References
Electromagnetic Wave Propagation. (2003). Fixed Broadband Wireless. 25-70. Retrieved from http://bibliotecavirtual.unad.edu.co:2048/login?url=http://search.ebscohost.com/login. aspx?direct=true&db=aci&AN=14505422&lang=es&site=ehost-live
Chen, W. (2005). The Electrical Engineering Handbook. Boston: Academic Press, 519-524. Retrieved fromhttp://bibliotecavirtual.unad.edu.co:2048/login?url=http://search.ebscohost.com/login. aspx?direct=true&db=nlebk&AN=117152&lang=es&site=ehostlive&ebv=EB&ppid=pp_519 The following video explains how to use the Snell Law to analyze phenomenon associated with transmission systems when there is a medium change, like refraction and reflection. Gutiérrez (2017). Snell Law [Video]. Retrieved fromhttp://hdl.handle.net/10596/13140
