TEORIA ELECTROMAGNETICA Y ONDAS 203058A_364
STEP 3 – 3 – UNDERSTAND UNDERSTAND THE WAVES BEHAVIOR IN OPEN AND ENCLOSURE MEDIUMS
TUTOR:
WILMER HERNAN GUTIERREZ
PRESENTED BY:
KATHERIN LIZETH VALEGA 1.045.701.215 LEONARDO QUINTERO BEJARANO 16461422 HERNAN PORTILLO TRILLOS 13.436.865 LUIS ENRIQUE PEREZ RAVELO 1.007.126.098 ORLANDO ANDRES RUIZ
GROUP: 243006_38
NOVEMBER, 2017
INTRODUCTION
In this group work, the aim is to understand and assimilate the principles that explain electromagnetic waves, to explore the theoretical applications related to el ectromagnetic waves and to investigate and develop related complementary exercises. With the development of the following exercises we can strengthen our knowledge in the area of electromagnetic waves, in addition to the use of snell's law, we will apply other very important topics given in unit 3 for the calculation of electromagnetic systems, for that reason then to define some terms such as the laws of reflection and refraction, critical angle, transmission of ionospheric waves, among others, we will proceed t o the realization of the exercises.
ACTIVITIES TO DEVELOP
Each student in the group has to answer the following questions using academic references to support the research:
1. What is the difference between specular reflection and diffuse reflection? R// The difference is that specular reflection occurs when the waves that fall on a very flat reflecting surface are reflected so that the incident an gle is equal to the reflected angle while diffuse reflection occurs when parallel rays falling on a rough surface reflect the waves with scattered angles, so that you can not see an image on the surface. This phenomenon occurs because the macro or micro-roughness deflects the waves at different angles. In an y case, in this case it is also true that the individual incident rays are reflected with identical angles to the incident.
2. What is the critical angle? Using a graphic show what happened when the critical angle is reached. R// When a wave passes from a medium with a lower index to a medium with a higher index of refraction, the refracted ray app roaches normal. For example, light passes from air to glass.If the step is from a medium of higher index to one of lower index, the refracted ray moves away from the normal one. This happens when the light passes from the glass to the air. In this case, t here is a limit value for the angle of incidence, above which the ray does not leave that medium because the angle would pass 90 ° The critical angle or limit angle is also the minimum angle of incidence at which the total internal reflection occurs. The angle of incidence is measu red with respect to the normal separation of the media. The critical angle is given by:
= 2 1
Where n1 and n2 are the refractive indices of the media with n2
3. Explain the following terms used when we are analyzing ionospheric wave propagation: Skip distance, maximum usable frequency and critical frequency. R// The electromagnetic waves that are transmitted through the horizon are called celestial waves, normally the celestial waves are transmitted in a direction which forms a relatively large angle with the earth, the y radiate towards the sky where they are reflected or refracted towards the earth b y the ionosphere, for this reason the propagation of celestial waves is also called ionosphere propagation -The critical frequency is defined as the maximum frequency that can be transmitted upwards and is reflected by the ionosphere towards the earth, this depends on the ionization density and consequently varies with the time of day or with the station. -The maximum useful frequency (MUF) is the highest frequency that c an be used to transmit celestial waves between two specific points on the surface of the earth, therefore there can be an infinite number of these, due to the instability of the ionosphere the maximum frequency usable is usually used less than the MUF, meaning that the MUF is only used as a reference. -The jump distance is the minimum distance of transmission to which a celestial wave of a certain frequency (which must be less than the MUF) will return to the earth.
4. What is the Brewster's angle? Give a practical application where it can be used.
R// In optical physics, the angle of Brewster (named after the Scottish physicist Sir David Brewster) corresponds to the angle of incidence of light on a surface that cancels the component with polarization parallel to the plane of incidence. The result when a non-polarized light beam is applied on a surface under the Brewster angle is the obtaining of a reflected beam of polarized light in one direction (whose polarization vector is perpendicular to the plane of incidence).
A practical application would be serious in the field of optics with respect to the manufacture of lenses to polarize light as in the manufacture of glasses
5. What relation is defined by the Snell law and how is the Snell law used in devices like refractometers? R// 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. It is called refractometry, the optical method that determines the speed of propagation of light in a medium, compound, substance, which is directly related to the density of this. The refraction of light is used to employ this principle, and it is here that Snell's law mentioned above used to measure the refractive index comes into play. Refractometers are the instruments that use this principle of refraction, either refraction (using several prisms), or the critical angle (using only one prism), and its primary scale of measurement is the refractive index , from from which the different specific scales are constructed, Brix (sugar), Specific Density,% salt, etc.
Refractometers are used to measure liquids, solids and gases, such as glasses or gems. Snell´s law
sin = Refractometer
6. Using a graphical representation, explain the difference between the polarization modes TE, TM y TEM. R// The modes of propagation depend on the wavelength, the polarization and the dimensions of the guide. The longitudinal mode of a waveguide is a particular type of standing wave formed by waves confined in the cavity. The transversal modes are classified in different types: -Mode TE (Electric Transversal), the component of the elect ric field in the direction of propagation is zero. -Mode TM (Magnetic Transversal), the component of the magnetic field in the direction of propagation is zero. -TEM (Electromagnetic Transversal) mode, the component of bo th the electric and the magnetic field in the direction of propagation is zero.
Choose one of the following problems, solve it and share the solution in the forum. Perform a critical analysis on the group members’ co ntributions and reply this in the forum.
1. There are two airplanes flying over the sea, separated by a horizontal distance of 1500m, one of the airplanes located at an altitude of 250m send a signal using a 600MHz channel to communicate with the other airplane. What would be the suitable altitude for the other airplane to avoid the noise produced by the refection effect with the sea surface? Make a diagram to support your answer. Solution: From the above information we can make the following diagram:
Signal using a 600MHz A2 A1
a2= X m
a1= 250m b2
b1
Dt = 1.500m
Data obtained: A1 = Airplane 1 A2 = Airplane 2 a1 = height of the plane A1 = 250m a2 = height of the plane A2 = X b1 = horizontal distance between the plane A1 and the normal. b2 = horizontal distance between the plane A2 and the normal. Dt = horizontal distance between the plane A1 and the plane A2. The problem asks us to find b 2 where Airplane 2 does not experience interference when communicating with Aircraft 1. For this we must initially find the angle of incidence β by initially taking the value of b1 = 750m:
=arctan() ) =arctan(250 750 =arctan0.33 =18.43 Knowing that =18.43 we could deduce that the reflected angle is the same and that the height of a2 should be equal to a1 to have a communication without interference, However the problem asks us to check different heights to observe the behavior of the waves, so we will vary b1 taking into account that the maximum distance between the aircraft Dt will be 1500m and see where there is interference or noise:
1.
=
=arctan() ) =arctan(250 600 =22.61
tan ′ = tan 22.61 = 900 tan 22.61 ∗900= =
Noise
a2 375m 250m 22.61⁰
β’
β
22.61⁰
600m
900m
= =arctan() ) =arctan(250 800 =17.35 2.
tan ′ = tan 17.35 = 700 tan 17.35 ∗700= .= Noise
250m a2 218.7m 17.35⁰ 800m
β
β’
17.35⁰ 700m
3.
=
=arctan() ) =arctan(250 900 =15.52
tan ′ = tan 15.52 = 600 tan 15.52 ∗600= .= Noise
250m a2 166.6m 15.52⁰
β
β’
900m
= =arctan() 250 ) =arctan(1000 =13.49
15.52⁰ 600m
4.
tan ′ = tan 13.49 = 500 tan 13.49 ∗500= .= Nois
250 a2 119.9m 13.49⁰ β 1000
β’ 13.49⁰ 500m
According to the results we could conclude that plane 2 must have an altitude approximately to avoid noise between its communication.
The exercise can be summarized with the following graph:
>
2. A uniform plane wave perpendicularly polarized from an antenna, incident onto a freshwater lake at the Brewster angle
= . Draw a graphic that shows the
scenario and calculate the reflection and transmission coefficients.
Unpolarized
Completely plane
light
polarized
AIR
WATER
tan = tan = 1.0002926 1.3330
90° Refracted light
tan =0.750 =0.932° 2182.0618815692248114 Γ = + = 1202559.053 = 120+2559.053 2936.0441184307751886 = 0.74319791990573678 =1+Γ=0.25680208009426322 3. In a transmission system, a wave incident at
= 36° on the ground. Explain if the
wave is reflected or refracted supporting your answer with the mathematical analysis. Solution:
Snell´s law
sin = ==1 == = c = speed of light v = speed of light in the medium (ground), in this case the speed of light on the ground will be ten times less than in the vacuum
= = 300.000/ 30.000/ = 10 Parameters:
= 1 = 36° = 10 = =arcsin .sin
=arcsin101 .sin36° =arcsin0,10,99) =arcsin0,099) =5,68 = 5°40´ The answer is that due to the difference in the refractive indices the light is refracted
4. A monitoring system is using a plane wave incide nt on a lake covered by ice to communicate with a device located 3m under the lake, if the horizontal distance is 15m. What would be the suitable incident angle to guarantee total reflection? Is it necessary to change the given distances to guarantee the total reflection condition? Solution:
top layer - ice 0⁰ n2 n1 3m
15m
bottom - liquid water 4⁰ Frozen wáter Medium 1 n1. n1 has a refractive index of 1,31 Air Medium 2 n2. N2 has a refractive index of 1,00
θc critical angle or angle of incidence. In this case we look for a total internal reflection or limit angle, and we get it through the following Snell formula and law:
=arcsin =arcsin 1,00 1,31 =49,7° ⁰
In this exercise we need a critical angle of 49.7 to obtain a total reflection,
Known data:
β 3m
49,7⁰ x
α
y Now we have to calculate the correct distance between the two points so that the angle of incidence causes such reflection, this we can do based on one of the principles of the right triangles:
The sum of its three angles is 180 ⁰ , and if one is 90⁰ the sum of the two missing ones must also be 90⁰ .
Then we can find α by the following formula:
=90⁰49,7⁰ =40,3⁰ Now we have to find x and we do it by the following formula: =∗ =3∗40,3⁰ =3∗1,18 =3,54 Knowing we can finally calculate the total distance between the two points: 2 =3,54 =2 =2∗3,54 =7.08 We can conclude that for a total reflection to take place we must have an angle of incidence of and a length of between the two points. For the proposed exercise it is necessary to change the distances given.
=49,7°
=7.08
top layer - ice 0⁰
n2 n1 3m
49,7⁰
7.08m
bottom - liquid water 4⁰
5. A security system uses a beam that has to impact in a receptor like is shown in the
following image to guarantee that there isn’t interference in an entrance:
According to the image, what would be the suitable incidence angle to guarantee the communication between the transmitter and the receiver? Explain your answer.
Medium 1= Glass Medium 2= Air
refractive index = 1.5
Refractive index = 1
= Incident angle
To obtain de required incident angle, we aplly the Snell Law, where it is necessary that
> ∗ = ∗ ∗ = 90° 1.5 ∗ =1∗90° = ∗.° = ∗ . = −1.51 = 41.8°
In group solve the following practical exercise
a. In an Excel Workbook design a Sheet where the user will be able to select two different mediums from a group of 6 mediums, where a wave travels from the first medium to the second and shows the incidence angle to guarantee total reflection and the incidence angle to guarantee total transmission.
b. In the same Sheet, the user could give and horizontal distance from the incidence point and automatically it has to appear the vertical length where the source has to be set.
Select the medium
Index of refraction (n)
Orientation
meters
n2
Fused quartz
1,46
Horizontal distance
6
n1
Heaviest flint glass
1,89
Vertical distance source
4,93
Critical Angle
θ c= arcsin n2/n1
Critical Angle 50,6 50,6
°
4
n2
2
Brewster Angle θ b= arctan n2/n1
37,7
0 -6
-4
-2
0 -2
x
y
6
-4,932362939
0
0
-6
-4,932362939
Título -4
Critical Angle 50,6°
n1 -6
2
4
6
Material
n
Air (STP)
1,000
Water (20⁰C)
1,330
Fused quartz
1,460
Sodium chloride
1,540
Heavy flint glass
1,650
Heaviest flint glass
1,890
Diamond
2,420
Note: the exercise in excel is in the collaborative forum
CONCLUSIONS
Thanks to the realization of this work we were able to consolidate even more our knowledge in the area of telecommunications, especially in the theory of electromagnetic waves, during the resolution of the exercises the u se of the laws of reflection and refraction was taken into account the snell law, for this reason we can say that we meet the objectives of unit 2 satisfactorily acquiring knowledge which will be ver y helpful especially in future courses and even in daily life when we have problems related to these phenomena.
BIBLIOGRAPHIC 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.as px?direct=true&db=aci&AN=14505422&lang=es&site=ehost-live
Woodwell, G. R. (2016). Reflection And Refraction. Salem Press Encyclopedia Of Science, Retrieved from http://bibliotecavirtual.unad.edu.co:2051/login.aspx?direct=true&db=ers&AN=893 17193&lang=es&site=eds-live
Quesada-Pe rez, M., & Maroto-Centeno, J. A. (2014). From Maxwell's Equations to
́
Free and Guided Electromagnetic Waves: An Introduction for First-year Undergraduates. New York: Nova Science Pu blishers, Inc, 49-80 Retrieved from http://bibliotecavirtual.unad.edu.co:2051/login.aspx?direct=true&db=nlebk&AN=7 46851&lang=es&site=eds-live&ebv=EB&ppid=pp_49
Magnetic field. From Wikipedia, the free encyclopedia https://en.wikipedia.org/wiki/Magnetic_field
Resolution of rectangle triangles. http://recursostic.educacion.es/descartes/web/materiales_didacticos/resolver_tri_rect angulos_pjge/Triangulos_rectangulos1.htm
Permeability (electromagnetism). From Wikipedia, the free encyclopedia https://en.wikipedia.org/wiki/Permeability_(electromagnetism)
APA 2017 : http://normasapa.net/2017-edicion-6/
