Competitive Electromagnetic Theory MCQs – EM Wave Propagation MCQs ( Electromagnetic Theory ) MCQs
Latest Electromagnetic Theory MCQs
By practicing these MCQs of EM Wave Propagation MCQs ( Electromagnetic Theory ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to “EM Wave Propagation MCQs ( Electromagnetic Theory ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Electromagnetic Theory to approach the success. Tab this page to check “EM Wave Propagation MCQs ( Electromagnetic Theory )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.
Electromagnetic Theory MCQs – EM Wave Propagation MCQs ( Electromagnetic Theory ) MCQs
The most occurred mcqs of EM Wave Propagation MCQs ( Electromagnetic Theory ) in past papers. Past papers of EM Wave Propagation MCQs ( Electromagnetic Theory ) Mcqs. Past papers of EM Wave Propagation MCQs ( Electromagnetic Theory ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related EM Wave Propagation MCQs ( Electromagnetic Theory ) Mcqs. The Important series of EM Wave Propagation MCQs ( Electromagnetic Theory ) Mcqs are given below:
Lossy and Lossless Dielectrics
1. For a dielectric, the condition to be satisfied is
a) σ/ωε > 1
b) σ/ωε < 1
c) σ = ωε
d) ωε = 1
Answer: b
Explanation: In a dielectric, the conductivity will be very less. Thus the loss tangent will be less than unity. This implies σ/ωε < 1 is true.
2. For a perfect dielectric, which parameter will be zero?
a) Conductivity
b) Frequency
c) Permittivity
d) Permeability
Answer: a
Explanation: The conductivity will be minimum for a dielectric. For a perfect dielectric, the conductivity will be zero.
3. Calculate the phase constant of a wave with frequency 12 rad/s and velocity 3×108 m/s(in 10-8 order)
a) 0.5
b) 72
c) 4
d) 36
Answer: c
Explanation: The phase constant is given by β = ω√(με), where ω is the frequency in rad/s and 1/√(με) is the velocity of wave. On substituting √(με) = 3×108 and ω = 12, we get β = 12/(3×108) = 4 x 10-8m/s.
4. For a lossless dielectric, the attenuation will be
a) 1
b) 0
c) -1
d) Infinity
Answer: b
Explanation: The attenuation is the loss of power of the wave during its propagation. In a lossless dielectric, the loss of power will not occur. Thus the attenuation will be zero.
5. Calculate the velocity of a wave with frequency 2 x109 rad/s and phase constant of 4 x 108 units.
a) 0.5
b) 5
c) 0.2
d) 2
Answer: b
Explanation: The velocity of a wave is the ratio of the frequency to the phase constant. Thus V = ω/β. On substituting the given values, we get V = 2 x109/ 4 x 108 = 5 units.
6. Which of the following is the correct relation between wavelength and the phase constant of a wave?
a) Phase constant = 2π/wavelength
b) Phase constant = 2π x wavelength
c) Phase constant = 1/(2π x wavelength)
d) Phase constant = wavelength/2π
Answer: a
Explanation: The phase constant is the ratio of 2π to the wavelength λ. Thus β = 2π/λ is the correct relation.
7. In lossy dielectric, the phase difference between the electric field E and the magnetic field H is
a) 90
b) 60
c) 45
d) 0
Answer: d
Explanation: In a lossy dielectric, the E and H component will be in phase. This implies that the phase difference between E and H will be 0.
8. The intrinsic impedance is the ratio of square root of
a) Permittivity to permeability
b) Permeability to permittivity
c) Phase constant to wavelength
d) Wavelength to phase constant
Answer: b
Explanation: The intrinsic impedance is the impedance of a particular material. It is the ratio of square root of the permeability to permittivity. For air, the intrinsic impedance is 377 ohm or 120π.
9. Calculate the skin depth of a material with attenuation constant of 2 units.
a) 2
b) 1
c) 0.5
d) 4
Answer: c
Explanation: The skin depth of a material is the reciprocal of the attenuation constant. Thus δ = 1/α. On substituting for α = 2, we get δ = ½ = 0.5 units.
10. Calculate the phase constant of a wave with skin depth of 2.5 units.
a) 5/2
b) 5
c) 2
d) 2/5
Answer: d
Explanation: The skin depth is the reciprocal of the phase constant and the attenuation constant too. Thus δ = 1/β. On substituting for δ = 2.5, we get β = 1/δ = 1/2.5 = 2/5 units.
11. An example for lossless propagation is
a) Dielectric waveguide propagation
b) Conductor propagation
c) Cavity resonator propagation
d) It is not possible
Answer: d
Explanation: There are many techniques employed to achieve zero attenuation or maximum propagation. But it is not achievable practically. Thus lossless propagation is not possible practically.
12. Skin depth phenomenon is found in which materials?
a) Insulators
b) Dielectrics
c) Conductors
d) Semiconductors
Answer: c
Explanation: Skin depth is found in pure conductors. It the property of the conductor to allow a small amount of electromagnetic energy into its skin, but not completely. This is the reason why EM waves cannot travel inside a good conductor.
Dielectric vs Conductor Wave Propagation
1. In conductors, which condition will be true?
a) σ/ωε > 1
b) σωε > 1
c) σ/ωε < 1
d) σωε < 1
Answer: a
Explanation: For conductors, the conductivity will be maximum. Thus the loss tangent is greater than unity. This is given by σ/ωε >1.
2. For metals, the conductivity will be
a) 0
b) 1
c) -1
d) Infinity
Answer: d
Explanation: Metals are pure conductors. Examples are iron, copper etc. Their conductivity will be very high. Thus the metal conductivity will be infinity. Practically the conductivity of conductors will be maximum.
3. In conductors, which two parameters are same?
a) Wavelength and phase constant
b) Phase and attenuation constant
c) Attenuation constant and skin depth
d) Skin depth and wavelength
Answer: b
Explanation: In conductors, which are considered to be lossy, the attenuation and the phase constant are the same. It is given by α=β= √(ωμσ/2).
4. Calculate the velocity of wave propagation in a conductor with frequency 5 x 108 rad/s and phase constant of 3 x 108 units.
a) 3/5
b) 15
c) 5/3
d) 8
Answer: c
Explanation: The velocity of wave propagation is the ratio of the frequency to the phase constant. It is given by V = ω/β. On substituting the given values, we get V = 5/3 units.
5. Calculate the wavelength of the wave with phase constant of 3.14 units.
a) 1
b) 2
c) 0.5
d) 4
Answer: b
Explanation: The wavelength is the ratio of 2π to the phase constant β. On substituting for β = 3.14, we get λ = 2π/β = 2π/3.14 = 2 units.
6. For dielectrics, which two components will be in phase?
a) E and wave direction
b) H and wave direction
c) Wave direction and E x H
d) E and H
Answer: d
Explanation: In dielectrics, the electric and magnetic components E and H will be in phase with each other. This is due the variation in the permittivities and the permeabilities of the dielectric surfaces. The phase difference between E and H will be 0.
7. In perfect conductors, the phase shift between the electric field and magnetic field will be
a) 0
b) 30
c) 45
d) 90
Answer: c
Explanation: For perfect conductors, the electric and magnetic field E and H respectively vary by a phase of 45 degree. This is due to the polarisation phenomenon in the conductors, unlike dielectrics.
8. The expression for phase constant is given by
a) Phase constant β = ωμε
b) Phase constant ω = με
c) Phase constant β = ω√(με)
d) Phase constant β = 1/ωμε
Answer: c
Explanation: The phase constant is represented as β. It is a complex quantity representing the constant angle of the wave propagated. It is given by β = ω√(με).
9. In waveguides, which of the following conditions will be true?
a) V > c
b) V < c
c) V = c
d) V >> c
Answer: a
Explanation: In waveguides, the phase velocity will always be greater than the speed of light. This enables the wave to propagate through the waveguide. Thus V > c is the required condition.
10. The attenuation constant in lossless dielectrics will be
a) 0
b) 1
c) -1
d) ∞
Answer: a
Explanation: In lossless dielectrics, the attenuation constant will not be same as the phase constant, unlike conductors. Also, due to the lossless behaviour, the attenuation will be nearly zero. Practically, zero attenuation is not possible.
Plane Waves in Free Space
1. In free space, the charge carriers will be
a) 0
b) 1
c) 100
d) Infinity
Answer: a
Explanation: Free space is not a conductor. Thus the charge carrier in free space is assumed to be zero. But the free space consists of particles or ions that get ionized during conduction.
2. In free space, which parameter will be unity?
a) Permittivity
b) Absolute permittivity
c) Relative permittivity
d) Permeability
Answer: c
Explanation: The relative permittivity is a constant for a particular material. It is unity for free space or air. The absolute permittivity is a constant given by 8.854 x 10-12 C/m2.
3. Which parameter is unity in air medium?
a) Permittivity
b) Absolute permeability
c) Relative permeability
d) Permeability
Answer: c
Explanation: In free space or air medium, the relative permeability is also unity, like relative permittivity. The absolute permeability is given by 4π x 10-7 units.
4. The conductivity in free space medium is
a) Infinity
b) Unity
c) Zero
d) Negative
Answer: c
Explanation: As the charge carriers are not available in free space, the conductivity will be very low. For ideal cases, the conductivity can be taken as zero.
5. Zero permeability/permittivity implies which state?
a) No ions are allowed in the medium
b) No current is generated in the medium
c) No magnetic or electric energy is permitted in the medium
d) No resistivity
Answer: c
Explanation: The zero permittivity in an electric field refers to the ability of the field/medium to permit electric charges in it. Similarly, zero permeability in a magnetic field refers to the ability of the field/medium to permit the magnetic energy into the field.
6. The intrinsic impedance of free space is
a) 489
b) 265
c) 192
d) 377
Answer: d
Explanation: The intrinsic impedance is the square root of ratio of the permeability to the permittivity. In free space, the permeability and the permittivity is same as the absolute permeability and permittivity respectively. This is due to unity permeability and permittivity in free space. Thus η = √(μ/ε), where absolute permeability is given by 4π x 10-7 and absolute permittivity is given by 8.854 x 10-12. The intrinsic impedance will be 377 ohms.
7. In free space, the condition that holds good is
a) Minimum attenuation and propagation
b) Minimum attenuation and maximum propagation
c) Maximum attenuation and minimum propagation
d) Maximum attenuation and propagation
Answer: b
Explanation: The free space does not have any barrier for attenuation. Thus it enables minimum attenuation and maximum propagation. This technique is employed in line of sight communication.
8. In free space, the ratio of frequency to the velocity of light gives the phase constant. State True/False.
a) True
b) False
Answer: a
Explanation: The phase constant is given by the ratio of the frequency in radian/sec to the velocity of the wave propagating. In free space, the velocity is considered to be the velocity of light. Thus the statement is true.
9. The velocity of a wave travelling in the air medium without transmission lines or waveguides(wireless) is
a) 6 x 108
b) 3 x 108
c) 1.5 x 108
d) 9 x 108
Answer: b
Explanation: In free space or air medium, the velocity of the wave propagating will be same as that of the light. Thus the velocity is the speed of light, V = c. It is given by 3 x 108m/s.
10. The vectors of the electromagnetic wave propagation can be expressed in
a) Dot product
b) Cross product
c) Unit vector
d) Perpendicular vector
Answer: b
Explanation: In an EM wave, the electric and the magnetic fields will be perpendicular to each other and with the direction of the propagation. Thus it can be expressed in cross product where iE x iH = iw. Here iE is the electric vector component, iH is the magnetic vector component and iw is the vector of the wave propagating.
Plane Waves in Good Conductor
1. For conductors, the loss tangent will be
a) Zero
b) Unity
c) Maximum
d) Minimum
Answer: c
Explanation: In conductors, the conductivity will be more. Thus the loss tangent σ/ωε will be maximum.
2. In metals, the total permittivity is
a) Absolute permittivity
b) Relative permittivity
c) Product of absolute and relative permittivity
d) Unity
Answer: a
Explanation: The total permittivity is the product of the absolute and the relative permittivity. For metals or conductors, the relative permittivity is unity. Thus the permittivity is simply the absolute permittivity.
3. The total permeability in a conductor is
a) Absolute permeability
b) Relative permeability
c) Product of absolute and relative permeability
d) Unity
Answer: c
Explanation: The total permeability is the product of the absolute and the relative permeability. For metals or conductors, the relative permittivity is not unity. Thus the permittivity is the product of absolute and relative permeability.
4. Calculate the phase constant of a conductor with attenuation constant given by 0.04 units.
a) 0.02
b) 0.08
c) 0.0016
d) 0.04
Answer: d
Explanation: The phase constant and the attenuation constant are both the same in the case of conductors. Given that the attenuation constant is 0.04, implies that the phase constant is also 0.04.
5. Calculate the attenuation constant of a conductor of conductivity 200 units, frequency 1M radian/s in air.
a) 11.2
b) 1.12
c) 56.23
d) 5.62
Answer: a
Explanation: The attenuation constant of a conductor is given by α = √(ωμσ/2). On substituting ω = 106, σ = 200 and μ = 4π x 10-7, we get α = 11.2 units.
6. The skin depth of a conductor with attenuation constant of 7 neper/m is
a) 14
b) 49
c) 7
d) 1/7
Answer: d
Explanation: The skin depth is the measure of the depth upto which an EM wave can penetrate through the conductor surface. It is the reciprocal of the attenuation constant. On substituting for α = 7, we get δ = 1/α = 1/7 units.
7. The expression for velocity of a wave in the conductor is
a) V = √(2ω/μσ)
b) V = √(2ωμσ)
c) V = (2ω/μσ)
d) V = (2ωμσ)
Answer: a
Explanation: The velocity is the ratio of the frequency to the phase constant. In conductors, the phase constant is given by √(ωμσ/2). On substituting for β,ω in v, we get v = √(2ω/μσ) units.
8. In conductors, the E and H vary by a phase difference of
a) 0
b) 30
c) 45
d) 60
Answer: c
Explanation: The electric and magnetic component, E and H respectively have a phase difference of 45 degrees. This is due to the wave propagation in conductors in the air medium.
9. EM waves do not travel inside metals. State True/False.
a) True
b) False
Answer: a
Explanation: The conductors or metals do not support EM wave propagation onto them due the skin effect. This is the reason why mobile phones cannot be used inside lifts.
10. The propagation constant of the wave in a conductor with air as medium is
a) √(ωμσ)
b) ωμσ
c) √(ω/μσ)
d) ω/μσ
Answer: a
Explanation: The propagation constant is the sum of the attenuation constant and the phase constant. In conductors, the attenuation and phase constant both are same and it is given by √(ωμσ/2). Their sum will be √(ωμσ), is the propagation constant.
11. An example for electromagnetic wave propagation is
a) refrigerator
b) electric fan
c) mobile transponder
d) relays in actuators
Answer: c
Explanation: The refrigerator, electric fan and relays are electrical devices. They do not use electromagnetic energy as medium of energy transfer. The mobile transponder is an antenna, which uses the EM waves for communication with the satellites.
12. The phase shift in the electric and magnetic fields in an EM wave is given by which parameter?
a) phase constant
b) attenuation constant
c) propagation constant
d) intrinsic impedance
Answer: d
Explanation: The intrinsic impedance in a conductor is given by η = √(ωμ/2σ) x (1+j). The phase shift is represented by the 1+j term. In polar form it indicates 45 degree phase shift.
Plane Waves in Dielectrics
1. The loss tangent of a perfect dielectric will be
a) Zero
b) Unity
c) Maximum
d) Minimum
Answer: d
Explanation: Dielectrics have poor conductivity. The loss tangent σ/ωε will be low in dielectrics. For perfect dielectrics, the loss tangent will be minimum.
2. In pure dielectrics, the parameter that is zero is
a) Attenuation
b) Propagation
c) Conductivity
d) Resistivity
Answer: c
Explanation: There are no free charge carriers available in a dielectric. In other words, the charge carriers are present in the valence band, which is very difficult to start to conduct. Thus conduction is low in dielectrics. For pure dielectrics, the conductivity is assumed to be zero.
3. The total permittivity of a dielectric transformer oil (relative permittivity is 2.2) will be (in order 10-11)
a) 1.94
b) 19.4
c) 0.194
d) 194
Answer: a
Explanation: The total permittivity is the product of the absolute and the relative permittivity. The absolute permittivity is 8.854 x 10-12 and the relative permittivity(in this case for transformer oil) is 2.2. Thus the total permittivity is 8.854 x 10-12 x 2.2 = 1.94 x 10-11 units.
4. The permeability of a dielectric material in air medium will be
a) Absolute permeability
b) Relative permeability
c) Product of absolute and relative permeability
d) Unity
Answer: a
Explanation: The total permeability is the product of the absolute and the relative permeability. In air medium, the relative permeability will be unity. Thus the total permeability is equal to the absolute permeability given by 4π x 10-7 units.
5. The attenuation in a good dielectric will be non- zero. State True/False.
a) True
b) False
Answer: a
Explanation: Good dielectrics attenuate the electromagnetic waves than any other material. Thus the attenuation constant of the dielectric will be non-zero, positive and large.
6. Calculate the phase constant of a dielectric with frequency 6 x 106 in air.
a) 2
b) 0.2
c) 0.02
d) 0.002
Answer: c
Explanation: The phase constant of a dielectric is given by β = ω√(με). On substituting for ω = 6 x 106 , μ = 4π x 10-7, ε = 8.854 x 10-12 in air medium, we get the phase constant as 0.02 units.
7. The frequency in rad/sec of a wave with velocity of that of light and phase constant of 20 units is (in GHz)
a) 6
b) 60
c) 600
d) 0.6
Answer: a
Explanation: The velocity of a wave is given by V = ω/β. To get ω, put v = 3 x 108 and β = 20. Thus ω = vβ = 3 x 108 x 20 = 60 x 108 = 6 GHz.
8. The relation between the speed of light, permeability and permittivity is
a) C = 1/√(με)
b) C = με
c) C = μ/ε
d) C = 1/με
Answer: a
Explanation: The standard relation between speed of light, permeability and permittivity is given by c = 1/√(με). The value in air medium is 3 x 108 m/s.
9. The phase constant of a wave with wavelength 2 units is
a) 6.28
b) 3.14
c) 0.5
d) 2
Answer: b
Explanation: The phase constant is given by β = 2π/λ. On substituting λ = 2 units, we get β = 2π/2 = π = 3.14 units.
10.The expression for intrinsic impedance is given by
a) √(με)
b) (με)
c) √(μ/ε)
d) (μ/ε)
Answer: c
Explanation: The intrinsic impedance is given by the ratio of square root of the permittivity to the permeability. Thus η = √(μ/ε) is the intrinsic impedance. In free space or air medium, the intrinsic impedance will be 120π or 377 ohms.
11.The electric and magnetic field components in the electromagnetic wave propagation are in phase. State True/False.
a) True
b) False
Answer: a
Explanation: In dielectrics, the electric and magnetic fields will be in phase or the phase difference between them is zero. This is due to the large attenuation which leads to increase in phase shift.
12. The skin depth of a wave with phase constant of 12 units inside a conductor is
a) 12
b) 1/12
c) 24
d) 1/24
Answer: b
Explanation: The skin depth is the reciprocal of the phase constant. On substituting for β = 12, we get δ = 1/β = 1/12 units.
Power, Power Loss and Return Loss
1. The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is
a) 180
b) 90
c) 45
d) 120
Answer: b
Explanation: The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.
2. The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is
a) 1.96
b) 19.6
c) 196
d) 19600
Answer: c
Explanation: The power of a wave is given by P = V2/2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 1402/(2×50) = 196 units.
Magnetic Forces And Materials MCQs
3. The power reflected by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 2
b) 8
c) 6
d) 4
Answer: d
Explanation: The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R2xPinc. On substituting the given data, we get Pref = 0.52 x 16 = 4 units.
4. The power transmitted by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 12
b) 8
c) 16
d) 4
Answer: a
Explanation: The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R2) Pinc. On substituting the given data, we get Pref = (1- 0.52) x 16 = 12 units. In other words, it is the remaining power after reflection.
5. The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is
a) 1/3
b) 2/3
c) 1
d) 3
Answer: b
Explanation: The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.
6. The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.
a) -0.75
b) 0.25
c) -0.25
d) 0.75
Answer: d
Explanation: The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.
7. The attenuation is given by 20 units. Find the power loss in decibels.
a) 13.01
b) 26.02
c) 52.04
d) 104.08
Answer: a
Explanation: The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.
8. The reflection coefficient is 0.5. Find the return loss.
a) 12.12
b) -12.12
c) 6.02
d) -6.02
Answer: c
Explanation: The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.
9. The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is
a) 4.8
b) 9.6
c) 3.6
d) 1.8
Answer: a
Explanation: The power of an antenna is given by Prad = Ia2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia2 = 120/52 = 4.8 ohm.
10. The transmission coefficient is given by 0.65. Find the return loss of the wave.
a) 9.11
b) 1.99
c) 1.19
d) 9.91
Answer: a
Explanation: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.
11. The return loss is given as 12 decibel. Calculate the reflection coefficient.
a) 0.35
b) 0.55
c) 0.25
d) 0.75
Answer: c
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 12, we get R = 10(-12/20) = 0.25.
12. Find the transmission coefficient of a wave, when the return loss is 6 decibel.
a) 0.498
b) 0.501
c) 0.35
d) 0.65
Answer: a
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 6, we get R = 10(-6/20) = 0.501. The transmission coefficient will be T = 1 –
R = 1-0.501 = 0.498.
Refractive Index and Numerical Aperture
1. The expression for refractive index is given by
a) N = v/c
b) N = c/v
c) N = cv
d) N = 1/cv
Answer: b
Explanation: The refractive index is defined as the ratio of the velocity of light in a vacuum to its velocity in a specified medium. It is given by n = c/v. It is constant for a particular material.
2. Numerical aperture is expressed as the
a) NA = sin θa
b) NA = cos θa
c) NA = tan θa
d) NA = sec θa
Answer: a
Explanation: The numerical aperture is the measure of how much light the fiber can collect. It is the sine of the acceptance angle, the angle at which the light must be transmitted in order to get maximum reflection. Thus it is given by NA = sin θa.
3. For total internal reflection to occur, which condition must be satisfied?
a) N1 = N2
b) N1 > N2
c) N1 < N2
d) N1 x N2=1
Answer: b
Explanation: The refractive of the transmitting medium should be greater than that of the receiving medium. In other words, the light must flow from denser to rarer medium, for total internal reflection to occur.
4. Find the refractive index of a medium having a velocity of 1.5 x 108.
a) 0.5
b) 5
c) 0.2
d) 2
Answer: d
Explanation: The refractive index is given by the ratio of the speed of light to the velocity in a particular medium. It is given by n = c/v. On substituting for v = 1.5 x 108 and c = 3 x 108, we get n = 3/1.5 = 2. The quantity has no unit.
5. The refractive index of water will be
a) 1
b) 2.66
c) 5
d) 1.33
Answer: d
Explanation: The velocity of light in water as medium will be 2.25 x 108. On substituting for the speed of light, we get refractive index as n = 3/2.25 = 1.33(no unit).
6. The refractive index of air is unity. State True/False.
a) True
b) False
Answer: a
Explanation: The velocity of light in the air medium and the speed of light are both the same. Since light travels at maximum velocity in air only. Thus the refractive index n = c/v will be unity.
7. The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is
a) 3.73
b) 0.83
c) 3.46
d) 1.86
Answer: d
Explanation: The numerical aperture is given by NA = √(n12 – n22), where n1 and n2 are the refractive indices of core and cladding respectively. On substituting for n1 = 2.33 and n2 = 1.4, we get NA = √(2.332-1.42) = 1.86.
8. Find the acceptance angle of a material which has a numerical aperture of 0.707 in air.
a) 30
b) 60
c) 45
d) 90
Answer: c
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is unity in air. Thus NA = sin θa. To get θ= sin-1(NA), put NA = 0.707, thus θa = sin-1(0.707) = 45 degree.
9. The numerical aperture of a material with acceptance angle of 60 degree in water will be
a) 1.15
b) 2.15
c) 5.21
d) 1.52
Answer: a
Explanation: The numerical aperture is given by NA = n sin θa, where n is the refractive index. It is 1.33 for water medium. Given that the acceptance angle is 60, we get NA = 1.33 sin 60 = 1.15.
10. The core refractive index should be lesser than the cladding refractive index for a coaxial cable. State True/False
a) True
b) False
Answer: b
Explanation: The light should pass through the core region only, for effective transmission. When light passes through cladding, losses will occur, as cladding is meant for protection. Thus core refractive index must be greater than the cladding refractive index.
11. The refractive index is 2.33 and the critical angle is 350. Find the numerical aperture.
a) 2
b) 1.9
c) 2.33
d) 12
Answer: b
Explanation: The numerical aperture is given by NA = n cos θc, where θc is the critical angle and n is the refractive index. On substituting for n = 2.33 and θc = 35, we get NA = 2.33 cos 35 = 1.9(no unit).
12. Choose the optical fibre material from the given materials.
a) Glass
b) Plastic
c) Silica
d) Quartz
Answer: c
Explanation: Silica is the most dominant optical fibre material. This is because of its hardness, flexibility, melting point. Also it is an easily available material.
Brewster Angle
1. Brewster angle is valid for which type of polarisation?
a) Perpendicular
b) Parallel
c) S polarised
d) P polarised
Answer: b
Explanation: The parallel polarisation of the electromagnetic waves is possible only when the transmission occurs at the Brewster angle.
2. The Brewster angle is expressed as
a) Tan-1(n)
b) Tan-1(n1/n2)
c) Tan-1(n2/n1)
d) Tan (n)
Answer: c
Explanation: The tangent of the Brewster angle is the ratio of the refractive indices of the second medium to that of the first medium. It is given by tan θb= n2/n1. Thus the Brewster angle will be θb = tan-1(n2/n1).
3. The refractive index of a material with permittivity 16 is given by
a) 16
b) 256
c) 4
d) 8
Answer: c
Explanation: The refractive index is the square root of the permittivity. Thus n = √ε. Given that ε = 16, we get refractive index as n = 4. It has no unit.
4. The reflection coefficient in the wave propagation when it is transmitted with the Brewster angle is
a) 0
b) 1
c) -1
d) Infinity
Answer: a
Explanation: Brewster angle propagation refers to complete transmission. The wave transmitted at the Brewster angle will be completely transmitted without reflection. Thus the reflection coefficient will be zero.
5. The transmission coefficient of a wave propagating in the Brewster angle is
a) 0
b) 1
c) -1
d) Infinity
Answer: b
Explanation: The transmission coefficient is the reverse of the reflection coefficient. At Brewster angle, the reflection will be zero. Thus the transmission is T = 1-R. Since R = 0, T = 1. It is to be noted that T and R lies in the range of 0 to 1.
6. A circularly polarised wave transmitted at the Brewster angle will be received as linearly polarised wave. State True/False
a) True
b) False
Answer: a
Explanation: The Brewster angle is said to be the polarisation angle. When a circularly polarised wave is incident at the Brewster angle, the resultant wave will be linearly polarised.
7. An elliptically polarised wave transmitted at the Brewster angle will be received as an elliptically polarised wave. State True/False
a) True
b) False
Answer: b
Explanation: Any polarised wave transmitted at the Brewster angle will be linearly polarised. It can be a parallel, perpendicular, circular or elliptical polarisation. The resultant wave is always linearly polarised. This is the reason why the Brewster angle is called polarisation angle.
8. Find the Brewster angle of a wave transmitted from a medium of permittivity 4 to a medium of permittivity 2.
a) 35.26
b) 53.62
c) 26.35
d) 62.53
Answer: a
Explanation: The Brewster angle is given by θb = tan-1(n2/n1), where n = √ε. Thus we can express the formula in terms of permittivity as θb = tan-1√ (ε 2/ε 1). Here ε1 = 4 and ε2 = 2. Thus we get θb = tan-1√ (2/4) =
tan-1(0.707) = 35.26 degree.
9. Find the ratio of refractive index of medium 2 to that of medium 1, when the Brewster angle is 60 degree.
a) 0.707
b) 1.5
c) 0.866
d) 1.732
Answer: d
Explanation: The tangent of the Brewster angle is the ratio of the medium 2 permittivity to the medium 1 permittivity. Thus tan θb = (n2/n1). Given that θb = 60 degree, the ratio n2/n1 will be tan 60 = 1.732.
10. The Brewster angle is the angle of
a) Incidence
b) Reflection
c) Refraction
d) Transmission
Answer: a
Explanation: The Brewster angle is the angle of incidence at which complete transmission of the electromagnetic wave occurs.
Snell Law and Critical Angle
1. The Snell’s law can be derived from which type of incidence?
a) Incidence angle
b) Reflected angle
c) Refracted angle
d) Oblique incidence
Answer: d
Explanation: The oblique incidence refers to the interface between dielectric media. Consider a planar interface between two dielectric media. A plane wave is incident at an angle from medium 1 and reflected from medium 2. The interface plane defines the boundary between the media. This is the oblique medium.
2. The Snell’s law is given by
a) N1 sin θi = N2 sin θt
b) N2 sin θi = N1 sin θt
c) sin θi = sin θt
d) N1 cos θi = N2 cos θt
Answer: a
Explanation: The Snell law states that in an oblique medium, the product of the refractive index and sine of incidence angle in medium 1 is same as that of medium 2. Thus it is given by N1 sin θi = N2 sin θt.
3. Calculate the ratio of sine of incident angle to the sine of reflected angle when the refractive indices of medium 1 and 2 are given as 2.33 and 1.66 respectively.
a) 0.71
b) 1.4
c) 2
d) 3.99
Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get sin θi/sin θt, the ratio is N2/N1. On substituting for N1 = 2.33 and N2 = 1.66, we get 1.66/2.33 = 0.71.
4. Find the ratio of the refractive index of medium 1 to that of medium 2, when the incident and reflected angles are given by 300 and 450 respectively.
a) 0.5
b) 1
c) 2
d) 4
Answer: c
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. For getting N1/N2, the ratio is sin θt/sin θi. On substituting for θi = 30 and θt = 45, we get sin 45/sin 30 = 2.
5. The refractive index of a medium with permittivity of 2 and permeability of 3 is given by
a) 3.56
b) 2.45
c) 3.21
d) 1.78
Answer: b
Explanation: The refractive index is given by n = c √(με), where c is the speed of light. Given that relative permittivity and relative permeability are 2 and 3 respectively. Thus n = 3 x 108 √(2 x 4π x 10-7 x 3 x 8.854 x 10-12) = 2.45.
6. The critical angle is defined as the angle of incidence at which the total internal reflection starts to occur. State True/False.
a) True
b) False
Answer: a
Explanation: The critical angle is the minimum angle of incidence which is required for the total internal reflection to occur. This is the angle that relates the refractive index with the angle of reflection in an oblique incidence medium.
7. The critical angle for two media of refractive indices of medium 1 and 2 given by 2 and 1 respectively is
a) 0
b) 30
c) 45
d) 60
Answer: b
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. To get θc, put n1 = 2 and n2 = 1. Thus we get θc = sin-1(n2/n1) = sin-1(1/2) = 30 degree.
8. The critical angle for two media with permittivities of 16 and 9 respectively is
a) 48.59
b) 54.34
c) 60
d) 45
Answer: a
Explanation: The sine of the critical angle is the ratio of refractive index of medium 2 to that in medium 1. Thus sin θc = n2/n1. Also n = √ε, thus sin θc = √ε2/√ε1. Put ε1 = 16 and ε2 = 9, we get θc = sin-1(3/4) = 48.59 degree.
9. The angle of incidence is equal to the angle of reflection for perfect reflection. State True/False.
a) True
b) False
Answer: a
Explanation: For complete wave reflection, the angle of incidence should be same as the angle of the reflection. In such cases, the reflection coefficient is unity and the transmission coefficient is zero.
10. The angle of incidence of a wave of a wave with angle of transmission 45 degree and the refractive indices of the two media given by 2 and 1.3 is
a) 41.68
b) 61.86
c) 12.23
d) 27.89
Answer: a
Explanation: The Snell law is given by N1 sin θi = N2 sin θt. To get θi, put N1 = 2, N2 = 1.3, θt = 45 degree. Thus we get θi = sin-1(1.3 sin 45)/2 = 41.68 degree.
11. The angle at which the wave must be transmitted in air media if the angle of reflection is 45 degree is
a) 45
b) 30
c) 60
d) 90
Answer: a
Explanation: In air media, n1 = n2 = 1. Thus, sin θi=sin θt and the angle of incidence and the angle of reflection are same. Given that the reflection angle is 45, thus the angle of incidence is also 45 degree.
12. For a critical angle of 60 degree and the refractive index of the first medium is 1.732, the refractive index of the second medium is
a) 1
b) 1.5
c) 2
d) 1.66
Answer: b
Explanation: From the definition of Snell law, sin θc = n2/n1. To get n2, put n1 = 1.732 and θc = 60. Thus we get sin 60 = n2/1.732 and n2 = 1.5.
Types of Polarization
1. When the phase angle between the Ex and Ey component is 00 or 1800, the polarisation is
a) Elliptical
b) Circular
c) Linear
d) Perpendicular
Answer: c
Explanation: The phase angle between the Ex and Ey component is 00 and 1800 for linearly polarised wave. The wave is assumed to be propagating in the z direction.
2. The magnitude of the Ex and Ey components are same in which type of polarisation?
a) Linear
b) Circular
c) Elliptical
d) Perpendicular
Answer: b
Explanation: In circular polarisation, the magnitude of the Ex and Ey components are the same. This is a form of the elliptical polarisation in which the major and minor axis are the same.
3. When the Ex and Ey components of a wave are not same, the polarisation will be
a) Linear
b) Elliptical
c) Circular
d) Parallel
Answer: b
Explanation: In elliptical polarisation, the magnitude of Ex and Ey components are not same. This is due to the variation in the major and minor axes of the waves representing its magnitude.
4. Identify the polarisation of the wave given, Ex = Exo cos wt and Ey = Eyo sin wt. The phase difference is +900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: c
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand elliptical.
5. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = sin wt. The phase difference is -900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: d
Explanation: The magnitude of the Ex and Ey components are not same. Thus it is elliptical polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand elliptical.
6. Identify the polarisation of the wave given, Ex = 2 cos wt and Ey = 2 sin wt. The phase difference is +900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: a
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For +90 phase difference, the polarisation is left handed. In other words, the rotation is in clockwise direction. Thus the polarisation is left hand circular.
7. Identify the polarisation of the wave given, Ex = cos wt and Ey = sin wt. The phase difference is -900.
a) Left hand circularly polarised
b) Right hand circularly polarised
c) Left hand elliptically polarised
d) Right hand elliptically polarised
Answer: b
Explanation: The magnitude of the Ex and Ey components are the same. Thus it is circular polarisation. For -90 phase difference, the polarisation is right handed. In other words, the rotation is in anti-clockwise direction. Thus the polarisation is right hand circular.
8. For a non-zero Ex component and zero Ey component, the polarisation is
a) Parallel
b) Perpendicular
c) Elliptical
d) Circular
Answer: a
Explanation: When the Ex is non-zero and the Ey is zero, the polarisation is parallel. The parallel polarisation is classified under the linear polarisation type.
9. Identify the polarisation of the wave given that, Ex = 2 cos wt and Ey = cos wt.
a) Elliptical
b) Circular
c) Parallel
d) Linear
Answer: d
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.
10. The Snell law is applicable for perpendicular polarisation and the Brewster law is applicable for parallel polarisation. State True/False.
a) True
b) False
Answer: a
Explanation: The Snell law is calculated from the oblique incidence media. Thus it is applicable for perpendicular polarisation. The Brewster law is applicable for perpendicular polarisation.
11. When the polarisation of the receiving antenna is unknown, to ensure that it receives atleast half the power, the transmitted wave should be
a) Linearly polarised
b) Elliptically polarised
c) Circularly polarised
d) Normally polarised
Answer: c
Explanation: The polarisation of the transmitting and receiving antenna has to be the same. This is the condition for maximum power transfer to occur. This is possible only when the polarisation is circular.
12. Identify the polarisation of the wave given that, Ex = 2 sin wt and Ey = 3 sin wt.
a) Linear
b) Elliptical
c) Circular
d) Parallel
Answer: a
Explanation: The magnitude of the Ex and Ey components are not the same. Thus it cannot be circular polarisation. For a phase difference of 0, the polarisation is linear. In other words, the waves are in phase. Thus the polarisation is linear.
S and P Polarised Waves
1. The resultant electric field of a wave with Ex = 3 and Ey = 4 will be
a) 7
b) 1
c) 25
d) 5
Answer: d
Explanation: The resultant electric field of two electric components Ex and Ey is E = √(Ex2 + Ey2). On substituting for Ex = 3 and Ey = 4, we get E = 5 units.
2. In S polarisation, the electric field lies in the plane perpendicular to that of the interface. State True/False
a) True
b) False
Answer: a
Explanation: In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The S polarised wave is similar to the transverse magnetic (TM) wave, the electric field lies in the plane perpendicular to that of the interface.
3. In P polarisation, the electric field lies in the same plane as the interface. State True/False.
a) True
b) False
Answer: a
Explanation: In the EM wave propagation, the electric and magnetic fields are perpendicular to each other. The P polarised wave is similar to the transverse electric (TE) wave, the magnetic field lies in the plane perpendicular to that of the interface or the electric field lies in the same plane as the interface.
4. The group delay of a wave with phase constant 2.5 units and frequency of 1.2 radian/sec is
a) 3.7
b) 1.3
c) 3
d) 2.08
Answer: d
Explanation: The group delay is given by td = β/ω. On substituting for β = 2.5 and ω = 1.2, we get the group delay as td = 2.5/1.2 = 2.08 units.
5. The Brewster angle is valid for which type of polarisation?
a) S polarised
b) P polarised
c) Elliptical
d) Linear
Answer: b
Explanation: The Brewster angle is valid for perpendicular polarisation. The P polarised wave is also a type of perpendicular polarisation. In P polarisation, the electric field lies in the plane of the interface.
6. Find the reflection coefficient of a wave with an incident electric field of 5 units and reflected electric field of 2 units.
a) 2.5
b) 0.4
c) 0.8
d) 1.2
Answer: b
Explanation: The reflection coefficient is the ratio of the reflected electric field to the incident electric field. Thus τ = Er/Ei. On substituting for Ei = 5 and Er = 2, we get τ = 2/5 = 0.4(no unit).
7. The transmission coefficient of a wave with incident and transmitted electric field of 5 and 5 respectively is
a) 0
b) 1
c) 10
d) 5
Answer: b
Explanation: The transmission coefficient is the ratio of the transmitted electric field to the incident electric field. Thus T = Et/Ei. On substituting for Et = 5 and Ei = 5, we get T = 5/5 = 1(no unit). Simply, when the incident and transmitted field are same, no reflection occurs and the transmission is unity.
8. Find the relative permittivity of the medium having a refractive index of 1.6
a) 0.4
b) 2.56
c) 3.2
d) 4.8
Answer: b
Explanation: The refractive index is the square root of the relative permittivity. It is given by n = √εr. To get εr, put n = 1.6. We get εr = n2 = 1.62 = 2.56(no unit).
9. Calculate the transmission coefficient of a wave with a reflection coefficient of 0.6
a) 0.6
b) 1
c) 0
d) 0.4
Answer: d
Explanation: The transmission coefficient is the reverse of the reflection coefficient. Thus T + τ = 1. On substituting for τ = 0.6, we get T = 0.4. It has no unit.
10. The phase constant of a wave propagation with frequency of 35 radian/sec and time delay of 7.5 sec is
a) 4.66
b) 262.5
c) 46.6
d) 26.25
Answer: b
Explanation: The group delay expression is td = β/ω. To get β, put ω = 35 and td = 7.5. Thus we get β = td x ω = 7.5 x 35 = 262.5 units.