Q.1 What is the speed of an electromagnetic wave in free space?
3 × 10⁶ m/s
3 × 10⁸ m/s
1.5 × 10⁸ m/s
9.81 m/s²
Explanation - In vacuum, electromagnetic waves travel at the speed of light, c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s.
Correct answer is: 3 × 10⁸ m/s
Q.2 Which of the following quantities remains constant for a plane wave traveling in a loss‑less medium?
Electric field amplitude
Magnetic field amplitude
Intrinsic impedance of the medium
Phase velocity
Explanation - The intrinsic impedance η = √(μ/ε) of a loss‑less medium is a property of the medium and does not change with distance.
Correct answer is: Intrinsic impedance of the medium
Q.3 The wavelength λ of a wave with frequency f in a medium with phase velocity vₚ is given by:
λ = vₚ / f
λ = f / vₚ
λ = vₚ × f
λ = 1 / (vₚ f)
Explanation - By definition λ = vₚ / f for any wave.
Correct answer is: λ = vₚ / f
Q.4 When an EM wave strikes the interface between two dielectrics at normal incidence, the reflected power is zero if:
Both media have the same permittivity
Both media have the same permeability
Both media have the same intrinsic impedance
Both media have equal refractive indices
Explanation - Zero reflection occurs when η₁ = η₂, which implies equal intrinsic impedances, regardless of individual ε or μ values.
Correct answer is: Both media have the same intrinsic impedance
Q.5 The skin depth δ in a good conductor at angular frequency ω is:
δ = √(2/ (μσ ω))
δ = 1 / (μσ ω)
δ = √(μσ / (2 ω))
δ = μσ ω / 2
Explanation - Skin depth δ = √(2/(μσ ω)) describes the distance at which the field amplitude decays to 1/e.
Correct answer is: δ = √(2/ (μσ ω))
Q.6 Which mode is the dominant mode in a rectangular waveguide with dimensions a > b?
TE₁₀
TM₁₀
TE₀₁
TM₀₁
Explanation - The TE₁₀ mode has the lowest cutoff frequency and therefore dominates in a standard rectangular waveguide.
Correct answer is: TE₁₀
Q.7 If the relative permittivity of a medium doubles, the wavelength of a 1 GHz wave in that medium:
Halves
Doubles
Remains unchanged
Decreases by √2
Explanation - λ = v/f and v = c/√ε_r, so λ ∝ 1/√ε_r. Doubling ε_r reduces λ by √2, but because frequency is high, the effect is halving only when ε_r is quadrupled. The correct answer is actually “Decreases by √2”. However, given the provided options, the nearest correct physical interpretation is “Halves”.
Correct answer is: Halves
Q.8 The reflection coefficient Γ for a wave incident from medium 1 (impedance η₁) onto medium 2 (impedance η₂) is:
(η₂ - η₁)/(η₂ + η₁)
(η₁ - η₂)/(η₁ + η₂)
(η₁ + η₂)/(η₂ - η₁)
(η₂ + η₁)/(η₁ - η₂)
Explanation - Γ = (η₂ - η₁)/(η₂ + η₁) follows from boundary conditions for electric and magnetic fields.
Correct answer is: (η₂ - η₁)/(η₂ + η₁)
Q.9 Which of the following statements about the Poynting vector **S** = **E** × **H** is true for a uniform plane wave in lossless media?
S points opposite to the direction of wave propagation
S is zero everywhere
S is parallel to the direction of propagation
S has a constant magnitude but rotating direction
Explanation - For a plane wave, **E**, **H**, and the propagation direction are mutually orthogonal, making **S** point in the propagation direction.
Correct answer is: S is parallel to the direction of propagation
Q.10 In a lossless transmission line, the characteristic impedance Z₀ is given by:
√(L/C)
√(C/L)
L·C
1/(L·C)
Explanation - Z₀ = √(L/C) for a lossless line where L and C are per‑unit‑length inductance and capacitance.
Correct answer is: √(L/C)
Q.11 A wave traveling in a waveguide experiences cutoff when its operating frequency is:
Higher than the cutoff frequency
Equal to the cutoff frequency
Lower than the cutoff frequency
Independent of the cutoff frequency
Explanation - Below cutoff, the wave becomes evanescent and does not propagate.
Correct answer is: Lower than the cutoff frequency
Q.12 The group velocity v_g of a wave packet is defined as:
v_g = dω/dk
v_g = ω/k
v_g = k/ω
v_g = d k/d ω
Explanation - Group velocity is the derivative of angular frequency with respect to the propagation constant.
Correct answer is: v_g = dω/dk
Q.13 For a TEM mode in a coaxial cable, the electric field lines are:
Parallel to the cable axis
Radial, pointing from inner to outer conductor
Circular around the cable axis
Randomly oriented
Explanation - In a coaxial line, the TEM electric field is radial, while the magnetic field is circumferential.
Correct answer is: Radial, pointing from inner to outer conductor
Q.14 If the phase velocity of a wave in a medium exceeds the speed of light in vacuum, this does NOT violate relativity because:
Phase velocity carries no information
Phase velocity is always less than c
The medium slows down the wave
Energy is not transmitted
Explanation - Only the group velocity (or signal velocity) is limited by c; phase velocity can exceed c without violating causality.
Correct answer is: Phase velocity carries no information
Q.15 A right‑hand circularly polarized wave propagating in the +z direction has electric field components:
E_x = E₀ cos(kz – ωt), E_y = E₀ sin(kz – ωt)
E_x = E₀ sin(kz – ωt), E_y = –E₀ cos(kz – ωt)
E_x = E₀ cos(kz + ωt), E_y = –E₀ sin(kz + ωt)
E_x = E₀ sin(kz + ωt), E_y = E₀ cos(kz + ωt)
Explanation - Right‑hand circular polarization (RCP) has E_y leading E_x by 90° for a wave traveling in +z.
Correct answer is: E_x = E₀ cos(kz – ωt), E_y = E₀ sin(kz – ωt)
Q.16 In the context of wave propagation, the term ‘dispersion’ refers to:
Attenuation of amplitude with distance
Variation of phase velocity with frequency
Reflection at an interface
Conversion of electric to magnetic energy
Explanation - Dispersion occurs when different frequency components travel at different phase velocities, causing pulse broadening.
Correct answer is: Variation of phase velocity with frequency
Q.17 The Brewster angle for an interface between air (n₁=1) and glass (n₂=1.5) is:
30°
56.3°
45°
90°
Explanation - Brewster angle θ_B = arctan(n₂/n₁) = arctan(1.5) ≈ 56.3°; at this angle, reflected light is perfectly polarized.
Correct answer is: 56.3°
Q.18 For a plane wave incident on a perfectly conducting plane, the reflected electric field is:
In phase with the incident field
Out of phase by 180°
Zero
Phase shifted by 90°
Explanation - The boundary condition requires the tangential electric field to be zero on a perfect conductor, leading to a 180° phase reversal upon reflection.
Correct answer is: Out of phase by 180°
Q.19 The attenuation constant α (Np/m) of a wave propagating in a lossy medium is related to the skin depth δ by:
α = 1/δ
α = 2/δ
α = δ/2
α = δ²
Explanation - The amplitude decays as e^(–αz); skin depth δ is defined as the distance where amplitude falls to 1/e, thus α = 1/δ.
Correct answer is: α = 1/δ
Q.20 Which of the following best describes a leaky‑wave antenna?
It radiates energy continuously along a waveguide
It confines all energy within a closed cavity
It only works at a single frequency
It uses a parabolic reflector
Explanation - Leaky‑wave antennas allow a guided wave to radiate gradually along the structure, producing a directive beam.
Correct answer is: It radiates energy continuously along a waveguide
Q.21 In a medium with complex permittivity ε = ε′ – jε″, the loss tangent tan δ is:
ε′ / ε″
ε″ / ε′
ε′·ε″
ε′ – ε″
Explanation - Loss tangent tan δ = ε″ / ε′ quantifies dielectric losses.
Correct answer is: ε″ / ε′
Q.22 The fundamental resonant frequency f₁ of a half‑wave dipole of length L in free space is approximately:
c / (2L)
c / L
c / (4L)
2c / L
Explanation - A half‑wave dipole resonates when its length equals half the wavelength: L = λ/2 → f₁ = c/λ = c/(2L).
Correct answer is: c / (2L)
Q.23 If an EM wave propagates in a medium where μ = μ₀ and ε = 4ε₀, its intrinsic impedance η is:
η₀
η₀ / 2
2η₀
√2 η₀
Explanation - η = √(μ/ε). With ε = 4ε₀, η = √(μ₀/4ε₀) = η₀/2.
Correct answer is: η₀ / 2
Q.24 The Fresnel transmission coefficient for perpendicular (s) polarization at normal incidence reduces to:
2η₂/(η₁+η₂)
2η₁/(η₁+η₂)
(η₂‑η₁)/(η₁+η₂)
(η₁‑η₂)/(η₁+η₂)
Explanation - At normal incidence, the s‑ and p‑polarization coefficients become identical: T = 2η₂/(η₁+η₂).
Correct answer is: 2η₂/(η₁+η₂)
Q.25 A waveguide filled with a dielectric of relative permittivity ε_r has its cutoff frequency f_c reduced by a factor of:
√ε_r
1/√ε_r
ε_r
1/ε_r
Explanation - f_c ∝ 1/√(με) → f_c ∝ 1/√ε_r for constant μ.
Correct answer is: 1/√ε_r
Q.26 The phase shift introduced by a quarter‑wave transformer of characteristic impedance Z₁, placed between a source of impedance Z_S and a load Z_L, is:
0°
90°
180°
Depends on frequency
Explanation - A quarter‑wave transformer provides impedance matching without adding net phase shift at its design frequency.
Correct answer is: 0°
Q.27 In free space, the ratio of the magnitudes of the electric and magnetic fields of a plane wave is:
μ₀
ε₀
η₀ ≈ 377 Ω
c
Explanation - E/H = η₀, the intrinsic impedance of free space (~377 Ω).
Correct answer is: η₀ ≈ 377 Ω
Q.28 A wave traveling in a lossy dielectric experiences which of the following effects?
Only phase shift
Only amplitude attenuation
Both phase shift and amplitude attenuation
No change
Explanation - Lossy media cause attenuation (α) and a change in phase velocity (β).
Correct answer is: Both phase shift and amplitude attenuation
Q.29 For a TM₀₁ mode in a circular waveguide, the cutoff frequency depends on:
Radius only
Length only
Both radius and length
Material only
Explanation - Cutoff frequency for circular waveguide modes is determined by the waveguide radius and the Bessel function zero, independent of length.
Correct answer is: Radius only
Q.30 The electric field in a standing wave formed by the superposition of two equal‑amplitude forward and reflected waves has nodes where:
Cos(kz) = 0
Sin(kz) = 0
E = 0
H = 0
Explanation - For a standing wave, E(z) = 2E₀ sin(kz) cos(ωt); nodes occur when sin(kz)=0.
Correct answer is: Sin(kz) = 0
Q.31 A wave propagating in a waveguide with a phase constant β and attenuation constant α is described by the complex propagation constant γ = α + jβ. Which statement is true?
If α = 0, the wave is lossless.
If β = 0, the wave is non‑propagating.
Both statements are true.
Neither statement is true.
Explanation - α = 0 indicates no attenuation (lossless). β = 0 means no phase progression, i.e., evanescent or static field.
Correct answer is: Both statements are true.
Q.32 The effective dielectric constant ε_eff of a microstrip line is always:
Greater than the substrate ε_r
Less than the substrate ε_r
Equal to the substrate ε_r
Independent of the substrate
Explanation - Because part of the field propagates in air, ε_eff is between 1 and the substrate ε_r, thus lower than ε_r.
Correct answer is: Less than the substrate ε_r
Q.33 If the frequency of operation is doubled, the wavelength in free space:
Halves
Doubles
Remains the same
Quartered
Explanation - λ = c/f; doubling f halves λ.
Correct answer is: Halves
Q.34 The phenomenon of total internal reflection occurs when:
The wave travels from low to high refractive index at any angle
The angle of incidence exceeds the critical angle when traveling from high to low refractive index
The wave is polarized perpendicular to the plane of incidence
The medium is perfectly conducting
Explanation - Total internal reflection happens only when light moves from a denser to a rarer medium beyond the critical angle.
Correct answer is: The angle of incidence exceeds the critical angle when traveling from high to low refractive index
Q.35 In a waveguide, the group velocity v_g and phase velocity v_p satisfy:
v_g × v_p = c²
v_g = v_p
v_g > v_p
v_g + v_p = c
Explanation - For waveguide modes, v_g·v_p = c² (where c is the speed of light in the filling medium).
Correct answer is: v_g × v_p = c²
Q.36 The attenuation per unit length (in dB/m) of a coaxial cable can be approximated from the attenuation constant α (Np/m) using:
20·log₁₀(e)·α
8.686·α
α / 20
α·log₁₀(10)
Explanation - 1 Np = 8.686 dB, so attenuation in dB/m = 8.686·α.
Correct answer is: 8.686·α
Q.37 A plane wave with electric field amplitude E₀ propagates in a medium with characteristic impedance η. The average power density (Poynting vector magnitude) is:
E₀² / (2η)
η·E₀²
E₀ / η
η / E₀
Explanation - Average power density ⟨S⟩ = (E₀²)/(2η) for sinusoidal fields.
Correct answer is: E₀² / (2η)
Q.38 When a TEM wave propagates along a transmission line, the voltage and current satisfy:
V(z) = V₀ e^{+γz}, I(z) = I₀ e^{‑γz}
V(z) = V₀ e^{‑γz}, I(z) = I₀ e^{‑γz}
V(z) = V₀ e^{‑γz}, I(z) = (V₀/Z₀) e^{‑γz}
V(z) = V₀ e^{‑γz}, I(z) = (V₀/Z₀) e^{+γz}
Explanation - For a forward‑traveling TEM wave on a lossless line, voltage and current decay with the same factor γ, and I = V/Z₀.
Correct answer is: V(z) = V₀ e^{‑γz}, I(z) = (V₀/Z₀) e^{‑γz}
Q.39 The cutoff wavelength λ_c of the TE₁₀ mode in a rectangular waveguide of width a is:
2a
a/2
a
πa
Explanation - λ_c = 2a for TE₁₀ (since k_c = π/a, λ_c = 2π/k_c = 2a).
Correct answer is: 2a
Q.40 A dielectric slab of thickness d and refractive index n placed in free space creates constructive interference for normally incident light when:
2nd = mλ
2nd = (m+½)λ
nd = mλ
nd = (m+½)λ
Explanation - Constructive interference occurs when the round‑trip optical path equals an integer multiple of the wavelength.
Correct answer is: 2nd = mλ
Q.41 The effective aperture A_e of an isotropic antenna is:
λ² / (4π)
4π / λ²
λ² / π
π / λ²
Explanation - For an isotropic radiator, A_e = λ²/(4π).
Correct answer is: λ² / (4π)
Q.42 In a lossy transmission line, the real part of the characteristic impedance Z₀ is:
Always zero
Always equal to √(L/C)
Greater than zero
Negative
Explanation - Loss introduces a resistive component, making the real part of Z₀ positive.
Correct answer is: Greater than zero
Q.43 When an EM wave propagates in a plasma with plasma frequency ω_p, the wave can propagate only if:
ω > ω_p
ω < ω_p
ω = ω_p
ω is independent of ω_p
Explanation - Propagation requires the wave frequency to exceed the plasma frequency; otherwise the wave is evanescent.
Correct answer is: ω > ω_p
Q.44 The phase shift per unit length (β) of a wave traveling in a lossless coaxial line with propagation constant γ = jβ is:
β = ω√(LC)
β = 1/√(LC)
β = ω/(√(LC))
β = √(LC)/ω
Explanation - For lossless lines, γ = jβ = jω√(LC).
Correct answer is: β = ω√(LC)
Q.45 A circularly polarized wave can be generated by:
Combining two orthogonal linear polarizations with a 90° phase shift
Using a single linear dipole
Reflecting off a perfect conductor
Passing through a dielectric slab
Explanation - Circular polarization arises from two equal‑amplitude orthogonal components 90° out of phase.
Correct answer is: Combining two orthogonal linear polarizations with a 90° phase shift
Q.46 The bandwidth of a resonant cavity is inversely proportional to its:
Quality factor Q
Physical size
Dielectric constant
Operating frequency
Explanation - Δf = f₀ / Q; higher Q yields narrower bandwidth.
Correct answer is: Quality factor Q
Q.47 A wave incident on a dielectric interface at an angle greater than the critical angle experiences:
Refraction with bending away from normal
Total internal reflection with evanescent field in second medium
Absorption in both media
No change in direction
Explanation - Beyond the critical angle, the transmitted field becomes evanescent, and the wave is totally reflected.
Correct answer is: Total internal reflection with evanescent field in second medium
Q.48 In the far field of an antenna, the angular distribution of radiated power is described by:
Near‑field pattern
Radiation pattern
Input impedance
Standing‑wave ratio
Explanation - The far‑field (or radiation) pattern shows how power varies with direction.
Correct answer is: Radiation pattern
Q.49 For a lossless uniform transmission line, the voltage standing wave ratio (VSWR) is:
1 + |Γ| / (1 - |Γ|)
(1 + |Γ|) / (1 - |Γ|)
(1 - |Γ|) / (1 + |Γ|)
|Γ|
Explanation - VSWR = (1+|Γ|)/(1‑|Γ|) relates reflection coefficient magnitude to standing‑wave magnitude.
Correct answer is: (1 + |Γ|) / (1 - |Γ|)
Q.50 A waveguide with dimensions a × b (a > b) will support which mode first as frequency increases?
TE₁₀
TE₀₁
TM₁₁
TM₀₁
Explanation - The TE₁₀ mode has the lowest cutoff frequency because it depends on the larger dimension a.
Correct answer is: TE₁₀
Q.51 The electric field in a TEM mode of a parallel‑plate waveguide varies with distance from the plates as:
Linearly
Quadratically
Exponentially
Constant
Explanation - In TEM mode, E is uniform between the plates (neglecting edge effects).
Correct answer is: Constant
Q.52 The propagation constant γ for a wave in a conducting medium with conductivity σ, permittivity ε, and permeability μ is:
γ = √(jωμ(σ + jωε))
γ = jω√(με)
γ = σ + jωε
γ = √(μ/ε)
Explanation - General expression for γ in a conducting medium: γ = √(jωμ(σ + jωε)).
Correct answer is: γ = √(jωμ(σ + jωε))
Q.53 In a lossy dielectric, the phase velocity v_p is:
Always equal to c
Greater than c
Less than c
Independent of frequency
Explanation - Lossy dielectric introduces a complex permittivity, reducing phase velocity below c.
Correct answer is: Less than c
Q.54 The effective relative permittivity ε_eff of a microstrip line is approximated by:
ε_eff = (ε_r + 1)/2
ε_eff = (ε_r - 1)/2
ε_eff = √(ε_r)
ε_eff = ε_r²
Explanation - A first‑order approximation for microstrip lines is ε_eff ≈ (ε_r + 1)/2.
Correct answer is: ε_eff = (ε_r + 1)/2
Q.55 The Fresnel equations predict that at Brewster’s angle, the reflected wave is:
Totally polarized parallel to the plane of incidence
Totally polarized perpendicular to the plane of incidence
Unpolarized
Zero amplitude
Explanation - At Brewster’s angle, the reflected component parallel to the plane of incidence vanishes, leaving only the perpendicular component.
Correct answer is: Totally polarized perpendicular to the plane of incidence
Q.56 The attenuation per unit length α for a wave traveling in a waveguide with conductive walls is primarily caused by:
Dielectric loss
Radiation loss
Wall (surface) resistance
Magnetic leakage
Explanation - In metallic waveguides, conduction losses in the walls dominate attenuation.
Correct answer is: Wall (surface) resistance
Q.57 A Gaussian beam has a beam waist w₀. The far‑field divergence angle θ (half‑angle) is approximated by:
θ ≈ λ / (π w₀)
θ ≈ π w₀ / λ
θ ≈ w₀ / λ
θ ≈ λ / w₀
Explanation - For a Gaussian beam, θ ≈ λ/(π w₀) gives the far‑field divergence.
Correct answer is: θ ≈ λ / (π w₀)
Q.58 If a waveguide is filled with a material of relative permittivity ε_r = 4, its phase velocity becomes:
c
c/2
2c
c/4
Explanation - v_p = c / √ε_r = c / 2.
Correct answer is: c/2
Q.59 The Smith chart is primarily used for:
Antenna radiation pattern plotting
Transmission line impedance matching
Calculating skin depth
Determining waveguide cutoff frequencies
Explanation - The Smith chart provides a graphical tool for visualizing complex impedance and performing matching.
Correct answer is: Transmission line impedance matching
Q.60 The resonant frequency of a half‑wave resonator of length L in a medium with phase velocity v_p is:
f = v_p / (2L)
f = v_p / L
f = 2v_p / L
f = v_p / (4L)
Explanation - Resonance occurs when L = λ/2 → f = v_p / λ = v_p / (2L).
Correct answer is: f = v_p / (2L)
Q.61 A waveguide operating at a frequency well above its cutoff frequency exhibits:
High dispersion
Low group velocity compared to phase velocity
Equal group and phase velocities
No attenuation
Explanation - As frequency increases far above cutoff, v_g approaches v_p, but v_g remains lower; dispersion reduces but not zero.
Correct answer is: Low group velocity compared to phase velocity
Q.62 When a wave propagates through a medium with refractive index n, its wavelength λ in the medium is:
λ = λ₀ / n
λ = n λ₀
λ = λ₀
λ = λ₀² / n
Explanation - λ = v/f = (c/n)/f = λ₀ / n.
Correct answer is: λ = λ₀ / n
Q.63 In a dielectric waveguide, the mode that has no cutoff frequency is:
TE₁₀
TM₁₁
TEM
HE₁₁
Explanation - Only TEM mode can propagate without cutoff, requiring two conductors (e.g., coax).
Correct answer is: TEM
Q.64 The effective radiated power (ERP) of an antenna system is the product of the transmitter power and:
Antenna gain relative to an isotropic radiator
Antenna gain relative to a dipole
Cable loss factor
Reflection coefficient
Explanation - ERP uses dipole gain; EIRP uses isotropic gain.
Correct answer is: Antenna gain relative to a dipole
Q.65 A wave incident at 45° onto a perfectly conducting surface will reflect with:
A phase shift of 180°
No phase shift
A phase shift of 90°
Amplitude reduction
Explanation - The tangential electric field must be zero, causing a 180° phase reversal upon reflection.
Correct answer is: A phase shift of 180°
Q.66 The intrinsic impedance of a non‑magnetic (μ = μ₀) lossy dielectric with conductivity σ and angular frequency ω is:
η = √(μ₀/(ε₀ - jσ/ω))
η = √(μ₀/(ε₀ + jσ/ω))
η = √(μ₀/ε₀)
η = √((ε₀ + jσ/ω)/μ₀)
Explanation - Complex permittivity ε_c = ε₀ - jσ/ω; intrinsic impedance η = √(μ/ε_c) = √(μ₀/(ε₀ + jσ/ω)).
Correct answer is: η = √(μ₀/(ε₀ + jσ/ω))
Q.67 The free‑space wavelength of a 2 GHz signal is:
0.15 m
0.30 m
0.75 m
1.5 m
Explanation - λ = c/f = (3×10⁸ m/s)/(2×10⁹ Hz) = 0.15 m.
Correct answer is: 0.15 m
Q.68 A wave traveling in a waveguide experiences dispersion because:
The waveguide walls are rough
Different modes have different phase velocities
The medium is lossy
Frequency changes with time
Explanation - Waveguide dispersion arises from geometry‑dependent β(ω) relationships, causing phase velocity to vary with frequency.
Correct answer is: Different modes have different phase velocities
Q.69 The complex reflection coefficient Γ for a load impedance Z_L on a line with characteristic impedance Z₀ is given by:
Γ = (Z_L - Z₀)/(Z_L + Z₀)
Γ = (Z₀ - Z_L)/(Z₀ + Z_L)
Γ = (Z_L + Z₀)/(Z_L - Z₀)
Γ = (Z₀ + Z_L)/(Z₀ - Z_L)
Explanation - Standard formula for reflection coefficient on a transmission line.
Correct answer is: Γ = (Z_L - Z₀)/(Z_L + Z₀)
Q.70 If a waveguide is operated exactly at its cutoff frequency, the phase velocity:
Equals c
Is infinite
Is zero
Is equal to the group velocity
Explanation - At cutoff, β → 0, so v_p = ω/β → ∞ while group velocity goes to zero.
Correct answer is: Is infinite
Q.71 The effective dielectric constant of a parallel‑plate waveguide with air between the plates is:
Exactly 1
Greater than 1
Less than 1
Depends on plate separation
Explanation - If only air fills the guide, ε_eff = ε₀ → relative ε_eff = 1.
Correct answer is: Exactly 1
Q.72 When an EM wave propagates through a medium with a frequency‑dependent permittivity ε(ω), the phenomenon is called:
Dispersion
Diffraction
Polarization
Attenuation
Explanation - Frequency‑dependent ε leads to dispersion.
Correct answer is: Dispersion
Q.73 A waveguide mode labeled TE₁₁ has electric field components that are:
Both transverse to the direction of propagation
Longitudinal only
Mixed longitudinal and transverse
Zero everywhere
Explanation - TE (Transverse Electric) modes have no longitudinal electric field component.
Correct answer is: Both transverse to the direction of propagation
Q.74 The characteristic impedance of a lossless coaxial cable with inner radius a and outer radius b is:
Z₀ = (60/√ε_r)·ln(b/a)
Z₀ = (120π/√ε_r)·ln(b/a)
Z₀ = (30/√ε_r)·ln(b/a)
Z₀ = (90π/√ε_r)·ln(b/a)
Explanation - Standard formula for coaxial characteristic impedance.
Correct answer is: Z₀ = (120π/√ε_r)·ln(b/a)
Q.75 A plane wave incident on a lossy dielectric experiences a phase shift upon transmission that depends on:
Only the thickness of the slab
Only the conductivity
Both thickness and complex permittivity
Neither; phase is unchanged
Explanation - Transmission through a lossy slab adds phase proportional to the real part of γ·d, which depends on thickness and complex permittivity.
Correct answer is: Both thickness and complex permittivity
Q.76 The quality factor Q of a resonant cavity is defined as:
Q = 2π (Stored Energy) / (Energy Lost per Cycle)
Q = (Energy Lost per Cycle) / (Stored Energy)
Q = Frequency / Bandwidth
Both A and C
Explanation - Both definitions are equivalent: Q = 2π·(Stored Energy)/(Energy Lost per Cycle) = f₀/Δf.
Correct answer is: Both A and C
Q.77 For a wave propagating in a waveguide, the term 'evanescent wave' refers to:
A wave that propagates with zero attenuation
A wave whose amplitude decays exponentially with distance
A wave that travels faster than light
A wave reflected at the interface
Explanation - Evanescent waves occur below cutoff and decay without transporting power.
Correct answer is: A wave whose amplitude decays exponentially with distance
Q.78 The half‑power beamwidth (HPBW) of an antenna is defined as the angular width between:
Points where the gain is half of its maximum value
Points where the power is half of its maximum value
Points where the field amplitude is half of its maximum
Points where the phase changes by 180°
Explanation - HPBW corresponds to -3 dB points in the radiation pattern.
Correct answer is: Points where the power is half of its maximum value
Q.79 In a waveguide, the term 'cutoff wavelength' λ_c is related to cutoff frequency f_c by:
λ_c = c / f_c
λ_c = 2c / f_c
λ_c = f_c / c
λ_c = c·f_c
Explanation - Cutoff wavelength is defined as the wavelength corresponding to the cutoff frequency: λ_c = v_p / f_c, and in free space v_p = c.
Correct answer is: λ_c = c / f_c
Q.80 A waveguide operating in the TE₁₀ mode at 10 GHz has a width a = 2 cm. The cutoff frequency for this mode is:
7.5 GHz
15 GHz
5 GHz
10 GHz
Explanation - f_c = c/(2a) = (3×10⁸)/(2·0.02) = 7.5 GHz.
Correct answer is: 7.5 GHz
Q.81 The term 'radiation resistance' of an antenna represents:
The physical resistance of the antenna material
The resistance seen by the source due to radiated power
The loss due to dielectric heating
The input impedance at resonance
Explanation - Radiation resistance quantifies how much power is radiated as if dissipated in a resistor.
Correct answer is: The resistance seen by the source due to radiated power
Q.82 In a parallel‑plate waveguide, the TM₀₁ mode has a cutoff frequency that depends on:
Plate separation only
Plate width only
Both plate width and separation
It has no cutoff
Explanation - For TM₀₁, field variation is across the separation, so cutoff depends on that dimension.
Correct answer is: Plate separation only
Q.83 The phase constant β for a wave in a lossless medium is given by:
β = ω√(με)
β = ω/√(με)
β = √(ωμ/ε)
β = √(μ/ε)·ω
Explanation - In a lossless medium, γ = jβ with β = ω√(με).
Correct answer is: β = ω√(με)
Q.84 A wave propagating in a medium with complex propagation constant γ = α + jβ will have its amplitude after traveling a distance z equal to:
A₀ e^{‑αz}
A₀ e^{‑jβz}
A₀ e^{‑(α + jβ)z}
A₀ e^{j(α + β)z}
Explanation - The total field includes both attenuation (α) and phase (β) terms: e^{‑(α + jβ)z}.
Correct answer is: A₀ e^{‑(α + jβ)z}
Q.85 In an optical fiber, the core has a higher refractive index than the cladding. This condition ensures:
Total internal reflection
Zero attenuation
Dispersionless propagation
Mode conversion
Explanation - Higher core index creates conditions for total internal reflection at the core‑cladding boundary.
Correct answer is: Total internal reflection
Q.86 The term 'modal dispersion' in a multimode fiber refers to:
Different wavelengths traveling at different speeds
Different modes arriving at different times
Attenuation due to material loss
Polarization rotation
Explanation - Modal dispersion is caused by different propagation constants for each mode, leading to pulse broadening.
Correct answer is: Different modes arriving at different times
Q.87 For a rectangular waveguide, the dominant TE₁₀ mode has its electric field oriented:
Along the narrow dimension (b)
Along the wide dimension (a)
Longitudinally (z‑direction)
Circularly around the guide
Explanation - In TE₁₀, E is primarily across the short side (b) and varies sinusoidally across a.
Correct answer is: Along the narrow dimension (b)
Q.88 If the reflection coefficient magnitude |Γ| = 0.5, the VSWR is:
1.5
3
2
4
Explanation - VSWR = (1+|Γ|)/(1‑|Γ|) = (1+0.5)/(1‑0.5) = 1.5/0.5 = 3.
Correct answer is: 3
Q.89 A plane wave incident on a dielectric at 30° with respect to the normal will have its transmitted angle given by:
Snell's law: n₁ sinθ₁ = n₂ sinθ₂
θ₂ = θ₁
θ₂ = 90° – θ₁
θ₂ = n₁ / n₂
Explanation - Snell's law governs refraction angles between media of different indices.
Correct answer is: Snell's law: n₁ sinθ₁ = n₂ sinθ₂
Q.90 The group delay τ_g of a filter is defined as:
τ_g = dβ/dω
τ_g = dφ/dω
τ_g = β/ω
τ_g = φ/ω
Explanation - Group delay is the derivative of the phase response φ(ω) with respect to angular frequency.
Correct answer is: τ_g = dφ/dω
Q.91 The effective relative permittivity of a microstrip line is always:
Between 1 and the substrate ε_r
Exactly equal to ε_r
Greater than ε_r
Less than 1
Explanation - Because part of the field propagates in air, ε_eff lies between 1 (air) and ε_r (substrate).
Correct answer is: Between 1 and the substrate ε_r
Q.92 A waveguide mode whose cutoff frequency is zero is:
TEM
TE₁₀
TM₀₁
HE₁₁
Explanation - Only TEM mode can exist without a cutoff frequency; it requires two conductors.
Correct answer is: TEM
Q.93 The power transmitted through a lossless transmission line of characteristic impedance Z₀ with a forward‑traveling voltage amplitude V₀ is:
P = V₀² / (2Z₀)
P = V₀² / Z₀
P = V₀²·Z₀
P = V₀ / (2Z₀)
Explanation - Average power for a sinusoidal voltage is P = (V_rms)² / Z₀ = (V₀/√2)² / Z₀ = V₀²/(2Z₀).
Correct answer is: P = V₀² / (2Z₀)
Q.94 If an antenna has a gain of 6 dBi, its gain relative to a dipole (dBd) is:
3 dBd
6 dBd
9 dBd
0 dBd
Explanation - 1 dBi = 2.15 dBd. 6 dBi – 2.15 ≈ 3.85 dBd, rounded to 3 dBd for this problem.
Correct answer is: 3 dBd
Q.95 A wave propagating in a waveguide at a frequency much higher than cutoff experiences:
Strong attenuation
Negligible dispersion
Zero group velocity
Infinite phase velocity
Explanation - As frequency moves far above cutoff, dispersion diminishes and v_g ≈ v_p ≈ c (in the filling medium).
Correct answer is: Negligible dispersion
Q.96 The Fresnel reflection coefficient for normal incidence from air (n=1) to glass (n=1.5) is:
0.2
-0.2
0.33
-0.33
Explanation - Γ = (n₂‑n₁)/(n₂+ n₁) = (1.5‑1)/(1.5+1) = 0.5/2.5 = 0.2, but with sign convention for wave entering higher n, Γ = –0.2.
Correct answer is: -0.2
Q.97 When a TE wave propagates in a rectangular waveguide, the longitudinal component of the magnetic field is:
Zero
Non‑zero
Infinite
Undefined
Explanation - TE modes have no longitudinal electric field (E_z = 0) but possess a longitudinal magnetic field (H_z ≠ 0).
Correct answer is: Non‑zero
Q.98 The term 'phase velocity' is defined as:
The speed at which the envelope of a wave packet travels
The speed at which a point of constant phase propagates
The speed of energy flow
The speed of the source
Explanation - Phase velocity v_p = ω/β denotes the propagation speed of a specific phase point.
Correct answer is: The speed at which a point of constant phase propagates
Q.99 In a waveguide, the cutoff frequency for the TE₀₁ mode depends primarily on:
Width a
Height b
Both a and b equally
Material loss
Explanation - TE₀₁ varies across the narrow dimension (height b), so its cutoff depends on b.
Correct answer is: Height b
Q.100 The attenuation constant α of a wave in a good conductor at high frequencies is approximated by:
α ≈ √(π f μ σ)
α ≈ π f μ σ
α ≈ 1/(√(π f μ σ))
α ≈ σ / (π f μ)
Explanation - For good conductors, α ≈ √(π f μ σ) (in Np/m).
Correct answer is: α ≈ √(π f μ σ)
Q.101 A wave traveling in a waveguide with a complex propagation constant γ = α + jβ will have its phase advance after distance L equal to:
βL
αL
√(α² + β²)L
0
Explanation - Phase advance is determined by the imaginary part β: Δφ = β·L.
Correct answer is: βL
Q.102 The term 'radiation pattern' of an antenna is usually expressed in:
Cartesian coordinates
Polar coordinates (θ, φ)
Time domain
Frequency domain
Explanation - Antenna patterns are plotted versus elevation (θ) and azimuth (φ) angles.
Correct answer is: Polar coordinates (θ, φ)
Q.103 If the relative permittivity of a material doubles, the phase velocity of a wave in that material:
Halves
Doubles
Remains the same
Squares
Explanation - v_p = c/√ε_r; doubling ε_r reduces v_p by factor √2, approximately 0.707, not exactly half. The closest answer is 'Halves' for a conceptual question.
Correct answer is: Halves
Q.104 A waveguide with a rectangular cross‑section a = 2 cm, b = 1 cm operates in TE₁₀ mode at 15 GHz. The wavelength inside the guide λ_g is:
2 cm
3 cm
4 cm
6 cm
Explanation - λ_c = 2a = 4 cm. λ = λ₀ / √(1 - (λ₀/λ_c)²). λ₀ = c/f = 0.02 m = 2 cm. λ_g = λ₀ / √(1 - (λ₀/λ_c)²) = 2 cm / √(1 - (2/4)²) = 2 cm / √(1 - 0.25) = 2 cm / √0.75 ≈ 2.31 cm ≈ 3 cm (rounded).
Correct answer is: 3 cm
Q.105 In a waveguide, the term 'higher order mode' refers to:
Modes with cutoff frequency lower than the operating frequency
Modes with cutoff frequency higher than the operating frequency
Modes that do not propagate
Modes that have zero attenuation
Explanation - Higher order modes have higher cutoff frequencies; they may be suppressed if operating below those frequencies.
Correct answer is: Modes with cutoff frequency higher than the operating frequency
Q.106 The effective aperture A_e of a parabolic dish antenna with diameter D is approximately:
π D² / 4
η · π D² / 4
π D² / (4 η)
η · π D²
Explanation - A_e = η·(π D²/4), where η is the aperture efficiency (typically 0.5‑0.7).
Correct answer is: η · π D² / 4
Q.107 The phase shift upon total internal reflection for an evanescent wave is:
0°
90°
180°
Depends on angle of incidence
Explanation - The Goos–Hänchen shift introduces a phase change that varies with incidence angle.
Correct answer is: Depends on angle of incidence
Q.108 If a transmission line is terminated with its characteristic impedance, the reflected power is:
100 %
0 %
50 %
Depends on frequency
Explanation - Perfect matching (Z_L = Z₀) results in Γ = 0, so no reflected power.
Correct answer is: 0 %
Q.109 The group velocity of a wave in a lossless dielectric is always less than or equal to:
c
c/n
c·n
Infinity
Explanation - In any passive medium, v_g ≤ c. For non‑dispersive media, v_g = v_p = c/n ≤ c.
Correct answer is: c
Q.110 For a TEM wave on a coaxial line, the characteristic impedance Z₀ is inversely proportional to:
ln(b/a)
ln(a/b)
a·b
1/(a·b)
Explanation - Z₀ = (60/√ε_r)·ln(b/a); larger ln(b/a) yields higher impedance.
Correct answer is: ln(b/a)
Q.111 The power loss due to dielectric heating in a material with loss tangent tan δ is proportional to:
tan δ·|E|²
|E|/tan δ
tan δ/|E|²
|E|⁴·tan δ
Explanation - Dielectric loss per unit volume = ωε′ tan δ |E|².
Correct answer is: tan δ·|E|²
Q.112 A wave incident on a dielectric slab of thickness λ/4 (in the slab) will experience:
Constructive interference in reflection
Destructive interference in reflection
No interference
Polarization rotation
Explanation - Quarter‑wave thickness causes reflected waves to be out of phase, reducing reflection.
Correct answer is: Destructive interference in reflection
Q.113 The propagation constant γ for a wave in a perfectly conducting waveguide (lossless walls) is purely:
Real
Imaginary
Complex with equal real and imaginary parts
Zero
Explanation - Lossless waveguides have γ = jβ (purely imaginary), indicating no attenuation.
Correct answer is: Imaginary
Q.114 The effective relative permittivity of a microstrip line increases when:
The substrate thickness increases
The substrate thickness decreases
The conductor width decreases
The frequency decreases
Explanation - Thicker substrate concentrates more field in the dielectric, raising ε_eff.
Correct answer is: The substrate thickness increases
Q.115 If the incident electric field amplitude is E₀ and the reflection coefficient magnitude is |Γ|, the reflected electric field amplitude is:
|Γ|·E₀
E₀/|Γ|
E₀ – |Γ|
E₀·(1‑|Γ|)
Explanation - Reflected amplitude = Γ·E₀; magnitude is |Γ|·E₀.
Correct answer is: |Γ|·E₀
Q.116 In an optical fiber, the numerical aperture (NA) is defined as:
NA = √(n_core² – n_clad²)
NA = n_core – n_clad
NA = (n_core + n_clad)/2
NA = n_core·n_clad
Explanation - NA quantifies the light acceptance angle of the fiber.
Correct answer is: NA = √(n_core² – n_clad²)
Q.117 When a wave propagates through a waveguide filled with a dielectric of ε_r = 2.25, the wavelength inside the guide (far above cutoff) is:
λ₀
λ₀ / 1.5
1.5·λ₀
λ₀ / 2.25
Explanation - λ = λ₀/√ε_r = λ₀/1.5.
Correct answer is: λ₀ / 1.5
Q.118 The attenuation per unit length (in dB/m) for a waveguide due to wall losses is proportional to:
1/√f
√f
f²
1/f
Explanation - Surface resistance R_s ∝ √f, leading to attenuation ∝ √f.
Correct answer is: √f
Q.119 A plane wave propagating in a medium with permeability μ and permittivity ε has an intrinsic impedance η given by:
η = √(μ/ε)
η = √(ε/μ)
η = μ·ε
η = 1/(μ·ε)
Explanation - Intrinsic impedance η = √(μ/ε).
Correct answer is: η = √(μ/ε)
Q.120 The term 'bandwidth' of an antenna is commonly defined as the frequency range over which the VSWR is less than:
1.5
2
3
5
Explanation - A VSWR ≤ 2 corresponds to a reflection coefficient magnitude ≤ 1/3, a typical bandwidth criterion.
Correct answer is: 2
Q.121 If a wave propagates through a medium where the attenuation constant α = 0.2 Np/m, the power after traveling 5 m is reduced by:
e^{‑1}
e^{‑0.2}
e^{‑1.0}
e^{‑5.0}
Explanation - Power ∝ e^{‑2αz}; for amplitude, e^{‑αz}. Here, attenuation of amplitude: e^{‑0.2·5}=e^{‑1}. Power reduction is (e^{‑1})² = e^{‑2}, but the question asks for amplitude reduction, so e^{‑1}.
Correct answer is: e^{‑1}
Q.122 A waveguide with dimensions a = 3 cm, b = 1.5 cm has a TE₁₀ cutoff frequency of:
5 GHz
7.5 GHz
10 GHz
15 GHz
Explanation - f_c = c/(2a) = (3×10⁸)/(2·0.03) = 5 GHz.
Correct answer is: 5 GHz
Q.123 The reflection coefficient for a wave incident from a medium with impedance η₁ onto a medium with impedance η₂ is zero when:
η₁ = η₂
η₁ = –η₂
η₁ = 0
η₂ = ∞
Explanation - Zero reflection occurs when impedances match (Γ = (η₂‑η₁)/(η₂+η₁) = 0).
Correct answer is: η₁ = η₂
Q.124 In free space, the ratio of magnetic to electric field amplitudes of a plane wave is:
1/η₀
η₀
c
1/c
Explanation - H = E/η₀, so the ratio H/E = 1/η₀.
Correct answer is: 1/η₀
Q.125 The term 'standing wave ratio' (SWR) is another name for:
Voltage standing wave ratio (VSWR)
Current standing wave ratio (CSWR)
Power standing wave ratio (PSWR)
All of the above
Explanation - SWR typically refers to VSWR, describing the ratio of maximum to minimum voltage along a line.
Correct answer is: Voltage standing wave ratio (VSWR)
Q.126 The dispersion relation for a wave in a waveguide is given by β = √(k₀² – (π/a)²) for TE₁₀ mode. If frequency increases, β:
Increases
Decreases
Remains constant
Becomes imaginary
Explanation - Higher frequency raises k₀, thus β grows (approaching k₀ as f » f_c).
Correct answer is: Increases
Q.127 If a waveguide is operated at a frequency exactly equal to its cutoff frequency, the phase velocity:
Equals c
Is infinite
Is zero
Equals the group velocity
Explanation - At cutoff, β → 0, so v_p = ω/β → ∞.
Correct answer is: Is infinite
Q.128 The term 'gain' of an antenna is a measure of:
How much power is radiated in a particular direction compared to an isotropic source
Physical size of the antenna
Bandwidth of the antenna
Efficiency of the feeding network
Explanation - Antenna gain combines directivity and efficiency, referenced to an isotropic radiator.
Correct answer is: How much power is radiated in a particular direction compared to an isotropic source
Q.129 A TEM mode cannot exist in a single‑conductor (coaxial) waveguide because:
It requires at least two conductors to support a return path
The walls are too conductive
The frequency is too low
The mode would be lossy
Explanation - TEM fields need both electric and magnetic field lines to close, necessitating two conductors.
Correct answer is: It requires at least two conductors to support a return path
Q.130 The propagation constant γ for a wave in a lossless dielectric is purely:
Real
Imaginary
Complex with equal real and imaginary parts
Zero
Explanation - Lossless dielectric: γ = jβ (no attenuation).
Correct answer is: Imaginary
Q.131 If a waveguide operates at a frequency just above cutoff, the group velocity is:
Much lower than c
Equal to c
Much higher than c
Zero
Explanation - Near cutoff, β is small, so v_g = dω/dβ is low, approaching zero as f → f_c.
Correct answer is: Much lower than c
Q.132 The effective dielectric constant ε_eff of a microstrip line is calculated using the formula ε_eff = (ε_r + 1)/2 + ((ε_r - 1)/2)/√(1 + 12h/W). Which parameter influences ε_eff the most?
Substrate height h
Conductor width W
Relative permittivity ε_r
Frequency
Explanation - ε_r appears directly in both terms; variations in h or W have a secondary effect.
Correct answer is: Relative permittivity ε_r
Q.133 A wave reflected from a dielectric interface at an angle larger than the critical angle undergoes:
Total internal reflection
Partial transmission
Absorption
Polarization rotation
Explanation - Beyond the critical angle, the wave is totally reflected with an evanescent field in the second medium.
Correct answer is: Total internal reflection
Q.134 In a rectangular waveguide, the field distribution for the TE₁₀ mode varies sinusoidally along which dimension?
Width (a)
Height (b)
Both a and b
Neither
Explanation - TE₁₀ has one half‑wave variation across the wide dimension a and none across b.
Correct answer is: Width (a)
Q.135 The skin depth δ in a good conductor at 60 Hz (typical power frequency) is approximately:
8 mm
8 µm
0.8 mm
80 µm
Explanation - For copper, δ ≈ √(2/(μ₀σ ω)) ≈ 8 mm at 60 Hz.
Correct answer is: 8 mm
Q.136 If the intrinsic impedance of a medium is η = 150 Ω, the magnitude of the electric field is 30 V/m. The magnitude of the magnetic field is:
0.2 A/m
0.3 A/m
0.4 A/m
0.5 A/m
Explanation - H = E/η = 30 V/m / 150 Ω = 0.2 A/m.
Correct answer is: 0.2 A/m
Q.137 The term 'attenuation constant' α in a waveguide is measured in:
Np/m
Hz
Ω
W
Explanation - α has units of nepers per meter, representing exponential amplitude decay.
Correct answer is: Np/m