Electromagnetic Wave Propagation MCQ Quiz - Objective Question with Answer for Electromagnetic Wave Propagation - Download Free PDF
Last updated on May 21, 2025
Latest Electromagnetic Wave Propagation MCQ Objective Questions
Electromagnetic Wave Propagation Question 1:
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster’s angle. Then:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 1 Detailed Solution
Correct option is : (1) Reflected light is completely polarized and the angle of reflection is close to 60°
Using Brewster's law
μ = tan θp
⇒ 1.73 = tan θp
⇒ √3 = tan θp
⇒ θp = 60°
Electromagnetic Wave Propagation Question 2:
In the wave equation
y = 0.5 sin \(\frac{2 \pi}{\lambda}\) (400t - x)m
the velocity of the wave will be:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 2 Detailed Solution
Concept:
Wave Equation:
The general form of a wave equation is y = A sin(kx - ωt), where:
A = Amplitude of the wave
k = Wave number (k = 2π / λ)
ω = Angular frequency (ω = 2πf)
t = Time
x = Position
The wave velocity v can be calculated using the relation:
v = ω / k
Calculation:
Given the wave equation: y = 0.5 sin(2π / λ (400t - x)) m, we can identify the following:
ω = 2π × 400 = 800π rad/s
k = 2π / λ
The velocity of the wave is given by:
v = ω / k = (800π) / (2π / λ) = 400λ
Since the equation is in the standard wave form, we can conclude that the velocity of the wave is 400 m/s.
∴ The velocity of the wave is 400 m/s, which corresponds to Option 3.
Electromagnetic Wave Propagation Question 3:
If skin depth of a conductor with frequency f Hz is d, what will be the new skin depth if the frequency increased to 4f :
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 3 Detailed Solution
Concept:
Skin depth (δ) is the distance into a conductor at which the current density falls to 1/e of its value at the surface due to the skin effect in AC systems.
The formula gives it: \( \delta = \sqrt{\frac{2}{\omega \mu \sigma}} \)
Where,
- ω = 2πf (angular frequency)
- μ = permeability of the material
- σ = conductivity of the material
Explanation:
Since \( \delta \propto \frac{1}{\sqrt{f}} \)Skin depth is inversely proportional to the square root of the frequency.
If the frequency increases from f to 4f, then:
\( \delta_{\text{new}} = \frac{\delta}{\sqrt{4}} = \frac{\delta}{2} \)
Correct Option:
The new skin depth is: d/2
Correct Answer: 3) d/2
Electromagnetic Wave Propagation Question 4:
Light is incident on an interface between water (µ=4/3) and glass (µ=3/2). For total internal reflection, light should be travelling from:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 4 Detailed Solution
The correct answer is - glass to water and ∠ i > ∠ i_c
Key Points
- Total Internal Reflection
- Total internal reflection occurs when light travels from a medium with a higher refractive index (glass) to a medium with a lower refractive index (water).
- The critical angle (i_c) is the angle of incidence in the denser medium at which the refracted ray emerges along the interface.
- For total internal reflection to occur, the angle of incidence (i) must be greater than the critical angle (i_c).
Additional Information
- Refractive Index
- The refractive index (μ) is a measure of how much the speed of light is reduced inside a medium.
- In this context, glass has a refractive index of 3/2 and water has a refractive index of 4/3.
- Critical Angle Calculation
- The critical angle can be calculated using the formula: sin(i_c) = μ2 / μ1, where μ1 is the refractive index of the denser medium (glass) and μ2 is the refractive index of the less dense medium (water).
- Here, sin(i_c) = (4/3) / (3/2) = 8/9, giving a critical angle i_c for the glass to water interface.
- Angle of Incidence
- If the angle of incidence i is greater than i_c, total internal reflection occurs, and light is reflected entirely within the denser medium (glass).
- This phenomenon is utilized in optical fibers and other applications requiring efficient light transmission without loss.
Electromagnetic Wave Propagation Question 5:
An electromagnetic wave is propagating along X -axis. At x = 1cm and t = 18 s, its electric vector |E| = 8V/m, then the magnitude of its magnetic vector is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 5 Detailed Solution
Calculation:
Given:
Electric field magnitude, |E| = 8 V/m
Speed of light, c = 3 × 108 m/s
We know the relation between the electric and magnetic field in an electromagnetic wave:
E / B = c
Rearranging for B:
B = E / c
= (8) / (3 × 108)
= 266 × 10-8 T
∴ The magnitude of the magnetic vector is 266 × 10-8 T.
Top Electromagnetic Wave Propagation MCQ Objective Questions
Which of the following effect proves the wave nature of light?
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 6 Detailed Solution
Download Solution PDFThe correct answer is option 4) i.e. Polarization.
CONCEPT:
- A wave is an oscillation that carries energy from one place to another without transporting matter.
- Light is a combination of electric fields and magnetic fields that are perpendicular to each other.
- So, the two perpendicular planes are occupied by these fields. The electric and magnetic vibrations can simultaneously occur in a number of perpendicular planes.
- Therefore, light is an electromagnetic wave.
- A wave that oscillates in many numbers of planes is called an unpolarized wave.
- Using devices called polarizers, light can be made to vibrate along a single plane. Such light waves are called polarized light.
- Polarization of light occurs when light is reflected, refracted, and scattered.
EXPLANATION:
- The direction of vibration of particles is a property associated with waves. Since light shows the vibration of the electric and magnetic field through polarization, the wave nature of light is concluded from polarization.
Additional Information
Photoelectric effect |
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Compton effect |
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Pair production |
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The electric field intensity E and magnetic field intensity H are coupled and propagating in free space in x and y direction respectively, the Poynting vector is given by
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 7 Detailed Solution
Download Solution PDFConcept:
The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves.
mathematically, the Poynting vector states that:
\(P = \vec E \times \vec H\;Watt/{m^2}\)
Calculation:
Given,
\({\rm{\vec E}} = \left| {{\rm{\vec E}}} \right|{a_x}\)
\(\vec H = \left| {\vec H} \right|{a_y}\)
∴ \(P = \left| {{\rm{\vec E}}} \right|\left| {\vec H} \right|.{\vec a_x} \times {\vec a_y}\)
\(= \left| {{\rm{\vec E}}} \right|\left| {\vec H} \right|{\vec a_z}\)
The plane wave propagating through the dielectric has the magnetic field component as H = 20 e-ax cos (ωt – 0.25x) ay A/m (ax, ay, az are unit vectors along x, y, and z-axis respectively)
Determine the Polarization of the waveAnswer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 8 Detailed Solution
Download Solution PDFConcept:
1) The direction of the electric field is considered as the polarization of the electromagnetic wave.
2) The direction of propagation of EM wave is given by the cross-product of the vector product of direction of electric field and magnetic field, i.e.
\({\hat a_p} = {\hat a_E} \times {\hat a_H}\)
This is an application of the Poynting theorem.
Analysis:
Given:
\(\vec H = 20\;{e^{ - \alpha x}}\cos \left( {\omega t - 0.25x} \right){\hat a_y}\)
\({\hat a_H} = {\hat a_y}\)
\({\hat a_p} = {\hat a_x}\)
\({\hat a_p} = {\hat a_e} \times {\hat a_H} \Rightarrow {\hat a_x} = {\hat a_E} \times {\hat a_y}\)
\({\hat a_E} = - {\hat a_z}\)
Now, the polarization of the wave = direction of the electric field, i.e.
\(= - {\hat a_z}\)The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 9 Detailed Solution
Download Solution PDFWavelength (λ) is equal to the distance traveled by the wave during the time in which any one particle of the medium completes one vibration about its mean position. It is the length of one wave.
Frequency (f) of vibration of a particle is defined as the number of vibrations completed by the particle in one second. It is the number of complete wavelengths traversed by the wave in one second.
The relation between velocity, frequency, and wavelength:
c = f × λ
Where,
c is the speed of light in vacuum = 3 x 108 m/s
\(\lambda = \frac{3\times10^8}{f}\)
But since it is given that the frequency is in MHz, we can write:
\(\lambda = \frac{300\times10^6}{f(MHz)}\)
Since 1 MHz = 106 Hz, the above can be written as:
\(\lambda = \frac{300~MHz}{f(MHz)}\)
\(\lambda = \frac{300}{f}\)
A 50 MHz uniform plane wave is propagating in a material with relative permeability and relative permittivity as 2.25 and 1 respectively. The material is assumed to be loss less. Find the phase constant of the wave propagation
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 10 Detailed Solution
Download Solution PDFConcept:
A uniform plane wave is propagating in the material which is lossless i.e. there is no loss.
Hence R = G = 0, σ = 0 only L and C of the material is considered.
L is represented by μ0μr
And C represented by ε0 εr in the wave.
Propagation velocity, \({V_p} = \frac{\omega }{\beta }\)
And \({V_p} = \frac{C}{{\sqrt {{\mu _r}{\varepsilon _r}} }}\;\;\;\;\;\;\;\left( {C = 3 \times {{10}^8}\;m/s} \right)\)
Calculation:
Now, given that: freq. f = 50 × 106 Hz
Relative permeability μr = 2.25
Relative permittivity εr = 1
Since the material is lossless, σ = 0
To find β (propagation constant):
\({V_p} = \frac{{3\; \times \;{{10}^8}}}{{\sqrt {2.25\; \times \;1} }} = \frac{{2\pi \left( {50\; \times \;{{10}^6}} \right)}}{\beta }\)
\(\Rightarrow \beta = \frac{\pi }{2}rad/s\)
Brewster angle is the angle when a wave is incident on the surface of a perfect dielectric at which there is no reflected wave and the incident wave is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 11 Detailed Solution
Download Solution PDFThe Brewster law states the relationship between light waves and polarized light. The polarized light vanishes at this maximum angle.
Brewster angle is the angle when a wave is an incident on the surface of a perfect dielectric at which there is no reflected wave and the incident wave is partially polarized.
The refracted ray is oriented at a 90-degree angle from the reflected ray and is only partially polarized
Brewster angle is given by:
\(\theta = {\tan ^{ - 1}}\sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}}\)
The British Physicist David Brewster found the relationship between Brewster angle (ip) and refractive index (μ).
μ = tan ip
Brewster's Angle-an angle of incidence at which there is no reflection of p-polarized light at an uncoated optical surface.
Brewster Angle:
EP = parallel polarized component (reflection = 0)
ES = perpendicular polarized compenent
r = reflection coefficient
τ = transmission coefficient
θB = Brewster Angle \(= {\tan ^{ - 1}}\left( {\frac{{{n_2}}}{{{N_1}}}} \right)\)
where n2 = Refractive index of medium 2)
n1 = Refractive index of medium (1)
Important Point-
When referring to polarization states, the p-polarization refers to the polarization plane parallel to the polarization axis of the polarizer being used ("p" is for "parallel"). The s-polarization refers to the polarization plane perpendicular to the polarization axis of the polarizer.
The group velocity of matter waves associated with a moving particle is:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 12 Detailed Solution
Download Solution PDFConcept:
The group velocity of a wave is the velocity with which the overall envelope shape of the wave's amplitudes known as the modulation or envelope of the wave propagates through space.
The group velocity is defined by the equation:
\({v_g} = \frac{{d\omega }}{{dk}}\)
Where ω = wave’s angular frequency
k = angular wave number = 2π/λ
Wave theory tells us that a wave carries its energy with the group velocity. For matter waves, this group velocity is the velocity u of the particle.
Energy of a photon is given by the planck as:
E = hν
With ω = 2πν
ω = 2πE/h ----- (1)
Wave number is given by:
k = 2π/λ = 2πp/h ----(2)
where λ = h/p (de broglie)
Now from equations 1 and 2, we get:
\(d\omega = \frac{{2\pi }}{h}dE;\)
\(dk = \frac{{2\pi }}{h}dp;\)
\(\frac{{d\omega }}{{dk}} = \frac{{dE}}{{dp}}\)
By definition: \({v_g} = \frac{{d\omega }}{{dk}}\)
vg = dE/dp ---- (3)
If a particle of mass m is moving with a velocity v, then
\(E = \frac{1}{2}m{v^2} = \frac{{{p^2}}}{{2m}}\)
\(\frac{{dE}}{{dp}} = \frac{p}{m} = {v_p}\) ---(4)
Now from equations 3 and 4:
vg = vp
A long cylindrical wire of radius r and length l is carrying a current of magnitude i. When the ends are across potential difference V, the pointing vector on the surface of the wire will be
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 13 Detailed Solution
Download Solution PDFPoynting vector (S):
It states that the cross product of electric field vector (E) and magnetic field vector (H) at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point that is
\( \vec S = \vec E \times \vec H\)
S = Poynting vector
E = Electric field and
H = Magnetic field
The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves per unit volume.
Application:
Given,
length = l
radious = r
current = i
Potential = V
Since, electric field (E) is the potential per unit length,
Hence, \(E=\frac{V}{l}\)
For a long straight wire conductor, the magnetic field intensity (H) is given by,
\(H=\frac{i}{2\pi r}\)
Hence, the magnitude of pointing vector (S) will be,
S = EH = \(\frac{V}{l}\times \frac{i}{2\pi r}=\frac{Vi}{2\pi rl}\)
In electromagnetic spectrum visible light lies in between
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 14 Detailed Solution
Download Solution PDFThe electromagnetic spectrum is the range of frequencies (the spectrum) of electromagnetic radiation and their respective wavelengths and photon energies.
The electromagnetic spectrum can be shown in the figure below:
The wavelength and frequency of different colors are shown in the following table:
Sl no. | Colour | Wavelength | Frequency |
---|---|---|---|
1 | Violet | 400 to 440 | 668 THz to 789 THz |
2 | Blue | 460 to 500 | 606 THz to 668 THz |
3 | Green | 500 to 570 | 526 THz to 606 THz |
4 | Red | 620 to 720 | 400 THz to 484 THz |
The coefficient of reflection of voltage for a short circuited line is:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 15 Detailed Solution
Download Solution PDFwhen a transmission line is loaded with impedance it is represented as follows:
V = incident voltage
V’= reflected voltage
V’’= refracted or transmitted voltage.
Transmission line impedance is surge impedance Zs.
Load impedance is ZL.
Short circuited is line is considered when ZL = 0
Coefficient of reflection is given by the expression \({V_{reflection}} = \frac{{V'}}{V}\)
\({V_{reflection}} = \frac{{{Z_L} - {Z_s}}}{{{Z_L} + {Z_S}}}\)
Calculations:
\({{\rm{V}}_{{\rm{reflection}}}} = \frac{{0 - {Z_S}}}{{0 + {Z_S}}}\)
V reflection = -1
Therefore coefficient of reflection of voltage for a short circuited line is -1.