Electromagnetic Wave Propagation MCQ Quiz - Objective Question with Answer for Electromagnetic Wave Propagation - Download Free PDF

Last updated on Jul 3, 2025

Latest Electromagnetic Wave Propagation MCQ Objective Questions

Electromagnetic Wave Propagation Question 1:

 

- guacandrollcantina.com A plane electromagnetic wave is propagating in the +x direction in a isotropic homogeneous free space. If the electric field vector E oscillates along the +z direction, what is the direction of the associated magnetic field vector B?

  1. +x direction
  2. +y direction
  3. −x direction
  4. −y direction

Answer (Detailed Solution Below)

Option 4 : −y direction

Electromagnetic Wave Propagation Question 1 Detailed Solution

Calculation:

In a plane electromagnetic wave:

EB, and the direction of propagation are mutually perpendicular.

The direction of B is determined using the right-hand rule: E × B = direction of wave.

Direction of wave propagation: +x

Electric field E direction: +z

To satisfy +z × ? = +x, the correct direction for B is -y.

Correct Option: (4)

Electromagnetic Wave Propagation Question 2:

All electromagnetic wave is transporting energy in the negative z direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be the direction of the magnetic field of the wave at that point and instant?

  1. Positive direction of x 
  2. Positive direction of z 
  3. Negative direction of x 
  4. Negative direction of y 

Answer (Detailed Solution Below)

Option 1 : Positive direction of x 

Electromagnetic Wave Propagation Question 2 Detailed Solution

Calculation:

As, Poynting vector

S = E × H

Given: energy transport = negative z-direction

Electric field = positive y-direction

(−) = (+ĵ) × [î]

Hence, according to the vector cross product, the magnetic field should be in the positive x-direction.

Electromagnetic Wave Propagation Question 3:

When a photographic film is exposed to light, the electric field of light causes the film to turn dark after chemical processing. A photographic film of thickness 50 nm is kept inclined to a shiny metal surface at an angle of 𝜃=0.01 radian, as shown in the figure. 
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After exposing this film to a linearly polarized beam of light of wavelength 500 nm incident normally to the metal surface, it developed periodic bright bands. We can explain this observation as the proof of 

  1. Interference between the incident wave and the wave reflected from the surface of the metal.
  2. Diffraction pattern produced by the photographic film. 
  3. Interference of light due to the presence of photographic film.
  4. Polarization of light due to photographic film.

Answer (Detailed Solution Below)

Option 1 : Interference between the incident wave and the wave reflected from the surface of the metal.

Electromagnetic Wave Propagation Question 3 Detailed Solution

Explanation:

The light beam reflects off the metal surface, and some of it interacts with the film on the way down and back up.

Because the film is inclined, the optical path difference between the direct beam through the film and the reflected beam from the metal varies linearly with position. This causes interference fringes — constructive and destructive interference — which are recorded as periodic bright and dark bands on the film. This is a classic interference setup, akin to the Lloyd's mirror or Newton's rings phenomena.

Electromagnetic Wave Propagation Question 4:

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster’s angle. Then:

  1. reflected light is completely polarized and the angle of reflection is close to 60°
  2. reflected light is partially polarized and the angle of reflection is close to 30°
  3. both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to 60° and 30°, respectively
  4. transmitted light is completely polarized with angle of refraction close to 30°

Answer (Detailed Solution Below)

Option 1 : reflected light is completely polarized and the angle of reflection is close to 60°

Electromagnetic Wave Propagation Question 4 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°

1 (6)

Electromagnetic Wave Propagation Question 5:

In the wave equation

y = 0.5 sin \(\frac{2 \pi}{\lambda}\) (400t - x)m

the velocity of the wave will be:

  1. 200 m/s
  2. 200√2 m/s
  3. 400 m/s
  4. 400√2 m/s

Answer (Detailed Solution Below)

Option 3 : 400 m/s

Electromagnetic Wave Propagation Question 5 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.

Top Electromagnetic Wave Propagation MCQ Objective Questions

Which of the following effect proves the wave nature of light?

  1. Photoelectric effect
  2. Compton effect
  3. Pair production
  4. Polarization

Answer (Detailed Solution Below)

Option 4 : Polarization

Electromagnetic Wave Propagation Question 6 Detailed Solution

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The 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
  • The photoelectric effect is the phenomenon where electrons from a metal surface eject when the light of a particular frequency is incident on it.
  • This explains the particle nature of light.
Compton effect
  • Compton effect is the scattering of photons when it collides with a charged particle such as an electron.
Pair production
  • The conversion of a photon into pairs of electron and positron, when it interacts with a strong electric field present around a nucleus is called pair production.
  • Pair production explains the conversion of radiant energy to matter.

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

  1. \(EH~\hat {a_x}\)
  2. \(EH~\hat {a_y}\)
  3. \(EH~\hat {a_x}\hat {a_y}\)
  4. None of the above

Answer (Detailed Solution Below)

Option 4 : None of the above

Electromagnetic Wave Propagation Question 7 Detailed Solution

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Concept:

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 wave

  1. ax
  2. -az
  3. \(\frac{{\left( {{a_x} + {a_y}\;} \right)}}{{\sqrt 2 }}\)
  4. ay

Answer (Detailed Solution Below)

Option 2 : -az

Electromagnetic Wave Propagation Question 8 Detailed Solution

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Concept:

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:

  1. \(\lambda = \frac{3\times10^8}{f}\)
  2. \(\lambda = \frac{3\times10^{10}}{f}\)
  3. \(\lambda = \frac{300}{f}\)
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : \(\lambda = \frac{300}{f}\)

Electromagnetic Wave Propagation Question 9 Detailed Solution

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Wavelength (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

  1. π rad/m
  2. \(\frac{\pi }{4}rad/m\)
  3. \(\frac{\pi }{2}rad/m\)
  4. 2π rad/m

Answer (Detailed Solution Below)

Option 3 : \(\frac{\pi }{2}rad/m\)

Electromagnetic Wave Propagation Question 10 Detailed Solution

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Concept:

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

  1. parallely polarized
  2. Perpendicularly polarized
  3. normally polarized
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : parallely polarized

Electromagnetic Wave Propagation Question 11 Detailed Solution

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The 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:

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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.

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The group velocity of matter waves associated with a moving particle is:

  1. the same as phase velocity
  2. less than the particle velocity
  3. equal to the particle velocity
  4. more than the particle velocity

Answer (Detailed Solution Below)

Option 3 : equal to the particle velocity

Electromagnetic Wave Propagation Question 12 Detailed Solution

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Concept:

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

  1. \(\frac {Vi}{2\pi r l}\)
  2. \(\frac {Vi}{\pi r^2 l}\)
  3. \(\frac {Vi}{2\pi r^3 + 2 \pi rl}\)
  4. \(\frac {Vi}{2\pi r^2 l}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac {Vi}{2\pi r l}\)

Electromagnetic Wave Propagation Question 13 Detailed Solution

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Poynting 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 = \(\frac{V}{l}\times \frac{i}{2\pi r}=\frac{Vi}{2\pi rl}\)

Sun appears red at sunset because

  1. The sun has only red color
  2. Red color is scattered more than other colors
  3. Red color scatters less than other colors
  4. Our eyes are more sensitive to red

Answer (Detailed Solution Below)

Option 3 : Red color scatters less than other colors

Electromagnetic Wave Propagation Question 14 Detailed Solution

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CONCEPT:

  • When light rays travel from one medium to another then there is bending in the light ray. This phenomenon is called as refraction of light.
  • The refraction occurs due to different refractive indexes of the medium for different colours of light.
  • When the light ray strikes on a small molecule of any medium particles then it gets spread in different directions then this phenomenon is called as a scattering of light.
  • White light coming from the sun consists of seven constituent colours.
  • These are – Violet, Indigo, blue, green, yellow, orange and red
  • The red colour light has the longest wavelength among them and Violet has the shortest wavelength.

EXPLANATION:

  • During the sunrise and sunset time, the light ray coming from the sun travel a longer distance in the atmosphere because the sun is on the horizon.
  • The light having the least wavelength will be scattered most and light having the longest wavelength will be scattered least.
  • Since the red and orange colour light has maximum wavelength among all other colours so it will be least scattered in the atmosphere and will come directly to our eye.
  • As the combination of red and orange colours will come directly to our eyes so the sun appears reddish-orange. This is due to the least scattering by the atmosphere. Hence option 3 is correct.

In electromagnetic spectrum visible light lies in between

  1. X-rays and UV
  2. Infrared and microwave
  3. Microwaves and radio waves
  4. UV and infrared

Answer (Detailed Solution Below)

Option 4 : UV and infrared

Electromagnetic Wave Propagation Question 15 Detailed Solution

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The 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:

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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

 

 

 

 

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