Text Book Back Questions and Answers - Chapter 5 Electromagnetic Waves 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Electromagnetic Waves
TextualEvaluation Solved
Multiple Choice Questions
Question 1.
The dimension of $\frac{1}{\mu_0 \varepsilon_0}$ is
(a) $\left[\mathrm{LT}^{-1}\right]$
(b) $\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$
(c) $\left[\mathrm{L}^{-1} \mathrm{~T}\right]$
(d) $\left[\mathrm{L}^{-2} \mathrm{~T}^2\right]$
Answer:
(b) $\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$.
Question 2.
If the amplitude of the magnetic field is $3 \times 10^{-6} \mathrm{~T}$, then amplitude of the electric field for a electromagnetic waves is
(a) $100 \mathrm{~V} \mathrm{~m}^{-1}$
(b) $300 \mathrm{~V} \mathrm{~m}^{-1}$
(c) $600 \mathrm{~V} \mathrm{~m}^{-1}$
(d) $900 \mathrm{~V} \mathrm{~m}^{-1}$
Answer:
(d) $900 \mathrm{~V} \mathrm{~m}^{-1}$.

Question 3.
Which of the following electromagnetic radiation is used for viewing objects through fog
(a) microwave
(b) gamma rays
(c) X- rays
(d) infrared
Answer:
(d) infrared.
Question 4.
Which of the following are false for electromagnetic waves
(a) transverse
(b) mechanical waves
(c) longitudinal
(d) produced by accelerating charges
Answer:
(c) longitudinal.
Question 5.
Consider an oscillator which has a charged particle and oscillates about its mean position with a frequency of $300 \mathrm{MHz}$. The wavelength of electromagnetic waves produced by this oscillator is
(a) $1 \mathrm{~m}$
(b) $10 \mathrm{~m}$
(c) $100 \mathrm{~m}$
(d) $1000 \mathrm{~m}$
Answer:
(a) $1 \mathrm{~m}$.
Question 6.
The electric and the magnetic field, associated with an electromagnetic wave, propagating along $X$ axis can be represented by
(a) $\vec{E}=\mathrm{E}_0 \hat{j}$ and $\vec{B}=\mathrm{B}_0 \hat{k}$

(b) $\vec{E}=\mathrm{E}_0 \hat{k}$ and $\vec{B}=\mathrm{B}_0 \hat{j}$
(c) $\vec{E}=\mathrm{E}_0 \hat{i}$ and $\vec{B}=\mathrm{B}_0 \hat{j}$
(d) $\vec{E}=\mathrm{E}_0 \hat{j}$ and $\vec{B}=\mathrm{B}_0 \hat{j}$
Answer:
(b) $\vec{E}=\mathrm{E}_0 \hat{k}$ and $\vec{B}=\mathrm{B}_0 \hat{j}$.
Question 7.
In an electromagnetic wave in free space the rms value of the electric field is $3 \mathrm{~V} \mathrm{~m}^{-1}$. The peak value of the magnetic field is .........
(a) $1.414 \times 10^{-8} \mathrm{~T}$
(b) $1.0 \times 10^{-8} \mathrm{~T}$
(c) $2.828 \times 10^{-8} \mathrm{~T}$
(d) $2.0 \times 10^{-8} \mathrm{~T}$
Answer:
(a) $1.414 \times 10^{-8} \mathrm{~T}$.
Question 8.
A During the propagation of electromagnetic waves in a medium ..........
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density.
Question 9.
If the magnetic monopole exists, then which of the Maxwell's equation to be modified
(a) $\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}_{\text {enclosed }}}{\varepsilon_0}$
(b) $\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=0$
(c) $\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=\mu_0 \mathrm{I}_{\text {enclosed }}+\mu_0 \varepsilon_0 \frac{d}{d t} \int \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}$
(d) $\oint \overrightarrow{\mathrm{E}} \cdot d \vec{l}--\frac{d}{d t} \Phi_{\mathrm{B}}$
Answer:
(b) $\oint \vec{E} \cdot \mathrm{d} \vec{A}=0$

Question 10.
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
(a) $\frac{E}{c}$
(b) $2 \frac{E}{c}$
(c) $\mathrm{Ec}$
(d) $\frac{E}{c^2}$
Answer:
(b) $2 \frac{E}{c}$.
Question 11.
Which of the following is an electromagnetic wave
(a) $\alpha$-rays
(b) $\beta$ - rays
(c) $\gamma$-rays
(d) all of them
Answer:
(c) $\gamma$-rays.
Question 12.
Which one of them is used to produce a propagating electromagnetic wave
(a) an accelerating charge
(b) a charge moving at constant velocity
(c) a stationary charge
(d) an uncharged particle
Answer:
(a) an accelerating charge.

Question 13.

et $\mathrm{E}=\mathrm{E}$
$0 \sin \left[10^6 \mathrm{x}-\omega \mathrm{t}\right]$ be the electric field of plane electromagnetic wave, the value of to is
(a) $0.3 \times 10^{-14} \mathrm{rad} \mathrm{s}^{-1}$
(b) $3 \times 10^{-14} \mathrm{rad} \mathrm{s}^{-1}$
(c) $0.3 \times 10^{14} \mathrm{rad} \mathrm{s}^{-1}$
(d) $3 \times 10^{14} \mathrm{rad} \mathrm{s}^{-1}$
Answer:
(d) $3 \times 10^{14} \mathrm{rad} \mathrm{s}^{-1}$.
Question 14.
Which of the following is NOT true for electromagnetic waves
(a) it transport energy
(b) it transport momentum
(c) it transport angular momentum
(d) in vacuum, it travels with different speeds which depend on their frequency
Answer:
(d) in vacuum, it travels with different speeds which depend on their frequency.
Question 15 .
The electric and magnetic fields of an electromagnetic wave are
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other.

Short Answer Questions
Question 1 .
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.
Question 2.
What are electromagnetic waves?
Answer:
Electromagnetic waves are non-mechanical waves which move with speed equals to the speed of light (in vacuum).
Question 3.
Write down the integral form of modified Ampere's circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
$\oint \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_0 \mathrm{I}_{\text {enclosed }}+\mu_0 \varepsilon_0 \frac{d}{d t} \int_s \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}} \quad$ (Ampere-Maxwell's law)
Question 4.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, $\mathrm{I}=\langle u\rangle \mathrm{c}$
Question 5.
What is meant by Fraunhofer lines?

Answer:
When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.
Long Answer Questions
Question 1.
Write down Maxwell equations in integral form.
Answer:
Maxwell's equations in integral form:
Electrodynamics can be summarized into four basic equations, known as Maxwell's equations. These equations are analogous to Newton's equations in mechanics. Maxwell's equations completely explain the behaviour of charges, currents and properties of electric and magnetic fields. So we focus here only in integral form of Maxwell's equations:
(i) First equation is nothing but the Gauss's law. It relates the net electric flux to net electric charge enclosed in a surface. Mathematically, it is expressed as
$
\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}_{\text {enclosed }}}{\varepsilon_0}
$
Where $\vec{E}$ is the electric field and Qenclosed is the charge enclosed. This equation is true for both discrete or continuous distribution of charges. It also indicates that the electric field lines start from positive charge and terminate at negative charge. This implies that the electric field lines do not form a continuous closed path. In other words, it means that isolated positive charge or negative charge can exist.
(ii) Second equation has no name. But this law is similar to Gauss's law in electrostatics. So this law can also be called as Gauss's law in magnetism. The surface integral of magnetic field over a closed surface is zero. Mathematically, $\oint \vec{B} \cdot \mathrm{d} \vec{A}=0$
where $\vec{B}$ is the magnetic field. This equation implies that the magnetic lines of force form a continuous closed path. In other words, it means that no isolated magnetic monopole exists.
(iii) Third equation is Faraday's law of electromagnetic induction. This law relates electric field with the changing magnetic flux which is mathematically written as $\oint \vec{E} \cdot \mathrm{d} \vec{l}=\frac{d}{d t} \Phi_{\mathrm{B}}$

where $\vec{E}$ is the electric field. This equation implies that the line integral of the electric field around any closed path is equal to the rate of change of magnetic flux through the closed path bounded by the surface.
(iv) Fourth equation is modified Ampere's circuital law. This is also known as ampereMaxwell's law. This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
$
\oint \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_0 \mathrm{I}_{\text {enclosed }}+\mu_0 \varepsilon_0 \int_s \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d \mathrm{~A}}
$
Where $\vec{B}$ is the magnetic field. This equation shows that both conduction and also displacement current produces magnetic field. These four equations are known as Maxwell's equations in electrodynamics.
Question 2.
Write short notes on (a) microwave (b) X-ray (c) radio waves (d) visible spectrum.
Answer:
Microwaves:
It is produced by electromagnetic oscillators in electric circuits. The wavelength range is $1 \times 10^{-3} \mathrm{~m}$ to $3 \times 10^{-1} \mathrm{~m}$ and frequency range is $3 \times 10^{11} \mathrm{~Hz}$ to $1 \times 10^9 \mathrm{~Hz}$. It obeys reflection and polarization. It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.
X-rays:
It is produced when there is a sudden deceleration of high speed electrons at high- atomic number target, and also by electronic transitions among the innermost orbits of atoms.
The wavelength range $10^{-13} \mathrm{~m}$ to $10^{-8} \mathrm{~m}$ and frequency range are $3 \times 10^{21} \mathrm{~Hz}$ to $1 \times 10^{16}$ $\mathrm{Hz}$. X-rays have more penetrating power than ultraviolet radiation.

$\mathrm{X}$-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
Radio waves:
It is produced by oscillators in electric circuits. The wavelength range is $1 \times 10^{-1} \mathrm{~m}$ to $1 \mathrm{x}$ $10^4 \mathrm{~m}$ and frequency range is $3 \times 10^9 \mathrm{~Hz}$ to $3 \times 10^4 \mathrm{~Hz}$. It obeys reflection and diffraction. It is used in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.
Visible light:
It is produced by incandescent bodies and also it is radiated by excited atoms in gases. The wavelength range is $4 \times 10^{-7} \mathrm{~m}$ to $7 \times 10^{-7} \mathrm{~m}$ and frequency range are $7 \times 10^{14} \mathrm{~Hz}$ to $4 \times 10^{14} \mathrm{~Hz}$. It obeys the laws of reflection, refraction, interference, diffraction, polarization, photo-electric effect and photographic action. It can be used to study the
structure of molecules, arrangement of electrons in external shells of atoms and sensation of our eyes.
Question 3.
Discuss briefly the experiment conducted by Hertz to produce and detect electromagnetic spectrum.
Answer:
Production of electromagnetic waves-Hertz experiment: Maxwell's prediction was experimentally confirmed by Heinrich Rudolf Hertz in 1888. The experimental set up used is shown in figure.

It consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns. This is to produce very high electromotive force (emf). Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced. The gap between electrode (ring type - not completely closed and has a small gap in between) kept at a distance also gets spark.

This implies that the energy is transmitted from electrode to the receiver (ring electrode) as a wave, known as electromagnetic waves. If the receiver is rotated by $90^{\circ}$ - then no spark is observed by the receiver. This confirms that electromagnetic waves are transverse waves as predicted by Maxwell. Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light $\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)$.
Question 4.
Explain the Maxwell's modification of Ampere's circuital law.
Answer:
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ' $A$ ' be the area of the capacitor plates and ' $q$ ' be the charge on the plates at any instant ' $t$ ' during the charging process, then the electric field in the gap will be
$
\mathrm{E}=\frac{\sigma}{\varepsilon_0}=\frac{q}{\varepsilon_0 \mathrm{~A}} \text { (or) } \mathrm{EA}=\frac{q}{\varepsilon_0}
$
Flux $\Phi_{\mathrm{E}}=\frac{q}{\varepsilon_0}$
$
\therefore \frac{d \Phi_{\mathrm{E}}}{d t}=\frac{d}{d t}\left(\frac{q}{\varepsilon_0}\right)=\frac{1}{\varepsilon_0}\left(\frac{d q}{d t}\right)
$
But $\frac{d q}{d t}$ is the rate of change of charge on the capacitor plates. It is called displacement current and is given by $\mathrm{I}_{\mathrm{d}}=\frac{d q}{d t}=\varepsilon_0 \frac{d \Phi_E}{d t}$
This is the missing term in Ampere's Circuital Law. The total current must be the sum of the conduction current $\mathrm{Ic}$ and the displacement current ld. Thus, $\mathrm{I}=\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{d}}=\mathrm{I}_{\mathrm{c}}+\varepsilon_0 \frac{d \Phi_E}{d t}$. Hence, the modified form of the Ampere's Law is
$
\oint_c \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_0\left(\mathrm{I}_c+\varepsilon_0 \frac{d \Phi_{\mathrm{E}}}{d t}\right)
$
Question 5.
Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:
(i) Electromagnetic waves are produced by any accelerated charge.
(ii) Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
(iii) Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
(iv) Electromagnetic waves travel with speed which is equal to the speed of light in vacuum or free space, $\mathrm{c}=\frac{1}{\sqrt{\varepsilon_0 \mu_0}} 3 \times 10^8 \mathrm{~ms}^{-1}$

(v) The speed of electromagnetic wave is less than speed in free space or vacuum, that is, $\mathrm{v}<\mathrm{c}$. In a medium of refractive index,
$
\mu_0=\frac{c}{v}=\frac{\frac{1}{\sqrt{\varepsilon_0 \mu_0}}}{\frac{1}{\sqrt{\varepsilon \mu}}} \Rightarrow \mu=\sqrt{\varepsilon_r \mu_r},
$
(vi) Electromagnetic waves are not deflected by electric field or magnetic field.

(vii) Electromagnetic waves can show interference, diffraction and can also be polarized.
(viii) The energy density (energy per unit volume) associated with an electromagnetic wave propagating in vacuum or free space is
$$
\mathrm{u}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2+\frac{1}{2 \mu_0} \mathrm{~B} \varepsilon^2
$$
Where, $\frac{1}{2} \varepsilon_0 \mathrm{E}^2=\mathrm{u}_{\mathrm{E}}$ is the energy density in an electric field and $\frac{1}{2 \mu_0} \mathrm{~B} \varepsilon^2=\mathrm{u}_{\mathrm{B}}$ is the energy density in a managnetic field.
Since, $\mathrm{E}=\mathrm{Bc} \Rightarrow \mathrm{u}_{\mathrm{B}}=\mathrm{u}_{\mathrm{E}}$.
The energy density of the electromagnetic wave is $=\varepsilon \mathrm{E}^2=-\mathrm{B}^2$
(ix) The average energy density for electromagnetic wave, $\langle u\rangle=\frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{1}{2} \frac{1}{\mu_0} \mathrm{~B}^2$
(x) The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, $\mathrm{I}=\langle u\rangle \mathrm{c}$
(xi) Like other waves, electromagnetic waves also carry energy and momentum. For the electromagnetic wave of energy U propagating with speed c has linear momentum which is given by $=\frac{\text { Energy }}{\text { speed }}=\frac{U}{c}$. The force exerted by an electromagnetic wave on unit area speed of a surface is called radiation pressure.
(xii) If the electromagnetic wave incident on a material surface is completely absorbed, then the energy delivered is $U$ and momentum imparted on the surface is $\mathrm{p}=\frac{U}{c}$.
(xiii) If the incident electromagnetic wave of energy $\mathrm{U}$ is totally reflected from the surface, then the momentum delivered to the surface is $\Delta \mathrm{p}=\frac{U}{c}-\left(-\frac{U}{c}\right)=2 \frac{U}{c}$.
(xiv) The rate of flow of energy crossing a unit area is known as pointing vector for electromagnetic waves, which is $\vec{S}=\frac{1}{\mu_0}(\vec{E} \times \vec{B})=\mathrm{c}^2 \varepsilon_0(\vec{E} \times \vec{B})$.The unit for pointing

vector is $\mathrm{W} \mathrm{m}^{-2}$. The pointing vector at any point gives the direction of energy transport from that point.
(xv) Electromagnetic waves carries not only energy and momentum but also angular momentum.
Question 6.
Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves:
Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows.

If the charged particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves.

Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
$
\begin{aligned}
& \mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \sin (\mathrm{kz}-\omega \mathrm{t}) \\
& \mathrm{B}_{\mathrm{x}}=\mathrm{B}_0 \sin (\mathrm{kz}-\omega \mathrm{t})
\end{aligned}
$
Where, $E_0$ and $B_0$ are amplitude of oscillating electric and magnetic field, $k$ is a wave number, $\omega$ is the angular frequency of the wave and $\hat{k}$ (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.
Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between $E_0$ and $B_0$ is equal to the speed of electromagnetic wave, which is equal to speed of light $\mathrm{c}$.
$
\mathrm{c}=\frac{E_0}{B_0}
$
In any medium, the ratio of $\mathrm{E}_0$ and $\mathrm{B}_0$ is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
$
\mathrm{v}=\frac{E_0}{B_0}<\mathrm{c}
$
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.
Question 7.
What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:
(i) Continuous emission spectra (or continuous spectra):
If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum):
Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies.

Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.

(iii) Band emission spectrum (or band spectrum):
Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end.
Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

Question 8.
What is absorption spectra? Give their types.
Answer:
Absorption spectra: When light is allowed to pass through a medium or an absorbing substance then the spectrum obtained is known as absorption spectrum. It is the characteristic of absorbing substance. Absorption spectrum is classified into three types:
(i) Continuous absorption spectrum:
When the light is passed through a medium, it is dispersed by the prism, we get continuous absorption spectrum. For instance, when we pass white light through a blue glass plate, it absorbs everything except blue. This is an example of continuous absorption spectrum.
(ii) Continuous absorption spectrum:
When light from the incandescent lamp is passed through cold gas (medium), the
spectrum obtained through the dispersion due to prism is line absorption spectrum. Similarly, if the light from the carbon arc is made to pass through sodium vapour,-a continuous spectrum of carbon arc with two dark lines in the yellow region of sodium vapour is obtained.

(iii) Band absorption spectrum:
When the white light is passed through the iodine vapour, dark bands on continuous bright background is obtained. This type of band is also obtained when white light is passed through diluted solution of blood or chlorophyll or through certain solutions of organic and inorganic compounds.