8.

ELECTROMAGNETIC WAVES
INTERIORS OF TEXTBOOK
1. What is displacement current?
The displacement current is produced when in an electric circuit (say across the
plates of a parallel plate capacitor) the electric field changes with time.
𝑑𝛷
Formula: id= 𝜀𝑜 𝑑𝑡𝐸
2. A variable frequency ac source is connected to a capacitor. Will the
displacement current increase or decrease with the increase in frequency?
If we increase ω, the reactance of the capacitor will decrease and consequently
conduction current will increase. Since the displacement is equal to the conduction
current the, the displacement current will increase with the increase in frequency of
ac source.
3. Name the electromagnetic radiation to which the following wavelengths
belong : (a) 10–2 m (b) 1 Å
(a) Microwaves (b) X – rays or ϒ – rays.
4. What is the name given to that part of electromagnetic spectrum which is used
for taking photographs of earth under foggy conditions from great heights?
Infrared rays
5. What is the name given to that part of electromagnetic spectrum which is used
in ‘Radar’?
Microwave
6. ’Microwaves are used in Radar’. Why?
Due to short wavelength and less diffraction.
7. Name the electromagnetic waves that have frequencies greater than those of
ultra violet light but less than those of gamma rays.
X – Rays
8. Name the scientist who first (i) predicted the existence of electromagnetic
waves (ii) experimentally demonstrated the existence of electromagnetic
waves.
(i) James Clerk Maxwell (ii) Heinrich Rudolf Hertz.

9. Why is shortwave band used for long distance radio broadcast?

Short waves are used in long distance broadcast because they are reflected by the
ionosphere back to the surface of the earth. This way long distance can be reached.
10. Arrange the given electromagnetic radiations in the descending order of their
frequencies: Infra – red, X – rays, Ultraviolet and Gamma rays.
Gamma rays, X – rays, ultraviolet. Infrared.
11. Why is short wave bands used for long distance transmission of signals?
It is so because ionosphere reflects the waves of short wave bands.
12. Why are T.V. signals not transmitted using sky waves?
T.V signals are not reflected by ionosphere.
13. What is the ratio of speed of infra – red rays and ultra – violet rays in vacuum?
1:1
14. What is the ratio of speed of gamma rays and radio waves in vacuum?
1:1
15. State the reason why ground waves cannot be transmitted above the
frequency range of 1500 kHz.
Due to ‘Attenuation’ Or, Absorption of wave increases with increase in frequency.
16. Transmission of T.V. signals is not possible using sky waves. Why ?
They do not get reflected by ionosphere. Sky waves are not used for the
transmission of T.V signals of range 50MHz to 890MHz because the ionosphere does
not reflect them back to the earth’s surface.
17. Give a reason to show that microwaves are better carriers of signals for long
range transmission than radio waves.
Microwaves are better carriers of signals for long range transmission than radio
waves because
(i) They posses shorter wavelength.
(ii) They show directional properties.
18. Why stationary charges & constant currents do not produce electromagnetic
waves?
A stationary charge & constant current produce a constant electric field & constant
magnetic field respectively. A constant electric field can't generate a magnetic field
likewise a constant magnetic field cannot generate a electric field. Hence, EM waves
can't be produced.
19. If the electric field that constitutes an electromagnetic wave conservative?
Justify your answer.
No, the electric field produced by a time varying magnetic field is non conservative.
So that electric field that constitutes the EM waves is non-conservative.
20. The radio waves, the infrared, the visible ray are EM radiations. Then how are
they different from each other?
They are different because the way they interact with matter is different. Interaction
depends on the energy of the EM waves, which in turn depends upon its frequency
(E=hv).
21. Although in an electromagnetic wave the ratio of the electric field to the
magnetic field is a constant still we say that the vision of our eye is due to only
electric field.
The vision of our eye is due to the force experienced by the moving charge on our
retina. The moving charge experiences force both due to electric & magnetic fields.
FE = qE,
FB = qVB
FE/FB = E/VB
= C/V.
C/V>108 . Therefore moving particle oscillates primarily due to the electric field.
22. Why is the transmission of signals using ground waves restricted to
frequencies less than about 1500 kHz.
Ground wave transmission is restricted to frequencies less than 1500 kHz because
at higher frequencies intensity of the wave decreases.
23. Why is the transmission of signals using sky wave restricted to frequencies
upto 30 MHz?
Frequencies on higher side are not reflected by ionosphere.
24. Why are microwaves used in RADAR?
Because microwaves posses greater energy and least angular speed.

25. Which part of electromagnetic spectrum has largest penetrating power?

Gamma rays.
26. Which part of electromagnetic spectrum has highest frequency?
Gamma rays.
27. Write the frequency limit of visible range of electromagnetic spectrum in kHz.
Frequency limit of visible range of electromagnetic spectrum is 3.9 x 1011 kHz to
7.5x 1011kHz approximately.
28. Write the following radiations in an ascending order in respect of their
frequencies: X – rays, microwaves, ultraviolet rays and radio waves.
Given radiations in the ascending order of their frequencies are: radiowaves,
microwaves, and ultraviolet rays, X – rays.
29. Name the electromagnetic radiations used for studying crystal structure of
solids.
X – rays are used for studying crystal structure of solids.
30. Name the electromagnetic radiations used for viewing objects through haze
and fog.
IR radiations.
31. Why do long distance radio broadcasts use shortwave bands?
Energy of a wave is inversely proportional to wavelength of wave. Hence TV
transmission always uses short wave.
32. What is the range of frequencies used for TV transmission?
Range of frequencies used for TV transmission is 100 MHz – 200MHz.
33. On what factors does the velocity of light in vacuum depend?
It does not depend on anything. It is constant.
34. In terms of the constants µ0 and ε0 what is the expression for the speed of the
light in vacuum.
1
c= 𝜇 𝜀
√ 𝑜 𝑜
𝟏
35. The velocity of e.m. wave in free space is given by: c= How does this
√𝝁𝒐 𝜺𝒐
relation indicate that the velocity of all types of electromagnetic waves is same
in free space?
As constants εo and μo do not depend on the frequency or wavelength of the e.m
waves, velocity of all types of e.m waves is same.
36. What is common between different types of e.m. radiations ?
(i) All e.m waves have transverse nature.
(ii) All e.m waves travel with the same speed i.e, 3 x 108 ms-1 in vaccum.
37. Is the light emitted by an ordinary electric lamp an electromagnetic wave?
Yes, it is an e.m wave.
38. What is the phase relation between electric and magnetic oscillation in an
electromagnetic wave?
They are in the same phase.
39. What type of waves is used in telecommunication?
Microwaves.
40. Radiowaves diffract pronouncedly around building while light waves, which
are also electromagnetic waves, do not. Why?
For diffraction wavelength of a wave should be comparable to the side of the
obstacle.
As wavelength of light waves is very small in comparison to the size of the buildings
and other obstacles, diffraction does not occur.
41. What is the order of magnitude of the frequency of vibration of the longest and
shortest wave in the electromagnetic spectrum?
Vibrations of the longest waves (radiowaves) are of the order of 105 Hz. Frequency
of vibration of the shortest waves is of the order of 1019 to 1024 Hz.
42. For which frequency of light, the human eye is most sensitive?
5.405 x 1014 Hz
43. Which layer of the earth’s atmosphere is useful in long distance radio
transmission?
Ionosphere
44. Give the wavelength range and frequency range of γ – rays.
(i) Wavelength range is 10-4Å to 1.4 Å
(ii) Frequency range is 1018Hz to 1024 Hz.
45. What is the time period of visible light for which human eye is most sensitive?
1 𝜆 5550 ×10−10
T = 𝜈 = 𝑐 = 3×108 = 1.85 x 10-15 s.
46. Why is it necessary to use satellites for long distance T.V. transmission?
(a) Ionosphere does not reflect TV signals because these are of high frequency.
(b) Ground wave transmission is possible only upto a limited range.
Thus, only satellites can be used for long distance TV transmission.
47. What is the approximate wavelength of X – rays?
Wavelength of X – ray is approximately 1 Å.
48. A T.V. tower has a height of h meter. What is the area of earth’s surface over
which its signals can be received?
Area of earth’s surface covered by an antenna of height h is πd2 where d= √2𝑅ℎ.
Here R is earth’s radius.
49. Which part of the electromagnetic spectrum does the wavelength 10–10 m
correspond to?
X – ray spectrum.
50. What oscillates in an electromagnetic wave of frequency 10 MHz and at what
frequency?
Electric and magnetic fields and with frequency 10MHz.
51. An electric charge is oscillating with a frequency of 3 × 1010 Hz. Calculate the
wavelength of the E.M. waves emitted by the oscillating electric charge in
vacuum.
𝑐 3×108
Wavelength = 𝜈 = 3×1010 = 0.01m
52. What physical quantity is same for X – rays of wavelength 1 Å, green light of
wavelength 5500 Å and radiowaves of wavelength 21 cm?
All these travel with the same speed of 3 x 108 ms-1 in vaccum.
53. What does an electromagnetic wave consist of? On what factors does its
velocity in vacuum depend?
An electromagnetic wave consists of electric and magnetic fields varying both in
space and time. The two fields are perpendicular to each other and to the direction
of propagation of the wave. The velocity of electromagnetic wave in vacuum
depends upon the values of its absolute permeability (μo) and permittivity (εo).
54. State the principle of production of e.m waves. What is the value of velocity of
these waves?
The accelerated charge produces electric and magnetic fields, which vary both in
space and time. These varying electric and magnetic fields give rise to
electromagnetic waves. The velocity of em waves in free space is 3 x 108 ms-1.
55. A plane electromagnetic wave travels in vacuum along the Y direction. Write
the (i) ratio of the magnitudes and (ii) the directions of its electric and
magnetic field vectors.
𝐸
➢ 𝐵 = 𝐶, the speed of electromagnetic waves
➢ In an electromagnetic wave propagat5ing along Y direction the electric and
magnetic field vectors vary sinusoidally along X direction and Z direction
respectively.
56. State any four properties of electromagnetic waves.
Properties of electromagnetic waves:
➢ They are transverse in nature.
➢ They do not require any material medium for propagation.
➢ They travel with the same speed of 3 x 108 ms-1 in vacuum.
➢ They consist of mutually perpendicular electric and magnetic fields.
➢ The energy is equally divided between electric and magnetic field.
➢ They are chargeless hence will not get deflected in an electric or magnetic field.
57. Identify the part of electromagnetic spectrum, which is
➢ Suitable for RADAR system used in aircraft navigation – Microwaves
➢ Adjacent to the low frequency end of the electromagnetic spectrum – Radiowaves
➢ Produced in nuclear reaction – Gamma rays
➢ Produced by bombarding a metal target with high speed electrons – X-rays
58. What are microwaves? Write their two uses.
Microwaves are electromagnetic in nature and their wavelength ranges from 10-3 m
to about 1m.
Uses:
➢ These are used in RADAR.
➢ These are used in the study of atomic and molecular structure.
59. What are infra red rays? Write their two uses.
Infra red rays are heat radiations. About 60% of the solar radiations are infra red in
nature. The frequency range of infra red rays is 3 x 1011 to 4 x 1014 Hz.
Uses:
➢ These are used in water heaters and cookers.
➢ These are used to take photographs during conditions of fog, smoke.
60. What are ultraviolet rays? Give their two uses.
The ultra violet rays are part of solar spectrum. Their frequency range is 8 x 1014 to
8 x 1016 Hz.
Uses:
➢ As they destroy bacteria they are used to sterilize surgical instruments.
➢ These are used in burglar alarms.
61. Why can light travel through vacuum, whereas sound cannot do so?
The light waves are electromagnetic in nature. They do not require a material
medium for propagation and can travel in vacuum. On the other hand, the sound
waves require a material medium for propagation. They are mechanical waves and
cannot travel in vacuum.
62. Static crashes are heard in radio, when a lightning flash occurs even if it
occurs far away. Why does this happen?
A lightning flash involves tremendous electrical fields and currents, which oscillate
between the earth and the clouds or between two groups of clouds. In this electrical
activity many charges oscillate and produce a wide variety of electromagnetic
waves. The flashes of light we see are emitted by atoms during this intense activity.
Those electromagnetic waves which have frequencies in radio wave range interfere
with radio signals. Since light and radiowaves travel with the same speed they
arrive at the same time as does the light.
63. Radio waves diffract pronouncedly around building, while light waves which
are electromagnetic waves do not. Why?
For a wave to suffer diffraction, its wavelength should be of the order of size of the
obstacle. The wavelength of radio waves is of the order of the size of the building
and the other obstacles coming in their path and hence they easily get diffracted.
The wavelength of the light waves is very small and as such they are not diffracted
by the buildings.
64. Why is it that induced electric fields due to changing magnetic flux are more
readily observable than the induced magnetic fields due to changing electric
fields?
The induced magnetic fields produced due to displacement currents are of
magnitude too small to be observed. The effect can be increased by increasing the
displacement current. For example, in an a.c circuit, it can be increased by
increasing the displacement current. For example, in an a.c circuit it can be
increased by increasing ω. On the other hand, the induced electric fields due to
changing magnetic flux can be easily increased by simply taking a coil of large turns.
Hence, it is readily observable.
 ⃗⃗ in a region of space, does it necessarily mean that
65. If you find closed loops of 𝑩
charges are flowing across the area bounded by the loops?
It is not necessary. For example, displacement current across the plates of a
capacitor can also produce loops of 𝐵⃗ , but charges are not flowing actually.
66. A closed loop of ⃗⃗⃗
𝑩 is produced by a changing electric field. Does it necessarily
⃗⃗
mean that 𝑩 ⃗⃗ and 𝒅𝑩 are non zero at all points on the loop and in the area
𝒅𝒕
enclosed by the loop?
⃗ , the total electric flux
It is not necessary. For production of the closed loops of 𝐵
through the area enclosed by the loop should change with time. Therefore, change in
flux may arise from a part of the area of the loop.
67. What should be the order of the magnitude of the minimum frequency of
electromagnetic waves that could be used to detect the presence of
(a) The planet Venus.
(b) An aircraft 50m long and
(c) A bird 0.1m long?
From What sources of electromagnetic radiation, would you be able to generate
radiation of these wavelengths?
In order to use a wave phenomenon to detect the presence of some object, the
wavelength of the radiation used must be comparable to or smaller then the
dimensions of the object to be detected.
(a) The planet Venus is about 107 m in diameter. The frequency of the electromagnetic
waves of wavelength 107 m is given by
𝐶 3 𝑥 108
𝜈 = 𝜆 = 107 = 30Hz
The radiation of frequency 30Hz corresponds to very low audio frequency
radiowaves. Practically, it will not be possible to detect the planet Venus by
employing such a radiation, because of the following reasons:
➢ It will be almost impossible to get much power into such a low frequency wave.
➢ Such low frequency radio waves will get absorbed completely in the upper layer of
the atmosphere.
➢ Even if we could send such radio waves to the Venus and receive an echo, its beam
would be so broad that it will not be possible to pinpoint the location of the planet.
(b) The wavelength of the waves required to detect the aircraft is 50m. The
corresponding frequency of the electromagnetic waves is given by,
𝐶 3 𝑥 108
𝜈 = 𝜆 = 50 = 6 x 106Hz = 6MHz
This frequency is higher than that used by AM broadcasting stations. Primitive
radars operated at frequencies about 20 times this frequency.
(c) A 0.1m long bird can be detected by the radiation of frequency,
𝐶 3 𝑥 108
𝜈 = 𝜆 = 0.1 = 3 x 109Hz = 3,000MHz
It is very close to the popular radar frequency. Indeed radars occasionally detect the
flying birds in the sky.
68. Suggest reasons, why (a) food in metal containers cannot be cooked in a
microwave oven, (b) an empty glass container does not get hot in a microwave
oven.
In a microwave oven, the frequency of microwaves is selected to match the
resonance frequency of water molecules, so that the energy from the waves is
transferred efficiently to the kinetic energy of the molecules. This raises the
temperature of any food containing water.
(a) The atoms of the metallic container are set into forced vibrations by the
microwaves. Due to this, energy of the microwaves is not efficiently transferred to
the metallic container. Owing to this, food in metallic containers cannot be cooked in
a microwave oven.
(b) The molecules of the glass container do not respond to the frequency of
microwaves. Due to this, energy from the microwaves is not transferred to the glass
container and hence it does not get hot in a microwave oven.
69. The electric field of a plane electromagnetic wave in vaccum is represented by
Ex = 0, Ey = 0.5 cos[2π x 108(t – x/c)], Ez = 0
i) What is the direction of propagation of electromagnetic wave?
ii) Determine the wavelength of the wave.
iii)Calculate the component of associated magnetic field.
Standard form of the equation of wave moving along +ve x – axis is
EY = EO cos ω(t – x/c)
Compare this equation with the given one
EY = 0.5 cos [2π × 108 (t – x/c)
We find
(a) the wave is moving along the positive direction of x – axis
(b) We have ω = 2π × 108
2πv = 2π × 108
v = 108 s–1
𝑐 3×108
λ = 𝑣 = 108 = 3m
(c) Since, wave is moving along x – axis and electric field is along y – axis the
magnetic field must be along z – axis.
𝐸
As BO = 𝑜
𝑐
0.5 𝑥 𝑥
BZ = 3×108 cos[2𝜋 × 108 (𝑡 − 𝑐 )] = 1.6 × 10-9 cos [2𝜋 × 108 (𝑡 − 𝑐 )]
70. The current in a circuit containing a capacitor is 0.15A. What is the
displacement current and where does it exist?
As conventional current is equal to displacement current. It is also 0.15 A. It exists
across the capacitor plates.
71. A variable frequency a.c source is connected to a capacitor. What will happen
to the displacement current with increase in the frequency of a.c? Explain.
1
We known XC = 2𝜋𝑣𝐶 . With increase in frequency capacitive reactance will decrease.
Hence the current in the circuit (Ic and Id) will decrease.
72. Calculate displacement current between the square plates of side 1cm of a
capacitor of electric field between the plates is changing at the rate of
3×106V/ms.
𝑑𝐸
We know displacement current, ID = 𝜀𝑜 A 𝑑𝑡
𝜀𝑜 = 8.85 × 10–12C2N–1 m–2
A = 10–4 m2
𝑑𝐸
= 3 × 106 V/ms
𝑑𝑡
ID = (8.85×10–12)×(10–4) × (3 × 106) = 2.7 × 10–9 A
73. Does the colour of radiation depend on its frequency or on wavelength.
Frequency.
74. What physical quantity is the same for X-rays of wavelength 1A, green light of
wavelength 5500A0 & radiation of wavelength 21cm?
Speed.
75. Electromagnetic radiations with wavelength:
1): λ1 are used to kill germs in water purifiers.
2): λ2 are used in T.V communication system.
1) λ1 corresponds to ultraviolet spectrum.
2) λ2 corresponds to radio waves.
76. Why stationary charges & constant currents do not produce electromagnetic
waves?
A stationary charge & constant current produce a constant electric field & constant
magnetic field respectively. A constant electric field can't generate a magnetic field
 likewise a constant magnetic field cannot generate a electric field. Hence, EM waves
can't be produced.
77. If the electric field that constitutes an electromagnetic wave conservative?
Justify your answer.
No, the electric field produced by a time varying magnetic field is non conservative.
So that electric field that constitutes the EM waves is non-conservative.
78. The radio waves, the infrared, the visible ray are EM radiations. Then how are
they different from each other?
They are different because the way they interact with matter is different. Interaction
depends on the energy of the EM waves, which in turn depends upon its frequency
(E=hv).
79. Suppose that the electric field of an electromagnetic radiation wave in vacuum
is E=(3.1N/C cos[1.8rad/m]y+5.4 × 106 rad/s)t]
a) What is wavelength, λ?
b). What is frequency, ν?
c). What is magnitude of the magnetic field of the wave?
1): λ=2 π /k= 3.5m.
2): ν=w/2π = 5.4x106/2π = 0.86MHg.
3): B0 = E0/C = 3.1/3x108 = 10nT.
80. Although in an electromagnetic wave the ratio of the electric field to the
magnetic field is a constant still we say that the vision of our eye is due to only
electric field.
The vision of our eye is due to the force experienced by the moving charge on our
retina. The moving charge experiences force both due to electric & magnetic fields.
FE = qE,
FB = qVB
FE/FB = E/VB
= C/V.
C/V>108 . Therefore moving particle oscillates primarily due to the electric field.
81. A parallel capacitor is being charged by an external source. Show that the sum
of conductor current and displacement current has the same value every
where in the circuit.
Electric field between the capacitor plates is given by
σ q
E =ε =ε A
0 0
Where q is the charge accumulated on the positive plate. The electric flux through
this plate is
q
фE = EA = ε A
0
. .. Displacement current,
dф d q dq
Id = ε0 dt = ε0 dt [ε ] = dt
0
But dq/dt = the rate at which charge flows to positive plate through the conducting
wire.
Hence ID = IC
i.e., Displacement current between capacitor plates = Conduction current in
connecting wires.
Thus the sum ID + IC has the same value every where in the circuit.
82. Electromagnetic waves travel in a medium at a speed of 2.0 × 108 ms-1. The
relative permeability of the medium is 1.0. Find the relative permittivity.
Speed of an e.m. wave in a medium is given by
1 1 1 1 c
υ = µ𝜀 = µ µ ε ε = µ ε = µ ε =
√ √ r 0 r 0 √ 0 0 µ ε √ r r √ r r
c2
. .. υ2 =µ
r εr
𝑐2 (3×108 )2
Hence relative permittivity = , εr µ = = 2.25
𝑟 𝑣 2 1.0×(2×108 )2
83. Tabulate the various components of electromagnetic spectrum with the
method of production.
Wavelength Frequency
S.No. Name Source
Range(m) Range (Hz)
1 γ-rays Radiactive nuclei, -4
10 – 10-10 3x1022 – 3x1018
nuclear reactions
2 x-rays High energy
electrons
1x10-10 – 3 x 10-8 3x1018 – 1x1016
suddenly stopped
by a metal target
3 Ultra-violet(UV) Atoms and
molecules in an
6x10-10 – 4x10-7 5x1017 – 8x1014
electrical
discharge
4 Visible light Incandescent
solids Fluorescent 4 x 10-7 – 8x10-7 8x1014 – 4x1014
lamps
5 Infra-red(IR) Molecules of hot
8x10-7 – 3x10-5 4x1014 – 1x1013
bodies
6 Microwaves Electronic device -3
10 – 0.3 3x1011 – 1x109
(Vacuum tube)
7 Radio Charges
frequency waves accelerated
10 – 104 3x107 – 3x104
through
coundcting wires

84. Mention the uses of the various components of electromagnetic spectrum.

Uses of electromagnetic spectrum:
Radio waves: These waves are used in radio and television communication
systems. Am band is from 530 kHz to 1710 kHz. Higher frequencies upto 54 MHz
are used for short waves bands.
Television waves range from 54 MHz to 890 MHz. FM band is from 88 MHz to 108
MHz. Cellular phones use radio waves in ultra high frequency (UHF) band.
Microwaves: Due to their short wavelengths, they are used in radar communication
system. Microwave ovens are an interesting domestic application of these waves.
Infra red waves:
i) Infrared lamps are used in physiotherapy
ii) Infrared photographs are used in weather forecasting
iii) As infrared radiations are not absorbed by air, thick fog, mist etc, they are used
to take photograph of long distance objects.
iv) Infra red absorption spectrum is used to study the molecular structure.
Visible light: Visible light emitted or reflected from objects around us provides
information about the world. The wavelength range of visible light is 4000Å to
8000Å.
Ultra-violet radiations:
i) They are used to destroy the bacteria and for sterilizing surgical instruments.
ii) These radiations are used in detection of forged documents, finger prints in
forensic laboratories.
iii) They are used to preserve the food items.
iv) They help to find the structure of atoms.
X rays:
i) X rays are used as a diagnostic tool in medicine
ii) It is used to study the crystal structure in solids.
γ-rays: Study of γ rays gives useful information about the nuclear structure and it is
used for treatment of cancer.
 TEXT BOOK EXERCISES
1. Suppose that the electric field part of an electromagnetic wave in vacuum is
E={(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}ˆi .
(a) What is the direction of propagation?
(b) What is the wavelength λ ?
(c) What is the frequency ν ?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Here, the electric field part of the electromagnetic wave is given by
𝐸⃗ = (3.1 𝑁 𝐶 −1 ) cos[(1.8 𝑟𝑎𝑑 𝑚−1 )𝑦 + (5.4 × 106 𝑟𝑎𝑑 𝑠 −1 )𝑡] 𝑖̂
It follows that
E0 = 3.1 N C-1; k = 1.8 rad m-1 and  = 5.4 x 106 rad s-1
(a) Since the argument of sine in the expression for electric field is of the type (k y +
 t), the direction of propagation of the electromagnetic wave is along negative Y-
axis.
(b) The wavelength of the wave,
2 2
𝜆 = 𝑘 = 1.8 = 𝟑. 𝟒𝟗 𝒎
(c) The frequency of the wave,
𝑐 3×108
=𝜆= 3.49
= 85.96 × 106 𝐻𝑧 = 𝟖𝟓. 𝟗𝟔 𝑴𝑯𝒛
(d) The amplitude of the magnetic field part of the wave,
𝐸 3.1
𝐵0 = 𝑐0 = 3×108 = 10.33 × 10−9 𝑇 = 𝟏𝟎. 𝟑𝟑 𝒏𝑻
(e) Now, the electric field and the direction of propagation of the e. m. wave are
along negative X-axis and negative Y-axis respectively. Therefore, the magnetic field
is along negative Z-axis and expression for it is given by
⃗⃗ = (𝟏𝟎. 𝟑𝟑 𝒏𝑻) 𝐜𝐨𝐬[(𝟏. 𝟖 𝒓𝒂𝒅 𝒎−𝟏 )𝒚 + (𝟓. 𝟒 × 𝟏𝟎𝟔 𝒓𝒂𝒅 𝒔−𝟏 )𝒕] 𝒌
2. 𝑩 ̂ About 5% of the
power of a 100 W light bulb is converted to visible radiation. What is the
average intensity of visible radiation
(a) at a distance of 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Here, P = 100 W and η = 5 %
Since the efficiency of the bulb is 5 %, the effective power of the bulb,
5
Peff = P x η = 100 x 100 = 5 W
At a distance r, the radiation coming from the bulb is distributed over the surface of
a sphere of radius r i.e. over an area,
A = 4  r2
Therefore, intensity of light at a distance of r,
𝑃 5
𝐼 = 𝑒𝑓𝑓 = 4  𝑟2 ….(i)
𝐴
(a) Here, r = 1 m
From the equation (i), we have
5
𝐼 = 4×12 = 𝟎. 𝟒 𝑾 𝒎−𝟐
(b) Here, r = 10 m
From the equation (i), we have
5
𝐼 = 4×102 = 𝟎. 𝟎𝟎𝟒 𝑾 𝒎−𝟐
3. Use the formula λm T = 0.29 cmK to obtain the characteristic temperature
ranges for different parts of the electromagnetic spectrum. What do the
numbers that you obtain tell you?
Here, b = 0.29 cm K = 0.0029 m K
(i) Wavelength of a -ray photon, λm = 5 x 10-12 m
Therefore, temperature required for the emission of a -ray photon,
 𝑏 0.0029
𝑇 = 𝜆 = 5×10−12 = 𝟓. 𝟖 × 𝟏𝟎𝟖 𝑲
𝑚
(ii) Wavelength of an X-ray photon, λm = 5 x 10-9 m
Therefore, temperature required for the emission of an X-ray photon,
𝑏 0.0029
𝑇 = 𝜆 = 5×10−9 = 𝟓. 𝟖 × 𝟏𝟎𝟓 𝑲
𝑚
(iii) Wavelength of an UV-light photon, λm = 5 x 10-8 m
Therefore, temperature required for the emission of an UV-light photon,
𝑏 0.0029
𝑇 = 𝜆 = 5×10−8 = 𝟓. 𝟖 × 𝟏𝟎𝟒 𝑲
𝑚
(iv) Wavelength of a visible light photon, λm = 5 x 10-7 m
Therefore, temperature required for the emission of a visible light photon,
𝑏 0.0029
𝑇 = 𝜆 = 5×10−7 = 𝟓𝟖𝟎𝟎 𝑲
𝑚
(v) Wavelength of an IR-light photon, λm = 5 x 10-6 m
Therefore, temperature required for the emission of an IR-light photon,
𝑏 0.0029
𝑇 = 𝜆 = 5×10−6 = 𝟓𝟖𝟎 𝑲
𝑚
(vi) Wavelength of a microwave photon, λm = 0.1 m
Therefore, temperature required for the emission of a microwave photon,
𝑏 0.0029
𝑇 = 𝜆 = 0.1 = 𝟎. 𝟐𝟗 𝑲
𝑚
(vii) Wavelength of a radiowave photon, λm = 10 m
Therefore, temperature required for the emission of a radiowave photon,
𝑏 0.0029
𝑇 = 𝜆 = 10 = 𝟎. 𝟎𝟎𝟎𝟐𝟗 𝑲
𝑚
These numbers tell us the order of temperature for obtaining radiations in different
parts of the electromagnetic spectrum. At these temperatures, the radiation of
corresponding wavelength will be emitted with maximum intensity. The wavelength
in a particular case can be emitted at a lower temperature also, but its intensity will
not be maximum.
4. Given below are some famous numbers associated with electromagnetic
radiations in different contexts in physics. State the part of the
electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in
hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-
thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å - 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with
a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
(a)Radiation of wavelength of 21 cm: It corresponds to radiowaves of short
wavelength.
(b) Radiation of frequency of 1057 MHz: It also corresponds to radiowaves of
short wavelength (or of high frequency).
(c) Temperature of 2.7 K: As obtained in Q. 13 (vi), this temperature corresponds
to the emission of microwaves.
(d) Double lines of sodium (5890 Å – 5896 Å): These lines correspond to visible
light in yellow region.
(e) Energy of 14.4 keV: As obtained in Q. 9 (ii), this energy corresponds to X-rays
of low energy (or soft X-rays).
5. Answer the following questions:
(a) Long distance radio broadcasts use short-wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radiotelescopes are built on the ground but X-ray astronomy is
possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human
survival. Why?
(e) If the earth did not have an atmosphere, would its average surface
temperature be higher or lower than what it is now?
(f ) Some scientists have predicted that a global nuclear war on the earth
would be followed by a severe ‘nuclear winter’ with a devastating effect on life
on earth. What might be the basis of this prediction?
(a) The short waves (wavelength less than 200 m or frequencies greater than 1500
kHz) are absorbed by the earth due to their high frequency but are effectively
reflected by F-layer in the ionosphere. After reflection from the ionosphere, the
short waves reach the surface of the earth back only at a large distance from the
transmitter. For this reason, short waves are used in long distance transmission.
(b) Television signals are not reflected by the ionosphere. The TV signals from an
earth station are reflected back to the earth by making use of artificial satellite.
(c) The earth’s atmosphere is transparent to visible light and radiowaves, but
absorbs X-rays. Therefore, X-ray astronomy is possible only from the satellites
orbiting the earth.
(d) The ultraviolet radiation from the sun is harmful to the living cells and plants.
The ozone layer absorbs ultraviolet radiation and prevents it from reaching the
earth. It also keeps the earth warm by trapping infrared radiation.
(e) The infrared radiation emitted by the earth are retained by the earth’s
atmosphere due to the green house effect and this keeps the earth warm. If the earth
did not have atmosphere, its average temperature would have been low.
(f) Scientists estimate that in case of global nuclear war, the clouds produced will
cover probably whole of the sky. In that case, solar radiation would be prevented
from reaching the earth and it will result in, what they call as nuclear winter on the
earth.

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