Solutions-Class-11-Science-Physics-Chapter-13-Electromagnetic Waves and Communication System-Maharashtra Board

Electromagnetic Waves and Communication System

Maharashtra Board-Class-11-Science-Physics-Chapter-13


Question 1. Choose the correct option.

(i) The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to green house effect is

(A) Infra-red radiation

(B) X ray

(C) Microwave

(D) Visible light

Answer :

(A) Infra-red radiation

(ii) Earth’s atmosphere is richest in

(A) UV

(B) IR

(C) X-ray

(D) Microwaves

Answer :

(B) IR

(iii) How does the frequency of a beam of ultraviolet light change when it travels from air into glass?

(A) No change

(B) increases

(C) decreases

(D) remains same

Answer :

(D) remains same

(iv) The direction of EM wave is given by

(A) \(\vec{E}\) × \(\vec{B}\)

(B) \(\vec{E}\)·\(\vec{B}\)  

(C) along \(\vec{B}\)  

(D) along \(\vec{E}\)  

Answer :

(A) \(\vec{E}\) × \(\vec{B}\)

(v) The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to

(A) h1/2

(B) h

(C) h3/2

(D) h2

Answer :

(A) h1/2

(vi) The waves used by artificial satellites for communication purposes are..

(A) Microwave

(B) AM radio waves

(C) FM radio waves

(D) X-rays

Answer :

(A) Microwave

(vii) If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?

(A) 32000 m

(B) 53000 m

(C) 42000 m

(D) 55000 m

Answer :

(A) 32000 m

Question 2. Answer briefly.

(i) State two characteristics of an EM wave.

Answer :

  • EM waves are produced by accelerated electric charges.
  • EM waves can travel through free space as well as through solids, liquids and gases.
  • The self-sustained and mutually perpendicular electric field and magnetic field oscillations of an electromagnetic waves are always perpendicular to the direction of propagation of the wave. Hence, an electromagnetic wave is a transverse wave.

(ii) Why are microwaves used in radar?

Answer :

Microwaves can penetrate haze, light rain and snow, clouds and smoke. Hence, microwaves are used in radar for locating distant ships and aircrafts.

(iii) What are EM waves?

Answer :

An electromagnetic wave is a wave of oscillating electric and magnetic fields emitted by an accelerated electric charge, propagating in space at the speed of light. The fields oscillate in phase at right angles to each other and the direction of propagation, with the maximum and minimum values occurring simultaneously.

(iv) How are EM waves produced?

Answer :

According to Maxwell's electromagnetic theory, an accelerated charge radiates electromagnetic waves.

For example,

  • An electric charge at rest has an electric field in the region around it but has no magnetic field.
  • When the charge moves, it produces both electric and magnetic fields.
  • If the charge moves with a constant velocity, the magnetic field will not change with time, and hence, it cannot produce an EM wave.
  • But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
  • Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

(v) Can we produce a pure electric or magnetic wave in space? Why?

Answer :


Reason : In vacuum, an electric field cannot directly induce another electric field so a "pure" electric field wave cannot exist and same can be said for a "pure" magnetic wave.

(vi) Does an ordinary electric lamp emit EM waves?

Answer :

Yes. An incandescent lamp emits both visible and infrared radiations.

(vii) Why do light waves travel in vacuum whereas sound wave cannot?

Answer :

Light waves do not need a medium for propagation whereas sound waves are

mechanical progressive waves that require a material medium for propagation. Hence, sound cannot travel in vacuum while light can.

(viii) What are ultraviolet rays? Give two uses.

Answer :

UV radiation is electromagnetic radiation emitted in atomic transitions of orbital electrons. It is produced when the substance is at very high temperature.

  • The wavelengths extend from long X-rays up to visible violet light : about 10 nm to 390 nm.
  • Stars, e.g., the Sun, are the main sources of UV radiation. It is also produced by gas discharge tubes (e.g., deuterium, argon), mercury vapour lamps and UV laser diodes.
  • UV radiation is detected by a photographic plate, photoelectric effect (UV photomultipliers) and silicon photodiode.

Uses of ultraviolet radiation :

  • UV radiation is a bactericide and hence used for pest control, disinfecting drinking water, and sterilizing surgical instruments and workspaces.
  • They are used to analyze and authenticate gems.
  • They are used to detect counterfeit currency notes (which do not have the marker dyes or may fluoresce differently).

(ix) What are radio waves? Give its two uses.

Answer :

  • Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
  • Thus, by choosing suitable values of the inductance and the capacitance, radio waves of the desired frequency can be produced.

Uses :

  • Radio waves are used for wireless communication purpose.
  • They are used for radio broadcasting and transmission of TV signals.
  • Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

(x) Name the most harmful radiation entering the Earth's atmosphere from the outer space.

Answer :

High energy ultraviolet radiation. It can accelerate aging of skin and cause skin cancer.

(xi) Give reasons for the following:

         (i) Long distance radio broadcast uses short wave bands.

         (ii) Satellites are used for long distance TV transmission.

Answer :

(i) Long wave, medium frequency radio broadcasts use ground or surface wave propagation, but their range is limited due to energy losses from absorption by Earth and diffraction around obstacles. Therefore, long distance radio broadcasts use short wave high frequency bands and sky wave or ionospheric propagation.

 (ii) TV transmission uses the VHF (30 MHz to 300 MHz) and UHF (300 MHz to 3 GHz) ranges of the FM band. Ground wave propagation due to energy losses is very high, making long-distance communication impossible. Short wave communication is also not possible due to the frequencies being higher than critical frequencies for ionospheric reflection. Therefore, communication satellites are necessary for long-distance TV transmission.

(xii) Name the three basic units of any communication system.

Answer :

There are three basic (essential) elements of every communication system : (1) transmitter (2) communication channel (3) receiver.

  • Electrical transducers, such as a voice coder or microphone and video camera, which generates the basic information-carrying signal.
  • A transmitter first modulates the signal on a radio frequency carrier wave and then transmits it.
  • A communication channel is the path over which the signal is transmitted.
  • A receiver demodulates and recovers the information/message signal. This information signal is then passed to appropriate transducers for conversion into audio or video information for use by a user.

From the transducers at the transmitting end to those at the receiving end, noise can distort original information. The challenge is to keep the noise to the minimum.

(xiii) What is a carrier wave?

Answer :

The high-frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

(xiv) Why high frequency carrier waves are used for transmission of audio signals?

Answer :

Transmitting signal frequencies directly presents several challenges.

  • For efficient radiation and reception, transmitting and receiving antennas must have lengths comparable to a quarter of the wavelength of electromagnetic waves.
  • Vertical antennas are impractical for audio frequencies below 20 kHz, as they cannot radiate the entire spectrum.
  • For example, at 15 kHz, the antenna would have to be 5 km long, whereas at 1 MHz (in the broadcast band), it is 75 m. Even a 5 km long antenna will not be able to efficiently radiate the entire spectrum of music signals.
  • To confine signals into a narrow frequency band, modulation is used.
  • Audio signals have low frequencies, which cannot be transmitted over large distances.
  • All audible sound lies within the 20 Hz to 20 kHz range, so signals from different sources would be mixed up.

Hence to address these issues, high radio frequency (RF) carrier waves are used to modulate audio signals before transmission. This ensures efficient transmission of audio signals over a given channel.

(xv) What is modulation?

Answer :

  • Modulation is the process of superimposing a baseband or basic information-carrying signal, such as the output from a voice coder or video camera, on a radio frequency (RF) carrier wave for transmission.
  • In this, some characteristic of a high-frequency carrier wave, such as amplitude, frequency or phase, is varied in accordance with the baseband signal.

(xvi) What is meant by amplitude modulation?

Answer :

Amplitude modulation (AM) is a method where the amplitude of a radio frequency carrier wave is adjusted proportionally to the amplitude of a baseband signal at a frequency equal to the modulating baseband signal. The frequency of the modulated wave is the same as that of the carrier wave.

(xvii) What is meant by noise?

Answer :

Noise, as a term used in a communication system, is unwanted, randomly fluctuating electric energy that tends to distort a signal at any point during communication.

(xviii) What is meant by bandwidth?

Answer :

Bandwidth :- The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently. It is measured in hertz.

  • The information communicated by a signal in a communication system can be of various types, such as voice, music, picture or binary data.
  • Depending on the type of information, a signal has different range of frequencies. For example, voice or human speech ranges from about 300 Hz to about 3100 Hz. The audible range of frequencies in humans is from about 20 Hz to about 20000 Hz. Video signals to transmit visual or picture information require a bandwidth of about 4.2 MHz.

Communication systems have specific bandwidths for efficient operation, affecting various electronic circuits and channels.

Point-to-point communication lines have bandwidths ranging from 2 MHz to 100 GHz, while the International Telecommunication Union allocates and monitors different frequency bands for various communication types, including broadcasts, TV transmission, cellular communication, and satellite communication.

(xix) What is demodulation?

Answer :

The process of recovering the modulating baseband signal from a modulated wave is called demodulation.

(xx) What type of modulation is required for television broadcast?

Answer :

Frequency modulation.

(xxi) How does the effective power radiated by an antenna vary with wavelength?

Answer :

The maximum power radiated from a straight antenna of length L is proportional to \((\frac{L}{λ})^2\) where λ is the wavelength of the radiated radio wave. Therefore, the maximum power that can be radiated from an antenna of given length increases as λ decreases or frequency increases. The higher the power radiated, the greater is the transmission range. Hence, it is desirable that the radio waves should have short wavelength i.e., high frequency.

(xxii) Why should broadcasting programs use different frequencies?

Answer :

A receiver in a communication system selectively picks up a desired modulated wave

from a communication channel, detects the baseband signal, and feeds it after amplification to an appropriate electrical transducer for conversion into a nonelectrical signal, such as sound, light, etc. for the user of information.

Different broadcasting stations are allotted different, well separated, carrier frequencies for distinct reception.

(xxiii) Explain the necessity of a carrier wave in communication.

Answer :

High frequency carrier waves are used to modulate information signals to overcome transmission difficulties.

  • To achieve efficient radiation and reception, transmitting and receiving antennas should have lengths comparable to a quarter of the wavelength of em waves.
  • Radio frequency carrier waves allow for practical and convenient antennas.
  • The maximum power radiated from an antenna increases with frequency, increasing transmission range.
  • Modulated message signals can be assigned to appropriate passbands in the channel, allowing multiple messages to be transmitted without mixing.
  • By choosing the right carrier frequency and modulation technique, information signals can be translated into the appropriate communication channel passband.

(xxiv) Why does amplitude modulation give noisy reception?

Answer :

An amplitude modulated wave is susceptible to noise which distorts the transmitted signal at any point during communication. A receiver cannot distinguish between the amplitude variations representing a noise and those of a modulated carrier. Hence, amplitude modulation gives noisy reception.

(xxv) Explain why is modulation needed.

Answer :

  • Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters.
  • Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

Question 2. Solve the numerical problem.

(i) Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 108 m/s.

Answer :

Given : λ0 = 250 m, co = 3.0 x 108 m/s

The frequency of the radio wave,

ν = \(\frac{c_0}{λ_0}\) = \(\frac{3.0×10^8}{250}\) = 1.2 × 106 Hz  = 1.2 MHz

(ii) Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 1018 Hz.

Answer :

Given : ν = 2.0 × 1018 Hz, co = 3.0 x 108 m/s

The wavelength of the X-ray wave,

λ0 = \(\frac{c_0}{ν}\) = \(\frac{3.0×10^8}{2.0×10^{18}}\) = 1.5 × 1010 m  = 0.15 × 10−9 m = 0.15 nm

(iii) The speed of light is 3 × 108 m/s. Calculate the frequency of red light of wavelength of 6.5 × 10-7 m.

Answer :

Given : λ0 = 6.5 × 10-7 m, co = 3.0 x 108 m/s

The frequency of the red light,

ν = \(\frac{c_0}{λ_0}\) = \(\frac{3.0×10^8}{6.5×10^{-7}}\) = 4.615 × 1014 Hz 

(iv) Calculate the wavelength of a microwave of frequency 8.0 GHz.

Answer :

Given : ν = 80 GHz = 8.0 × 109 Hz, co = 3.0 x 108 m/s

The wavelength of the microwave,

λ0 = \(\frac{c_0}{ν}\) = \(\frac{3.0×10^8}{8.0×10^9}\) = 3.75 × 10−2 m = 3.75 cm

(v) In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 1010 Hz. What is the wavelength of the wave?

Answer :

Given : ν = 2.0 × 1010 Hz, co = 3.0 x 108 m/s

The wavelength of the microwave,

λ0 = \(\frac{c_0}{ν}\) = \(\frac{3.0×10^8}{2.0×10^{10}}\) = 1.5 × 10−2 m 

(vi) The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 × 10-7 T. What is the amplitude of the electric field part of the wave?

Answer :

Given :  B0 = 5 × 10-7 T, co = 3.0 x 108 m/s

The wavelength of the microwave,

c0 = \(\frac{E_0}{B_0}\), ∴ E0 = coB0 = (3.0 x 108)( 5 × 10-7) = 150 N/C = 150 V/m

150 V/m is the amplitude of the electric field part of the wave.

(vii) A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km2? (Radius of the Earth = 6.4 × 106 m)

Answer :

Given : h = 200 m, R = 6.4 x 106 m,

average population density = 1000/km2 = 10-3/m2

Range, d = \(\sqrt{2Rh}\) = (\sqrt{2×6.4×10^6×200}\) = 5.06 × 104 m

∴ Area covered = πd2 = 3.142 x (5.06 x 104)2 = 80.44 × 108 m2

∴ The population covered by TV transmission = 10-3 x 80.44 × 108 = 8.044 × 106

(viii) Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?

Answer :

Given : h = 600 m, R = 6.4 x 106 m,

(i) The coverage range of the TV tower,

d = \(\sqrt{2Rh}\) = \(\sqrt{2×6.4×10^6×600}\) = 8.764 × 104 m = 87.64 km

(ii) Coverage area, A = πd2

∴ A1 = πd12 and A2 = πd22

∴ \(\frac{A_2}{A_1}\) = \(\frac{d_2^2}{d_1^2}\) = \(\frac{2Rh_2}{2Rh_1}\) = \(\frac{h_2}{h_1}\)

For \(\frac{A_2}{A_1}\) = 2,

h2 = 2h1 = 2 x 600 = 1200 m

This is the required height of a TV tower.

(ix) A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode ? Given radius of Earth is 6.4 × 106 m.

Answer :

Given : h1 = 32 m, h2 = 50 m, R = 6.4 x 106 m

dmax = \(\sqrt{2Rh_1}\) + \(\sqrt{2Rh_2}\)

= \(\sqrt{2×6.4×10^6×32}\) + \(\sqrt{2×6.4×10^6×50}\)

= \(\sqrt{409.6}\) × 103 + \(\sqrt{640}\) × 103

=  20.24 × 103 m + 25.3 × 103 m

= 45.54 × 103 m = 45.54 km

This is the required maximum distance.

Rs 16

-Kitabcd Academy Offer-

Buy Notes(Rs.10)+ Solution(Rs.10) PDF of this chapter
Price : Rs.20 / Rs.16

Click on below button to buy PDF in offer (20% discount)

Useful Links

Main Page : – Maharashtra Board Class 11th-Physics  – All chapters notes, solutions, videos, test, pdf.

Previous Chapter : Chapter 12: Magnetism – Online Solutions

Next Chapter : Chapter 14: SemiconductorsOnline Solutions

Leave a Reply

Write your suggestions, questions in comment box

Your email address will not be published. Required fields are marked *

We reply to valid query.