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

(A) Infra-red radiation

(B) IR

(D) remains same

(A) \(\vec{E}\) × \(\vec{B}\)

(A) h^1/2

(A) Microwave

(A) 32000 m

• 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.

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

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.

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.

No.

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.

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

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.

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).

• 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.

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

(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.

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.

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

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.

• 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.

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.

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.

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.

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

Frequency modulation.

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.

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.

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.

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.

• 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.

Given : λ₀ = 250 m, c_o = 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

Given : ν = 2.0 × 10¹⁸ Hz, c_o = 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 × 10⁻¹⁰ m  = 0.15 × 10⁻⁹ m = 0.15 nm

Given : λ₀ = 6.5 × 10^-7 m, c_o = 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 

Given : ν = 80 GHz = 8.0 × 10⁹ Hz, c_o = 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

Given : ν = 2.0 × 10¹⁰ Hz, c_o = 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 

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

The wavelength of the microwave,

c₀ = \(\frac{E_0}{B_0}\), ∴ E₀ = c_oB₀ = (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.

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

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.

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

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

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

= \(\sqrt{409.6}\) × 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.