Chapter 8 Electromagnetic Waves Solutions
Question - 11 : - Suppose that the electric fieldpart of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y +(5.4 × 106 rad/s)t]}
(a) What isthe direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency ν?
(d) What is the amplitude of themagnetic field part of the wave?
(e) Write an expression for themagnetic field part of the wave..
Answer - 11 : -
(a) Fromthe given electric field vector, it can be inferred that the electric field isdirected along the negative x direction.Hence, the direction of motion is along the negative y directioni.e., 
.
(b) Itis given that,
The general equation for theelectric field vector in the positive x directioncan be written as:
On comparing equations (1) and(2), we get
Electricfield amplitude, E0 = 3.1 N/C
Angularfrequency, ω = 5.4 × 108 rad/s
Wavenumber, k = 1.8 rad/m
Wavelength, 
= 3.490 m
(c) Frequency of wave is given as:
(d) Magneticfield strength is given as:
Where,
c =Speed of light = 3 × 108 m/s
(e) Onobserving the given vector field, it can be observed that the magnetic fieldvector is directed along the negative z direction. Hence, the general equationfor the magnetic field vector is written as:
Question - 12 : - About 5% of the power of a 100 Wlight bulb is converted to visible radiation. What is the average intensity ofvisible radiation
(a) at a distance of 1 m from thebulb?
(b) at a distance of 10 m?
Assumethat the radiation is emitted isotropically and neglect reflection.
Answer - 12 : -
Power rating of bulb, P = 100 W
It isgiven that about 5% of its power is converted into visible radiation.
Power of visible radiation,
Hence, the power of visibleradiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence,intensity of radiation at that point is given as:
(b) Distanceof a point from the bulb, d1 = 10 m
Hence,intensity of radiation at that point is given as:
Question - 13 : - Use the formula λm T= 0.29 cm K to obtain the characteristictemperature ranges for different parts of the electromagnetic spectrum. What dothe numbers that you obtain tell you?
Answer - 13 : -
A body at a particulartemperature produces a continous spectrum of wavelengths. In case of a blackbody, the wavelength corresponding to maximum intensity of radiation is givenaccording to Planck’s law. It can be given by the relation,
Where,
λm =maximum wavelength
T = temperature
Thus, thetemperature for different wavelengths can be obtained as:
For λm = 10−4 cm; 
For λm = 5 ×10−5 cm;
For λm = 10−6 cm; 
and so on.
Thenumbers obtained tell us that temperature ranges are required for obtainingradiations in different parts of an electromagnetic spectrum. As the wavelengthdecreases, the corresponding temperature increases.
Question - 14 : - Given below are some famousnumbers associated with electromagnetic radiations in different contexts inphysics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted byatomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiationarising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associatedwith 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 ofsodium]
(e) 14.4 keV [energy of aparticular transition in 57Fenucleus associated with a famous high resolution spectroscopic method
Answer - 14 : -
(a) Radiowaves; it belongs to the short wavelength end of the electromagnetic spectrum.
(b) Radio waves; it belongs to the shortwavelength end.
(c) Temperature, T =2.7 °K
λm isgiven by Planck’s law as:This wavelength corresponds tomicrowaves.
(d) This is the yellow light of the visiblespectrum.
(e) Transition energy is given by therelation,
E = hν
Where,
h = Planck’s constant = 6.6× 10−34 Js
ν =Frequency of radiation
Energy, E = 14.4 K eV
This corresponds to X-rays.
Answer the following questions:
(a) Long distance radio broadcastsuse short-wave bands. Why?
(b) It is necessary to use satellitesfor long distance TV transmission. Why?
(c) Optical and radio telescopes arebuilt on the ground but X-ray astronomy is possible only from satellitesorbiting the earth. Why?
(d) The small ozone layer on top ofthe stratosphere is crucial for human survival. Why?
(e) If the earth did not have anatmosphere, would its average surface temperature be higher or lower than whatit is now?
(f) Some scientists havepredicted 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 thebasis of this prediction?
Answer - 15 : -
(a) Longdistance radio broadcasts use shortwave bands because only these bands can berefracted by the ionosphere.
(b) It is necessary to use satellitesfor long distance TV transmissions because television signals are of highfrequencies and high energies. Thus, these signals are not reflected by theionosphere. Hence, satellites are helpful in reflecting TV signals. Also, theyhelp in long distance TV transmissions.
(c) With reference to X-rayastronomy, X-rays are absorbed by the atmosphere. However, visible and radiowaves can penetrate it. Hence, optical and radio telescopes are built on theground, while X-ray astronomy is possible only with the help of satellitesorbiting the Earth.
(d) The small ozone layer on the top of theatmosphere is crucial for human survival because it absorbs harmful ultravioletradiations present in sunlight and prevents it from reaching the Earth’ssurface.
(e) In theabsenceof an atmosphere,there would be no greenhouse effect on the surface of the Earth. As a result,the temperature of the Earth would decrease rapidly, making it chilly anddifficult for human survival.
(f) A global nuclear war on thesurface of the Earth would have disastrous consequences. Post-nuclear war, theEarth will experience severe winter as the war will produce clouds of smoke thatwould cover maximum parts of the sky, thereby preventing solar light formreaching the atmosphere. Also, it will lead to the depletion of the ozonelayer.