Chapter 8: Electromagnetic Waves

Ans: From the given electric field vector, we can say that the electric field will be directed along the negative x direction. Hence, the direction of motion will be along the negative y direction i.e., $ - j$.

1. What is the Wavelength $\lambda $?

Ans: The general equation for electric field in the positive x direction is given as: \[\overrightarrow E  = {E_0}sin(kx - \omega t)\hat i\]. By comparing the given equation, \[\overrightarrow E  = \;\{ (3.1N/C)\;\left. {cos\;\left[ {(1.8rad/m)\;y\; + (5.4 \times {{10}^6}\;rad/s)t} \right]\;} \right\}\hat i\] with general equation, we get:

Electric field amplitude, ${E_0} = 3.1N/C$

Wave number, $k = 1.8{\kern 1pt} {\kern 1pt} rad/m$

Angular frequency, $\omega  = 5.4 \times {10^6}\;rad/s$

Wavelength, $\lambda  = \frac{{2\pi }}{k} = \frac{{2\pi }}{{1.8}} = 3.490m$.

1. What is the Frequency $\nu $?

Ans: The frequency, $\nu $ of the wave can be given as: $\nu  = \frac{\omega }{{2\pi }}$

$ = \frac{{5.4 \times {{10}^8}}}{{2\pi }} = 8.6 \times {10^7}Hz$.

1. What is the Amplitude of the Magnetic Field Part of the Wave?

Ans: Magnetic field strength can be given as:

${B_0} = \frac{{{E_0}}}{c}$

$\therefore {B_0} = \frac{{3.1}}{{3 \times {{10}^8}}} = 1.03 \times {10^{ - 7}}T$.

1. Write an Expression for the Magnetic Field Part of the Wave.

Ans: We can observe that the magnetic field vector will be directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as: \[\overrightarrow B  = \;{B_0}\cos (ky + \omega t)\hat k\], hence, \[\overrightarrow B  = \;\{ (1.03 \times {10^{ - 7}}T)\;\left. {cos\;\left[ {(1.8rad/m)\;y\; + (5.4 \times {{10}^6}\;rad/s)t} \right]\;} \right\}\hat k\].

Ans: Power rating of bulb, \[P = 100{\text{ }}W\]

About \[5\% \] of its power has been converted into visible radiation.

Therefore, power of visible radiation is given as:

${{p}^{'}} = \frac{5}{{100}} \times 100 = 5W$

Average intensity at distance, $d = 1m$ can be given as:

$I = \frac{{{P}^{'}}}{A} = \frac{{{{p}^{'}}}}{{4\pi {d^2}}}$  

$I = \frac{5}{{4 \times 3.14 \times {{(1)}^2}}} = 0.398{\kern 1pt} W/{m^2}$

1. At a Distance of 10m? Assume That the Radiation is Emitted Isotropically and Neglect Reflection.

Ans: Power of visible radiation can be given as:

${{p}^{'}} = \frac{5}{{100}} \times 100 = 5W$.

Average intensity at distance, $d = 10m$ can be given as:

$I = \frac{{{{p}^{'}}}}{A} = \frac{{{{p}^{'}}}}{{4\pi {d^2}}}$ 

$I = \frac{5}{{4 \times 3.14 \times {{(10)}^2}}} = 0.00398{\kern 1pt} W/{m^2}$

Ans: At a particular temperature, a body produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given by Planck’s law. It can be represented as:

${\lambda _m} = \frac{{0.29}}{T}{\kern 1pt} {\kern 1pt} cm{\kern 1pt} {\kern 1pt} K$.

Where,

${\lambda _m}$ = maximum wavelength

$T$ = temperature

Therefore, the temperature for different wavelengths can be given as:

For, ${\lambda _m} = {10^{ - 4}}cm$, $T = \frac{{0.29}}{{{{10}^{ - 4}}}} = 2900{}^ \circ K$

For, ${\lambda _m} = 5 \times {10^{ - 5}}cm$, $T = \frac{{0.29}}{{5 \times {{10}^{ - 5}}}} = 5800{}^ \circ K$

For, ${\lambda _m} = {10^{ - 6}}cm$, $T = \frac{{0.29}}{{{{10}^{ - 6}}}} = 290000{}^ \circ K$

We can see that, as the maximum wavelength decreases, the corresponding temperature increases. At lower temperature, wavelength produced will not have maximum intensity.

Ans: Wavelength of \[21{\text{ }}cm\] belongs to a short radio wave,

which is present at the end of the electromagnetic spectrum.

1. \[1057{\kern 1pt} {\kern 1pt} MHz\] (Frequency of Radiation Arising from Two Close Energy Levels in Hydrogen; Known as Lamb Shift).

Ans: Frequency of \[1057{\kern 1pt} {\kern 1pt} MHz\] belongs to radio waves. As, radio waves generally belong to the frequency range of $500kHz$ to \[1000MHz\]. The range of wavelength for radio waves is, $ > 0.1m$.

1. \[2.7K\] [Temperature Associated With the Isotropic Radiation Filling All Space-Thought to Be a Relic of the ‘Big-Bang’ Origin of the Universe].

Ans: Given, temperature,  \[T = 2.7K\].

According to Planck's law, \[\lambda  = 0.29\left( {cm} \right){\kern 1pt} {\kern 1pt} T\left( K \right)\]

Substituting the value of temperature, we get: $\lambda  = \frac{{0.29}}{{2.7}} = 0.11cm$.

Now, \[\lambda  = 0.11cm\] belongs to the Microwave region of electromagnetic spectrum.

1. $5890\mathop A\limits^ \circ   - 5896\mathop A\limits^ \circ  $ [double lines of sodium]

Ans: This wavelength belongs to yellow light of visible spectrum.

1. \[{\mathbf{14}}.{\mathbf{4}}{\text{ }}{\mathbf{keV}}\] [energy of a particular transition in \[^{{\mathbf{57}}}{\mathbf{Fe}}\] nucleus Associated With a Famous High Resolution Spectroscopic Method (Mossbauer Spectroscopy)].

Ans: Given: $14.4keV$

The transition energy can be given as: $E = h\nu $.

Where,

$h$ is Planck's constant ,

$\nu $ is frequency of radiation.

Substituting the values, we get:

$\nu  = \frac{E}{h} = \frac{{14.4 \times {{10}^3} \times 1.6 \times {{10}^{ - 19}}}}{{6.6 \times {{10}^{ - 34}}}} = 3.4 \times {10^{18}}Hz$.

Now, $3.4 \times {10^{18}}Hz$ belongs to range of X-ray frequencies. As, range of frequency for X-rays lies between \[1016Hz - 1020Hz\].

1. Answer the Following Questions:

1. Long-Distance Radio Broadcasts Use Short-Wave Bands. Why?

Ans: Short wave bands are used for long distance broadcasting because, after getting reflected from ionosphere present in our environment, these waves can reach to longer distances.

1. It Is Necessary to Use Satellites for Long Distance Tv Transmission. Why?

Ans: Yes, for long-distance TV transmission, it is compulsory to use satellites, because TV signal contains very high frequency of range \[54MHz - 890MHz\], and this much high frequency does not get reflected by ionosphere, and communication does not get stablished. Hence, we need to use satellites for TV transmission, over long distance.

1. Optical and Radio Telescopes Are Built on the Ground but X-Ray Astronomy Is Possible Only from Satellites Orbiting the Earth. Why?

Ans: Light and radio waves used in optical and radio telescopes respectively, penetrate through the atmosphere but X-rays used in X-ray astronomy, are absorbed by the atmosphere. Hence, optical and radio telescopes are built on earth's surface and for X-ray astronomy, satellites orbiting the earth are necessary.

1. The Small Ozone Layer on Top of the Stratosphere is Crucial for Human Survival. Why?

Ans: The ozone layer, present on the top of the stratosphere layer is essential for human survival because it prevents the harmful radiations, (like, ultraviolet rays, $\gamma $ rays, cosmic rays) coming from the sun to earth's atmosphere, which can cause many diseases like cancer and genetic damage in cells and plants. The ozone layer also traps the infrared radiations and maintains earth's warmth.

1. If the Earth Did Not Have an Atmosphere, Would Its Average Surface Temperature Be Higher or Lower Than What it Is Now?

Ans: If the earth would not have any atmosphere, then there will not

be any greenhouse effect. And if the greenhouse effect would not be present, then infrared radiations and $C{O_2}$would not retain in the atmosphere. The infrared radiations and $C{O_2}$ gets reflected from  the atmosphere and keep the earth's surface warm. But in the absence 

of an atmosphere, this would not be possible and the average surface 

temperature of the earth would become lower.

1. 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?

Ans: A global nuclear war on the earth would be followed by a severe ‘nuclear winter’, as the nuclear weapons would create clouds of smoke and dust, which would stop the solar light to reach the earth's surface and would stop the greenhouse effect. This would cause extreme winter conditions.