Electromagnetic Waves

Microwave over use the principle of giving vibrational energy to water molecule.

Energy is in terms of increasing wavelength, we have

$\lambda$_X-rays < $\lambda$_{Blue light} < $\lambda$_{Yellow light} < $\lambda$_{Radio waves}

Since $E = {{hc} \over \lambda }$

Therefore, order of increasing energy is

E_{Radio waves} < E_{Yellow light} < E_{Blue light} < E_X-rays

We know that

$ \begin{gathered} E_o=c B_o \Rightarrow B_o=\frac{E_o}{c}=\frac{27}{3 \times 10^8}=9 \times 10^{-8}(\mathrm{~T}) \\\\ \omega=2 \pi f=2 \pi \times 2 \times 10^{14} \mathrm{rad} / \mathrm{s} ; \\\\ K=\frac{2 \pi}{\lambda}=\frac{2 \pi f}{C}=2 \pi \times \frac{2 \times 10^{14}}{3 \times 10^8}=2 \pi \times\left(0.67 \times 10^{-6}\right) / \mathrm{m} \end{gathered} $

Since $\vec{B} \perp$ wave direction (x-axis)

$ \begin{aligned} B & =B_o \sin (K x-\omega t)=B_o \sin 2 \pi\left(\frac{x}{\lambda}-f t\right) \\\\ & =9 \times 10^{-8} \sin 2 \pi\left(\frac{x}{1.5 \times 10^6}-2 \times 10^{14} t\right) \end{aligned} $

Since $\vec{E}$ and $\vec{B}$ are mutually perpendicular. Therefore,

$ \begin{gathered} \vec{E} \times \vec{B}=\vec{c}=c \hat{k} \\\\ \Rightarrow(-2 \hat{i}-3 \hat{j}) \cdot(3 \hat{i}-2 \hat{j})=-6+6=0 \\\\ \Rightarrow(-2 \hat{i}-3 \hat{j}) \times(3 \hat{i}-2 \hat{j})=(6+9) \hat{k}=15 \hat{k} \end{gathered} $

The arrangement in ascending order of wavelength will be

$ \lambda_{\mathrm{G}}<\lambda_{\mathrm{X}}<\lambda_{\mathrm{U}}<\lambda_{\mathrm{v}}<\lambda_{\mathrm{I}}<\lambda_{\mathrm{M}}<\lambda_{\mathrm{R}} $

Since $\quad U_E \times c=I=$ Intensity

$ \begin{aligned} & \frac{1}{2} \varepsilon_0 E^2 \times c=I=\frac{P}{4 \pi R^2}=\frac{3 / 100 \times P}{4 \pi R^2} \\\\ E & =\sqrt{\frac{6 P}{\varepsilon_0 \times c \times 100 \times 4 \pi R^2}} \\\\ & =\sqrt{\frac{6 \times 100}{8.85 \times 10^{-12} \times 3 \times 10^8 \times 100 \times 4 \pi \times(5)^2}} \\\\ & =2.68 \mathrm{~V} / \mathrm{m} \end{aligned} $

(I) These have a wavelength of 589 nm – 589.6 nm that corresponds to visible light in the yellow region. The correct matching is (I)-(A), Visible radiation. The given wavelength range falls within the visible spectrum.

(II) The temperature is 2.7 K, which corresponds to the emission of microwaves. The correct matching is (II)-(B), Microwaves. The given temperature corresponds to the cosmic microwave background radiation.

(III) The wavelength associated is 21 cm, which corresponds to radiowaves of short wavelength. The correct matching is (III)-(C), Short radiowave. The given wavelength of 21 cm falls within the range of radiowaves.

(IV) This radiation has a frequency of 1,057 MHz, which corresponds to radiowaves of high frequency or short wavelength. The correct matching is (IV)-(B), Microwaves. The given frequency falls within the microwave range.

Therefore, the correct answer is Option B : (I)-(A), (II)-(B), (III)-(C), (IV)-(B).

We have

$ E=E_0 \sin (\omega t-k x) $

Since, energy density is $\mu_{\mathrm{E}}=\frac{1}{2} \varepsilon_0 E^2$

$ \begin{aligned} \Rightarrow \bar{\mu}_{\mathrm{E}} & =\frac{1}{2} \varepsilon_0 \bar{E}^2=\frac{1}{2} \varepsilon_0 E_{\mathrm{rms}}^2 \\\\ & =\frac{1}{2} \varepsilon_0 \frac{E_0^2}{2}=\frac{1}{4} \varepsilon_0 E_0^2 \\\\ & =\frac{1}{4} \times 8.8 \times 10^{-12} \times 4^2=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3 \end{aligned} $

Based on the given information, let's match the items from List I with their corresponding descriptions in List II :

(a) Range : 700 nm to 1 mm

Match : (i) Vibration of atoms and molecules

(b) Range : 1 nm to 400 nm

Match : (ii) Inner shell electrons in atoms moving from one energy level to a lower level

(c) Range : <10^-3 nm

Match : (iii) Radioactive decay of the nucleus

(d) Range : 1 mm to 0.1 m

Match : (iv) Magnetron valve

Therefore, the correct matches are :

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv).

$\left( 1 \right)\,\,\,\,\,\,$ Infrared rays are used to treat muscular strain because these are heat rays.

$\left( 2 \right)\,\,\,\,\,\,$ Radio waves are used for broadcasting because these waves have very long wavelength runging from few centimeters to few hundred kilometers

$\left( 3 \right)\,\,\,\,\,\,$ $X$-rays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like bones.

$\left( 4 \right)\,\,\,\,\,\,$ Ultraviolet rays are absorbed by ozone of the atmosphere.

${E_0} = C{B_0}$ and $C = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

Electric energy density $ = {1 \over 2}{\varepsilon _0}{E_0}^2 = {\mu _E}$

Magnetic energy density $ = {1 \over 2}{{{B_0}^2} \over {{\mu _0}}} = {\mu _B}$

Thus, ${\mu _E} = {\mu _B}$

Energy is equally divided between electric and magnetic field

From question,

${B_0} = 20nT = 20 \times {10^{ - 9}}T$

( as velocity of light in vacuum $C = 3 \times {10^8}\,\,m{s^{ - 1}}$ )

${\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C $

$\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \right|.\left| {\overrightarrow C } \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8}$

$ = 6\,\,V/m.$

as The $E.M.$ wave are transverse in nature i.e.,

$ = {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

where $\overrightarrow H = {{\overrightarrow B } \over \mu }$

and ${{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$\overrightarrow k $ is $ \bot \,\,\overrightarrow H $ and $\overrightarrow k $ is also $ \bot $ to $\overrightarrow E $

or In other words $\overrightarrow X ||\overrightarrow E $ and $\overrightarrow k ||\overrightarrow E \times \overrightarrow B $

${E_{rms}} = 720$

The average total energy density

$ = {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2$

$ = 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}$

$ = 4.58 \times {10^{ - 6}}\,J/{m^3}$

Frequency remains constant during refraction

${v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}$

${{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}$

$\therefore$ wavelength is halved and frequency remains unchanged

Momentum of photon $ = {E \over c}$

Change in momentum $ = {{2E} \over c}$

$=$ momentum transferred to the surface

(the photon will reflect with same magnitude of momentum in opposite direction)

The phenomenon of polarisation is shown only by transverse waves.

$\beta $ -rays are fast moving beam of electrons.