Electromagnetic Waves

$\overrightarrow E $ is electric field vector, $\overrightarrow B $ is magnetic field vector perpendicular to $\overrightarrow E $. The direction of propagation is $(\overrightarrow E \times \overrightarrow B )$. The direction of propagation of wave is along + y axis, then $\overrightarrow E $ is along + z axis and $\overrightarrow B $ is along + x axis.

$(E\widehat k \times B\widehat i) = EB(\widehat k \times \widehat i) = EB\widehat j$

As wave is travelling along + y axis with time, we will use (y $-$ ct) in wave equation.

Also, intensity is given by

$I = {1 \over 2}c{\varepsilon _0}E_0^2$

And $E = cB$

So, $|{E_0}| = \sqrt {{{2I} \over {c{\varepsilon _0}}}} $

Therefore, $\overrightarrow E = \sqrt {{{2I} \over {c{\varepsilon _0}}}} \cos \left[ {{{2\pi } \over \lambda }(y - ct)} \right]\widehat k$

And $\overrightarrow B = {E \over c}\widehat i \Rightarrow {1 \over c}E\widehat i$

Since $c = f\lambda \Rightarrow f = {c \over \lambda }$. Here, $\lambda = 2|{z_2} - {z_1}|$ and $c = 3 \times {10^8}$. Therefore,

$f = {{3 \times {{10}^8}} \over {2|{z_2} - {z_1}|}} = {{1.5 \times {{10}^8}} \over {|{z_2} - {z_1}|}}$

Given : Frequency $f = {3 \over {2\pi }} \times {10^{12}}$ Hz; direction $ = {{\widehat i + \widehat j} \over {\sqrt 2 }}$.

The beam is polarised along ${\widehat k}$ direction.

Direction of magnetic field $\overrightarrow B = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} \times \widehat k = {{ - \widehat j} \over {\sqrt 2 }} + {{\widehat i} \over {\sqrt 2 }}} \right)$

Therefore, direction of $\overrightarrow B = \left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$

Now from wave equation, we have

$B = {B_0}\cos (kr - \omega t)$

$\omega = 2\pi f = 2\pi \times {3 \over {2\pi }} \times {10^{12}}Hz = 3 \times {10^{12}}Hz$

Now, $k = {{\widehat i + \widehat j} \over {\sqrt 2 }}$;

${B_0} = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$

Therefore,

$B = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\,.\,r - (3 \times {{10}^{12}})t} \right]$

We know $B = {E \over c}$ or $E = Bc$

Now, $[E] = [B][c]$

We know that $[c] = [L{T^{ - 1}}]$

Therefore, $[E] = [B][L{T^{ - 1}}] = [B][L]{[T]^{ - 1}}$

$c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

${c^2} = {1 \over {{\mu _0}{\varepsilon _0}}}$

${\mu _0} = \varepsilon _0^{ - 1}\,.\,{c^{ - 2}}$

$[{\mu _0}] = {[{\varepsilon _0}]^{ - 1}}[{L^{ - 2}}{T^2}]$

We have

${{dE} \over {dz}} = {{ - dE} \over {dt}}$

${{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$

Therefore, $dB = + 2{E_0}k\sin kz\cos \omega t\,dt$

That is, $B = + 20{E_0}k\sin {k_z}\cos \omega t\,dt$

$B = + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt = + 2{E_0}{k \over \omega }\sin kz\sin \omega t} $

Now, ${{{E_0}} \over {{B_0}}} = {\omega \over k} = c$

Therefore, its magnetic field $\overrightarrow B $ is given by $B = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t$

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

The speed of light in vacuum is given by, $c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$ and The impedance (resistance) of free space is defined a $R = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} $. Using this check the dimensional correctness of equations.

(a) ${\mu _0}{I^2} = {\varepsilon _0}{V^2}$

${{{\mu _0}} \over {{\varepsilon _0}}} = {{{V^2}} \over {{I^2}}} = {R^2}$

$\therefore$ ${R^2} = {R^2}$ which is dimensionally correct.

(b) ${\varepsilon _0}I = {\mu _0}V \Rightarrow {{{\varepsilon _0}} \over {{\mu _0}}} = {V \over I} \Rightarrow {1 \over {{R^2}}} = R$

which is dimensionally incorrect.

(c) $I = {\varepsilon _0}cV \Rightarrow {I \over V} = {\varepsilon _0}C = {{{\varepsilon _0}} \over {\sqrt {{\varepsilon _0}{\mu _0}} }}$ = $\sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} $ = $ {1 \over R}$

${1 \over R} = \sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} \Rightarrow {1 \over R} = {1 \over R}$

which is dimensionally correct.

(d) ${\mu _0}cI = {\varepsilon _0}V \Rightarrow {{{\mu _0}c} \over {{\varepsilon _0}}} = {V \over I}$

${{{\mu _0}} \over {{\varepsilon _0}}}{1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} = R \Rightarrow {1 \over {{\varepsilon _0}}}\sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} = R$

${R \over {{\varepsilon _0}}} = R$, which is dimensionally incorrect.

(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).

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).

Duration of pulse, t = 100 ns = 100 $\times$ 10^$-$9 s.

Power of the pulse, P = 30 mW = 30 $\times$ 10^$-$3 W.

Speed of light, c = 3 $\times$ 10⁸ ms^$-$1

The final momentum of the object is

$p = {E \over c} = {{P \times t} \over c} = {{30 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 9}}} \over {3 \times {{10}^8}}} = {10^{ - 17}}$ kg-ms^$-$1