Physics Electromagnetic Waves

Infrared radiation is a part of the electromagnetic spectrum which is specifically used in night detection techniques to enhance visibility. They have a wavelength of around 700 nm to 1 mm and have a frequency range of around 10^12 Hz.

There are different types of electromagnetic waves and each of them have different wavelengths. Human eye can only see the light rays which have a minimum wavelength of around 400-700nm. Rays which don't meet this threshold will eventually not be visible to the humans.

It depends on the frequency of the wave. Rays such as Gamma rays, X-Rays etc. have higher frequencies which can lead to harmful impacts if exposed to the human body. These kind of rays cannot be even produced easily unlike low frequency rays.

$v=\frac{1}{\sqrt[]{\epsilon \mu }}=\frac{1}{\sqrt[]{2{\epsilon }_0.2{\mu }_0}}=\frac{1}{2\sqrt[]{{\epsilon }_0{\mu }_0}}=\frac{c}{2}=15*10^{7}m/s.$

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

Since

$\overrightarrow{E}.\overrightarrow{B}=0\left(\overrightarrow{E}\perp \overrightarrow{B}\right)$  

$\&\frac{E_0}{B_0}=C=3*10^{8}m/sec$  

for option D

$\frac{E_0}{B_0}=\frac{\sqrt[]{301.6^2+452.4^2}}{10^{-6}\sqrt[]{1.5^2+1^2}}\begin{array}{cc}\hfill =301.6*10^6\hfill & \hfill =3.01*10^{8}m/sec\hfill \end{array}$  

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$

$\Rightarrow \mathcal{l}=\frac{8.85*10^{-12}*40*10^{-4}*10^6}{4.425*10^{-6}}8*10^{-3}m$

$E_0=2.25\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$ Speed of declromagnetic signal, $v=\frac{E_0}{B_0}=\frac{2.25v/m}{1.5*10^{-8}T}$  

    $B_0=1.5*10^{-8}T\Rightarrow v=\frac{225}{150}*10^8=1.5*10^{8}m/s$  

Time to reach each to the same

radar,   $t=\frac{2*3*10^{3}m}{1.5*10^{8}m/s}=4*10^{-5}S$  

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$  

$\Rightarrow \mathcal{l}=\frac{8.85*10^{-12}*40*10^{-4}*10^6}{4.425*10^{-6}}8*10^{-3}m$  

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22*2*4*10^{-7}}{9*108}$

B₀ = 42.9 * 10^-9

42.9 Ans