91Ó°ÊÓ

Magnetic component of polarized wave

$B_x=\left(4\mathrm{μ}T\right)sin[ky+(2×{10}^{15}{s}^{-1})t]$

By equating theargument of the given magnetic component to zero, find thedirection of the wave. The direction of polarization is always perpendicular to the direction of propagation and magnetic field. Using this concept, find the direction of polarization of the wave. Using the formula for intensity, we can find the intensity of the wave.By substituting the given values in the general expression of electric field,find its expression. Usethe formula for wavelength,and determinethe region of the electromagnetic spectrum.

Formulae:

$c=\frac{E_m}{B_m}$

$I=\frac{E_{rms}^2}{c{\mathrm{μ}}_0}$

$E_z=\left(E_m\right)\sin [ky+\mathrm{Ó\neg }t]$

$\mathrm{λ}=\frac{2\mathrm{π}}{k}$

The magnetic component of polarized light wave is

$B_x=\left(4\mathrm{μ}T\right)sin[ky+(2×{10}^{15}{s}^{-1})t$

To find the direction of wave, we have to equate the argument to zero.

$ky+(2×{10}^{15}{s}^{-1})=0$

Solving for$y$we get

$y=−\left(\frac{2×{10}^{15}{s}^{-1}}{k}\right)$

The negative sign shows that the wave travels in negative$y$ direction.

The direction of propogation of wave is along the y-axis, and the magnetic field is along x-axis. The electric field vector $\stackrel{→}{E}$isperpenduicular to both of these directions.So the electric field vector isalong z-axis.

The electric field vector detemines the direction of polarization of the wave. Thus, the wave is polarized parallel to the z-axis.

Consider the wave speedis related totheelectric field and magnetic field components as

$c=\frac{E_m}{B_m}$

$E_m=C{B}_m$

The magnitude of the magnetic field is.$B_m=4\mathrm{μ}T$

Therefore,

$\begin{array}{c}{E}_m=3×{10}^8\text{ }\frac{\text{m}}{\text{s}}×4×{10}^{-6}\text{â€Í¿}\\ =1.2×{10}^3\text{ }\frac{\text{V}}{\text{m}}\end{array}$

By dividing this value of elecric field by,$\sqrt{2}$ determine the rms value oftheelectric field.

$\begin{array}{c}{E}_{rms}=\frac{E_m}{\sqrt{2}}\\ =\frac{1.2×{10}^3\text{ }\frac{V}{m}}{\sqrt{2}}\\ =848.5\text{ }\frac{V}{m}\end{array}$

Now the intensity is related to the electric field is obtained as:

$\begin{array}{c}I=\frac{E_{rms}^2}{c{\mathrm{μ}}_0}\\ =\frac{{\left(848.5\text{ }\frac{\text{V}}{\text{m}}\right)}^2}{(3×{10}^8\text{ }\frac{\text{m}}{\text{s}})(4\mathrm{Ï€}×{10}^{-7}\text{â€Í¿}â‹\dots \frac{\text{m}}{\text{A}})}\\ =1.91×{10}^3\text{ }\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Intensity of the wave is$1.91×{10}^3\text{ }\frac{\text{W}}{{\text{m}}^{\text{2}}}$

Since the electric field is along z direction, the general expression for the electric field is

$E_z=\left(E_m\right)\sin [ky+\mathrm{Ó\neg }t]$

We know that $\mathrm{Ó\neg }=kc$and

data-custom-editor="chemistry" $\mathrm{Ó\neg }=2×{10}^{15}{s}^{−1}$

Therefore,

$\begin{array}{c}k=\frac{2×{10}^{15}\text{ }{s}^{−1}}{3×{10}^8\text{ }\frac{m}{s}}\\ =6.67×{10}^6\text{ }{m}^{−1}\end{array}$

Substituting these values inthegeneral expression of electric field, we get

$E_z=(1.2×{10}^3\text{ }\frac{V}{m})sin[(6.67×{10}^6{m}^{−1})y+(2×{10}^{15}{s}^{−1})t]$

Expression for the electric field of the wave including the value for angular wave numberis.$E_z=(1.2×{10}^3\text{ }\frac{V}{m})sin[(6.67×{10}^6\text{ }{m}^{−1})y+(2×{10}^{15}{s}^{−1})t]$

The wavelength is given by

$\begin{array}{c}\mathrm{λ}=\frac{2\mathrm{π}}{k}\\ =\frac{2\mathrm{π}}{6.67×{10}^6{m}^{-1}}\\ =942nm\end{array}$

Wavelengthis$942nm.$

From the above calculated wavelength, we can say that the electromagnetic wave lies in the infrared region of light.