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The electric field part of an electromagnetic wave in vacuum is
$E = 3.1\,NC^{-1}\,cos\,[\,(1.8\,rad\,m^{-1})\,y + (5.4\times 18^8\,rad\,s^{-1})\,t\,]\,\hat i$
The wavelength of this part of electromagnetic wave is......$m$
$1.5$
$2$
$2.5$
$3.5$
Solution
Given $\mathrm{E}=3.1 \mathrm{\,N} \mathrm{C}^{-1} \times$
$\cos \left[\left(1.8 \mathrm{rad} \mathrm{\,m}^{-1}\right) \mathrm{y}+\left(5.4 \times 10^{8} \mathrm{rad} \mathrm{\,s}^{-1}\right) \mathrm{t}\right] \hat{\mathrm{i}}$ ……….$(i)$
Camparing $(i)$ with the equation
$\mathrm{E}=\mathrm{E}_{0} \cos (\mathrm{ky}+\omega \mathrm{t})$ ………$(ii)$
We get, $k = 1.8{\mkern 1mu} rad{\mkern 1mu} {m^{ – 1}}{{\rm{E}}_0} = 3.1{\mkern 1mu} {\rm{N}}{{\rm{C}}^{ – 1}},{\rm{c}} = 3 \times {10^8}{\mkern 1mu} {\rm{m}}{{\rm{s}}^{ – 1}},\omega = 5.4 \times {10^8}{\rm{rad}}{\mkern 1mu} {{\rm{s}}^{ – 1}}$
Now, $\quad \lambda=\frac{2 \pi}{k}=\frac{2 \times 22}{1.8 \times 7}=3.5 \mathrm{\,m}$
