Question

The electric field part of an electromagnetic wave in vacuum is
$E=3.1\frac{N}{C}cos\left[\left(1.8\frac{rad}{m}\right)y+\left(5.4\times 10^8\frac{rad}{s}\right)t\right]\hat{i}$
The wavelength of this part of electromagnetic wave is

• A. $1.5m$
• B. $2m$
• C. $2.5m$
• D. $3.5m$

Answer 1

Answer: (D)

Given $E=3.1N{C}^{-1}\times$
$cos\left[\left(1.8rad{m}^{-1}\right)y+\left(5.4\times 10^{8}rad{s}^{-1}\right)t\right]\hat{i}...\left(i\right)$
Comparing $\left(i\right)$ with the equation
$E=E_{0}cos\left(ky+\omega t\right)...\left(ii\right)$
We get, $k=1.8rad{m}^{-1}$,
$E_0=3.1N{C}^{-1}$,
$c=3\times 10^{3}m{s}^{-1}$,
$\omega =5.4\times 10^{8}rad{s}^{-1}$
Now, $\lambda =\frac{2\pi }{k}$
$=\frac{2\times 22}{1.8\times 7}=3.5m$