Question

The electric field component of plane wave travelling in a lossless dielectric medium is given by \({\rm{\vec E}}\left( {{\rm{z}},{\rm{f}}} \right) = {{\rm{\hat a}}_{\rm{y}}}2\cos \left( {{{10}^8}{\rm{t}} - \frac{{\rm{z}}}{{\sqrt 2 }}} \right){\rm{V}}/{\rm{m}}\). The wavelength (in m) for the wave is ____________.

Answer 1

Concept: General form of electrical field directed in Positive y-direction is given by \({\rm{\vec E}}\left( {{\rm{z}},{\rm{t}}} \right) = A\cos \left( {{\rm{\omega t}} - {\rm{β z}}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\) where \({\rm{β }} = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}} \)

Application:  Given The electric field

\({\rm{\vec E}} = \left( {{\rm{z}},{\rm{t}}} \right) = 2\cos \left( {{{10}^8}{\rm{t}} - \frac{{\rm{z}}}{{\sqrt 2 }}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\)

Comparing with

\({\rm{\vec E}}\left( {{\rm{z}},{\rm{t}}} \right) = 2\cos \left( {{\rm{\omega t}} - {\rm{β z}}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\)

We have, phase constant, \({\rm{β }} = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}} = \frac{1}{{\sqrt 2 }}\)

\(\Rightarrow {\rm{\lambda }} = 2\sqrt 2 {\rm{\pi }} = 8.88{\rm{m}}\)