A plane EM wave travelling along z - direction is desctribed `vec(E )=E_(0)sin(kz - omega t)hat(i)` and `veC(B)=B_(0)sin (kz - omega)hat(j)`. Show that
The time averaged intensity of the wave is given by `I_(av)=(1)/(2)c in_(0)E_(0)^(2)`.

Explanation

Radiant flux density,
`S=(1)/(mu_(0))(vec(E )xx vec(B))`
`therefore S=c^(2)in_(0)(vec(E )xx vec(B)) " " …(1) [because c=(1)/(sqrt(mu_(0)in_(0)))]`
Let electromagnetic waves propagate in x - direction. Electric field vector in y - direction and magnetic field vector in z - direction. Hence,
`vec(E )=vec(E )_(0)cos(k x-omega t)`
`vec(B)=vec(B)_(0)cos(kx - omega t)`
`vec(E )xx vec(B)=(vec(E )_(0)xx vec(B)_(0))cos^(2)(kx-omega t)`
`S=c^(2)in_(0)(vec(E )xx vec(B))` [`because` From equation (1)]
`c^(2)in_(0)(vec(E )_(0)xx vec(B)_(0))cos^(2)(kx-omega t)`
Average value of magnitude of radiant flux density over complete cycle is,
`S_("average")=c^(2)in_(0)|vec(E )_(0)xx vec(B)_(0)|(1)/(T)int_(0)^(T)cos^(2)(kx-omega t)dt`
`c^(2)in_(0)E_(0)B_(0)xx(1)/(T)xx(T)/(2) " " [because |vec(E )_(0)xx vec(B)_(0)|=E_(0)B_(0)sin 90^(@)=E_(0)B_(0)]`
and `int_(0)^(T)cos^(2)(kx-omega t)dt=(T)/(2)`
`therefore S_("average")=(c^(2))/(2).in_(0)E_(0)((E_(0))/(c )) " " [because c=(E_(0))/(B_(0))rArr B_(0)=(E_(0)^(2))/(c )]`
`=(c )/(2)in_(0)E_(0)^(2)`
`=(c )/(2)xx(1)/(c^(2)mu_(0))xx E_(0)^(2) " " [because c=(1)/(sqrt(mu_(0)in_(0)))rArr in_(0)=(1)/(c^(2)mu_(0))]`
`=(E_(0)^(2))/(2mu_(0)c)`