The electric field part of an electromagnetic wave in vacuum is
`E=3.1(N)/(C)cos[(1.8("rad")/("m"))y+(5.4xx10^(8)("rad")/(s))t]hati`
The wavelength of this part of electromagnetic wave is

Options

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

Answer

D

Explanation

Given `E=3.1 "N C"^(-1)xxcos[(1.8"rad m"^(-1))y+(5.4xx10^(8)"rad s"^(-1)t)]hati" ….(i)"`
Camparing (i) with the equation
`E=E_(0)cos(ky+omegat)" …(ii)"`
We get, `k=1.8"rad m"^(-1)`,
`E_(0)=3.1"N C"^(-1),c=3xx10^(8)"m s"^(-1)`,
`omega=5.4xx10^(8)"rad s"^(-1)` Now, `lambda=(2pi)/(k)=(2xx22)/(1.8xx7)=3.5m`