InterviewSolution
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InterviewSolution
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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 isA. 1.5 mB. 2 mC. 2.5 mD. 3.5 m |
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Answer» Correct Answer - D 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` |
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