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A plane electromagnetic wave travelling along X - direction has electric field of amplitude 300 Vm^(-1), directed along the Y - axis (a) What is the intensity of the wave ? (b) If the wave falls on a perfectly absorbing sheet of area 3.0 m^(2), at what rate is the momentum delivered to the sheet and what is the radiation pressure exerted on the sheet ? [epsilon_(0)=8.854xx10^(-12)C^(2)N^(-1)m^(-2), c=3xx10^(8)ms^(-1)] |
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Answer» Solution :For electromagnetic wave propagation in X - direction, `vec(E )_(0)=300 hat(J)V//m` `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)` `c=3xx10^(8)m//s` (a) Intensity of wave `I=epsilon_(0)c E_(rms)^(2)` `=(epsilon_(0)c E_(0)^(2))/(2)` `=(8.854xx3xx300xx300xx10^(-12)xx10^(8))/(2)` `=119.529 W//m^(2)` (B)Radiation pressure exerted on the sheet `P=("energy (U)")/(c )=(IA)/(c )=(IA)/(c ) ""` (Area `A=3m^(2)`) `=(119.529xx3)/(3xx10^(8))=1.195xx10^(-6)N` Radiation pressure `=("force")/("area")=("momentum"XX"time")/("area")` `therefore` Pressure `P=(1.19xx10^(-6))/(3)=3.98xx10^(-7)Pa` |
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