Electromagnetic Waves

JEE Main level, numerical answer, medium

77 of 165

Question 77 of 165

The electric field in a plane electromagnetic wave is given by
E⃗=200cos⁡[(0.5×103m−1)x−(1.5×1011rads−1)t]Vm−1j^\vec{E}=200\cos \left[\left(0.5\times {10}^{3}{m}^{-1}\right)x-\left(1.5\times {10}^{11}rad{s}^{-1}\right)t\right]V{m}^{-1}\hat{j}.
If this wave falls normally on a perfectly reflecting surface having an area of
100cm2100{cm}^{2}. If the radiation pressure exerted by the E.M. wave on the surface during a 10min⁡10\min exposure is k109Nm−2\dfrac{k}{{10}^{9}}N{m}^{-2}. Find the value of kk

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Answer: 354

E0=200{E}_{0}=200
I=12ϵ0E02cI=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}c
Radiation pressure

P=2lc=(2c)(12ϵ0E02C)=ϵ0E02P=(8.85×10−12)×2002=8.85×10−8×4=354109So,k=354P=\dfrac{2l}{c}=\left(\dfrac{2}{c}\right)\left(\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}C\right)={\epsilon }_{0}{E}_{0}^{2} P=\left(8.85\times {10}^{-12}\right)\times {200}^{2}=8.85\times {10}^{-8}\times 4=\dfrac{354}{{10}^{9}} So, k=354

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