The electric field in a plane electromagnetic wave is given by $\overrightarrow{\mathrm{E}}=200 \cos \left[\left(\frac{0.5 \times 10^{3}}{\mathrm{~m}}\right) \times-\left(1.5 \times 10^{11} \frac{\mathrm{rad}}{\mathrm{s}} \times \mathrm{t}\right)\right] \frac{\mathrm{V}}{\mathrm{m}} \hat{\mathrm{j}}$ If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $\frac{\mathrm{x}}{10^{9}} \frac{\mathrm{~N}}{\mathrm{~m}^{2}}$. Find the value of x

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · EM Waves

The electric field in a plane electromagnetic wave is given by

E = 200 cos [(\frac{0.5 \times 1 0 ^{3}}{m}) \times - (1.5 \times 1 0^{11} \frac{rad}{s} \times t)] \frac{V}{m} \hat{j}

If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is

\frac{x}{1 0 ^{9}} \frac{N}{m ^{2}}

. Find the value of x

Answer: 354

Solution

For a perfectly reflecting surface the radiation pressure is

p = \frac{2 I}{c}

, and the intensity of the wave is

I = \frac{1}{2} ε_{0} c E_{0}^{2}

, so

p = \frac{2}{c} \cdot \frac{1}{2} ε_{0} c E_{0}^{2} = ε_{0} E_{0}^{2}

The pressure depends only on the amplitude, not on the area struck or the length of the exposure. With

E_{0} = 200

V/m,

p = 8.85 \times 1 0^{- 12} \times (200)^{2} = 3.54 \times 1 0^{- 7} N m^{- 2} = \frac{354}{1 0 ^{9}} N m^{- 2}

x = 354

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 26 Aug 2021 · Shift 1 · EM Waves

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