JEE MainPhysicsElectromagnetic Waves
The electric field of an electromagnetic wave is given by E = 30 2 (10^7 t - k x - k y) ( i - j ) V/m. The wave is incident on a perfectly absorbing surface lying in the y-z plane. The radiation pressure exerted on the surface is
Options
- A450 ₀
- B1800 ₀
- C900 2 ₀
- D900 ₀
Correct answer
D. 900 ₀
Step-by-step solution
The amplitude of the electric field is E₀ = 30 2 1^2 + (-1)^2 = 30 2 2 = 60 V/m. The intensity of the electromagnetic wave is I = 1 2 ₀ c E₀^2 = 1 2 ₀ c (60)^2 = 1800 ₀ c . From the phase argument (10^7 t - k x - k y) , the wave propagates in the direction of the vector i + j . The unit vector along the direction of propagation is n = i + j 2 . The surface lies in the y-z plane, so its normal vector is N = i . The angle between the direction of propagation and the surface normal is given by = n N = ( i + j 2 ) i =