JEE Main201815 Apr 2018Evening ShiftPhysicsElectromagnetic WavesActual
A plane polarized monochromatic EM wave is travelling a vacuum along z direction such that at t = t ₁ it is found that the electric field is zero at a spatial point z₁ . The next zero that occurs in its neighbourhood is at z₂ . The frequency of the electromagnetic wave is:
Options
- A3 10^8 |z₂-z₁ |
- B6 10^8 | z ₂- z ₁ |
- C1.5 10^8 | z ₂- z ₁ |
- D1 t₁+ |z₂-z₁ | 3 10₈
Correct answer
A. 3 10^8 |z₂-z₁ |
Step-by-step solution
Using E=E₀-e^i(k z- t) Given, at t=t₁, z=z₁, E=0 the next zero that occurs in it's neighborhood is at z₂ , the frequency of the electromagnetic wave at t₂ aligned &e^ i (k z₁- t₁ ) =e^ i (k z₂- t₂ ) &k z₁- t₁=k z₂- t₂ & (t₂-t₁ ) =k (z-z₁ ) & where k= 2 =2 v & (t₂-t₁ )= 2 2 v (z₂-z₁ ) aligned aligned & = 1 x v (z₂-z₁ ) & v= (z₂-z₁ ) (t₂-t₁ ) =C & (t₂-t₁ )= (z₂-z₁ ) C & Frequency is f 1 t then 1 (t₂-t₁ ) = C (z₂-z₁ ) & Frequency, f= 3 10^8 (z₂-z₁ ) aligned