99 Percentile Qs Bank for JEE MainPhysicsElectromagnetic Waves
What is the amplitude of the electric field in a parallel beam of light intensity ( 15 ) W m ^2 ? [ . Assume, . 1 4 ₀ =9 10^9 Nm ^2 C ^2 ]
Options
- A60 ~N / C
- B50 ~N / C
- C40 ~N / C
- D30 ~N / C
Correct answer
A. 60 ~N / C
Step-by-step solution
Intensity of parallel beam of light is given as where, E₀= amplitude of electric field. Here, aligned I & = 15 W m ^2 , 1 4 ₀ =9 10^9 Nm ^2 C ^2 c & =3 10^8 ~ms ⁻¹ aligned From Eq. (i), we get aligned & E₀^2= 2 I ₀ c = 2 4 I 4 ₀ c = ( 2 4 15 9 10^9 3 10^8 ) & E₀^2=3600 E₀=60 ~N / C aligned