Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

The intensity of a plane monochromatic wave is 1380 Wm-2. What is the amplitude of electric field in this wave? (Take speed of light 3 × 108 ms-1; ε0 = 8.85 × 10-12 C2N-1m-2)

Options:

102 N C-1

10.2 N C-1

1020 N C-1

10200 N C-1

Correct Answer:

1020 N C-1

Explanation:

$ I = \frac{1}{2} \epsilon_0 c E_0^2 = 1380 $

$\Rightarrow E_0^=2 = \frac{2760}{8.85\times 10^{-12}\times 3\times 10^8} = 1.04\times 10^6$

$ E_0 = 1.02\times 10^3 = 1020 N/C$