The intensity of light from a source is $\left( \frac{500}{\pi} \right) \, W/m^2$. Find the amplitude of the electric field in this wave.

  • A
    $\sqrt{3} \times 10^2 \, N/C$
  • B
    $2\sqrt{3} \times 10^2 \, N/C$
  • C
    $\frac{\sqrt{3}}{2} \times 10^2 \, N/C$
  • D
    $2\sqrt{3} \times 10^1 \, N/C$

Explore More

Similar Questions

The property which is not of an electromagnetic wave travelling in free space is that:

For the plane electromagnetic wave given by $E=E_0 \sin (\omega t-kx)$ and $B=B_0 \sin (\omega t-kx)$,the ratio of average electric energy density to average magnetic energy density is

The speed of electromagnetic radiation in vacuum is

If a source is transmitting electromagnetic waves of frequency $8.196 \times 10^{6} \ Hz$,then the wavelength of the $EM$ waves transmitted from the source will be . . . . . . . (in $cm$)

Given below are two statements:
Statement $I$: Electromagnetic waves are not deflected by electric and magnetic fields.
Statement $II$: The amplitude of the electric field and the magnetic field in electromagnetic waves are related to each other as $E_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} B_0$.
In the light of the above statements,choose the correct answer from the options given below:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo