Q. The amplitude of electric field in a parallel beam of light of intensity $4\ Wm^{- 2}$ is
A) $40.5\ NC^{- 1}$
B) $45.5\ NC^{- 1}$
C) $50.5\ NC^{- 1}$
D) $55.5\ NC^{- 1}$
Explanation
(d) $I = \frac{1}{2}\ \varepsilon_{0}E_{0}^{2}c$
or $E_{2} = \sqrt{\frac{2I}{\varepsilon_{0}c}}$
$= \sqrt{\frac{2 \times 4}{\left( 8.85 \times 10^{- 12} \right) \times (3 \times 10^{8})}} = 55.5\ NC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.