Practicing Success
Practicing Success
CUET Preparation Today
The amplitude of electric field in a parallel beam of light of intensity 4 Wm−2 is |
40.5 NC−1 45.5 NC−1 50.5 NC−1 55.5 NC−1 |
55.5 NC−1 |
$I=\frac{1}{2}ε_0E^2_0 c$ $or \, E_2 = \sqrt{\frac{2I}{ε_0c}}$ $=\sqrt{\frac{2×4}{(8.85×10^{-12})×(3×10^8)}}$ $ = 55.5 \, NC^{-1}$ |
