Answer:
Option B
Explanation:
Consider the LED as a point source of light
Let the power of LED is P.
Intensity at r from the source
$I=\frac{P}{4\pi r^{2}}$ ..........(i)
As we know that
$I=\frac{1}{2}\epsilon_{0} E_{0}^{2}c$ ..........(ii)
From Eq (i) and (ii) , we can write
$\frac{P}{4\pi r^{2}}=\frac{1}{2}\epsilon_{0} E_{0}^{2}c$
or $ E_{0}^{2}=\frac{2P}{4\pi\epsilon_{0}r^{2}c}=\frac{2\times0.1\times9\times10^{9}}{1\times3\times 10^{8}}$
or $E_{0}^{2}=6\Rightarrow E_{0}=\sqrt{6}=2.45V/m$
