A red LED emits light at 0 . 1watt uniformly around it. The amplitude of the electric field of the light at a distance of 1m from the diode is:

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 2 . 45V m ^ - 1

For a point source of power = P, then intensity at a point at a separation x from the source is,&

I = \dfrac{Power}{Area} = \dfrac{P}{4\pix^{2}}∵

~The average intensity of E.M. wave is given by,&

I = \dfrac{1}{2}c{ε}_{0}{E}_o^{2}⇒ {E}_{0} = \sqrt{\dfrac{2P}{4\pi{ε}_{0}cx^{2}}}

∵ \dfrac{1}{4\pi{ε}_{0}} = 9 \times {10}^{9}, P = 0.1W, x = 1m

,~

c = 3 \times {10}^{8}ms^{- 1}

⇒ {E}_{0} = \sqrt{\dfrac{2 \times 0.1 \times 9 \times {10}^{9}}{3 \times {10}^{8} \times {1}^{2}}} = \sqrt{6} = 2.45Vm^{- 1}

A red LED emits light at 0 . 1watt uniformly around it. The...