A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

Answers (1)

As we learnt in

LED (Light emitting diode) -

It gives out light radiation when forward biased .

- wherein

LED's are made of GaAsP  ,Gap etc .

(Require low voltage ,long life )

 

 I=\frac{P}{4\pi r^{2}}=\frac{0.1}{4\pi}=U_{av}.C

Amplitude I = E0

U_{av}=\frac{1}{2}\epsilon_{0}E_{0}^{2}

\Rightarrow\ \; \frac{1}{2}\epsilon_{0}E_{0}^{2}=\frac{0.1}{4\pi C}

or  E_{0}^{2}=\frac{0.2}{\left(4\pi \epsilon_{0} \right )C}=\frac{0.2\times 9\times 10^{9}}{3\times 10^{8}}=6

\therefore\ \;E_{0}=2.45\ V/m

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