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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 :

  • Option 1)

    1.73 V/m

  • Option 2)

    2.45 V / m

  • Option 3)

    5.48 V / m             

  • Option 4)

    7.75 V / m

 

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

Correct option is 2.


Option 1)

1.73 V/m

This is an incorrect option.

Option 2)

2.45 V / m

This is the correct option.

Option 3)

5.48 V / m             

This is an incorrect option.

Option 4)

7.75 V / m

This is an incorrect option.

Posted by

Vakul

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