Question

A red LED emits light at $0.1W$ uniformly around it. The amplitude of the electric field of the light at a distance of $1m$ from the diode is:
(A) $1.73V/m$
(B) $2.45V/m$
(C) $5.48V/m$
(D) $7.75V/m$ ?

Answer 1

In this question, we have to find the amplitude of the electric field of the light at a given distance for which we will recall the concept of LED (light emitting diode) where we learnt that LED gives out light radiation when forward biased. According to the question, the LED emits lights at $P = 0.1W$ and the electric field of the light is at distance of $r = 1m$
In this question the two formulas used for Intensity are $I = \dfrac{P}{{4\pi {r^2}}}$ and $I = \dfrac{1}{2}\varepsilon \circ {E_0}^2 \times c$ .

Complete step by step solution:
According to the question, the LED emits lights at $P = 0.1W$ and the electric field of the light is at distance of $r = 1m$ from the diode,
Since we know that the intensity,
$I = \dfrac{P}{{4\pi {r^2}}}$ ,
Also, $I = \dfrac{1}{2}{\varepsilon _0}{E_0}^2 \times c$ ,
Hence after equating both the equations we have,
$\Rightarrow {E^2}_0 = \dfrac{{2P}}{{4\pi {\varepsilon _0}{r^2}c}}$ ,
$\Rightarrow \dfrac{{2 \times 0.1 \times 9 \times {{10}^9}}}{{1 \times 3 \times {{10}^8}}} = 6$ ,
$\Rightarrow {E_0} = \sqrt 6 = 2.45V/m$ .
Therefore, the amplitude of the electric field of the light is $2.45V/m$ .

Note:
We know that the LED (light emitting diode) is a PN junction diode which is specially made and doped of semiconductor. It can only emit light when it is in forward biased state. LED work on the phenomenon of electroluminescence. When the forward biased current is applied through the P-N junction of the diode, minority carrier electrons are injected into the P-region and corresponding minority carrier electrons are injected into the n-region. Due to electron hole recombination in the P-region the photo emission occurs.