Question

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

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Solution

Given dataThe power of the LED is $P=0.1watt.$The distance where the electric field has to be measured is $L=1m.$Intensity of lightIntensity is the energy emitted per unit area in unit time. For a point light source the intensity is $I=\frac{P}{4\pi {r}^{2}}$, where, P is the power of the light and r is the distance where the intensity has to measure.Intensity of a light source is defined by the form, $I=\frac{1}{2}{\epsilon }_{o}{E}^{2}×c$, where, ${\epsilon }_{o}$is the permittivity of free space, $E$ is the electric field, and $c$ is the velocity of light in free space.Finding the electric fieldAs we know, $\frac{1}{2}{\epsilon }_{o}{E}^{2}×c=$$\frac{P}{4\pi {r}^{2}}$So, ${E}^{2}=\frac{P×2}{4\pi {\epsilon }_{o}×{r}^{2}×c}\phantom{\rule{0ex}{0ex}}or{E}^{2}=\frac{0.1×2×9×{10}^{9}}{{1}^{2}×3×{10}^{8}}=6\left(\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}=Coulombcons\mathrm{tan}t=9×{10}^{9}{C}^{2}/N/{m}^{2}\right)\phantom{\rule{0ex}{0ex}}orE=\sqrt{6}=2.45\phantom{\rule{0ex}{0ex}}orE=2.45V/m.$ Therefore, the electric field of the light at a distance $1m$ from the diode is $2.45volt/m.$

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