Question

An LED with total power P tot = 960 mW emits UV light of wavelength 360 nm. Assuming the LED is 55% efficient and acts as an isotropic point source (i.e., emits light uniformly in all directions), what is the amplitude of the electric field, E 0 , at a distance of 2.5 cm from the LED?

Answer 1

Answer:

$E_0=225.09N/C$

Explanation:

We are given that

Power,Ptot=960mW=$960\times 10^{-3}W$

$1mW=10^{-3} W$

Wavelength,$\lambda=360 nm=360\times 10^{-9} m$

$1nm=10^{-9} m$

Distance,r=2.5 cm=$2.5\times 10^{-2} m$

1m=100 cm

Efficiency=55%

Power radiation emitted=$\frac{55}{100}\times 960\times 10^{-3}=0.528W$

Intensity,I=$\frac{P}{4\pi r^2}$

$I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2$

Intensity,$I=\frac{1}{2}c\epsilon_0E^2_0$

$E^2_0=\frac{2I}{c\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}$

$c=3\times 10^8 m/s$

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}$

$E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}$

$E_0=225.09N/C$