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sertanlavr [38]
3 years ago
5

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 iso

tropic 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?
Physics
1 answer:
Anit [1.1K]3 years ago
4 0

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

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