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

`E_0=225.09N/C`

Step-by-step explanation:

We are given that

Power,Ptot=960mW=
`960* 10^(-3)W`

`1mW=10^(-3) W`

Wavelength,
`\lambda=360 nm=360* 10^(-9) m`

`1nm=10^(-9) m`

Distance,r=2.5 cm=
`2.5* 10^(-2) m`

1m=100 cm

Efficiency=55%

Power radiation emitted=
`(55)/(100)* 960* 10^(-3)=0.528W`

Intensity,I=
`(P)/(4\pi r^2)`

`I=(0.528)/(4\pi(2.5* 10^(-2))^2)=67.26W/m^2`

Intensity,
`I=(1)/(2)c\epsilon_0E^2_0`

`E^2_0=(2I)/(c\epsilon_0)`

Where
`\epsilon_0=8.85* 10^(-12)`

`c=3* 10^8 m/s`

`E_0=\sqrt{(2I)/(c\epsilon_0)}`

`E_0=\sqrt{(2* 67.26)/(3* 10^8* 8.85* 10^(-12))}`

`E_0=225.09N/C`