Answer:
[tex]E_0=225.09N/C[/tex]
Explanation:
We are given that
Power,Ptot=960mW=[tex]960\times 10^{-3}W[/tex]
[tex]1mW=10^{-3} W[/tex]
Wavelength,[tex]\lambda=360 nm=360\times 10^{-9} m[/tex]
[tex] 1nm=10^{-9} m[/tex]
Distance,r=2.5 cm=[tex]2.5\times 10^{-2} m[/tex]
1m=100 cm
Efficiency=55%
Power radiation emitted=[tex]\frac{55}{100}\times 960\times 10^{-3}=0.528W[/tex]
Intensity,I=[tex]\frac{P}{4\pi r^2}[/tex]
[tex]I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2[/tex]
Intensity,[tex]I=\frac{1}{2}c\epsilon_0E^2_0[/tex]
[tex]E^2_0=\frac{2I}{c\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]c=3\times 10^8 m/s[/tex]
[tex]E_0=\sqrt{\frac{2I}{c\epsilon_0}}[/tex]
[tex]E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}[/tex]
[tex]E_0=225.09N/C[/tex]
