Asked by J
We can reasonaly model a 75 W incandescent lightbulb as a sphere 6 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible light intensity (in W/m^2) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
ok..once again I am totally confused.
ok..once again I am totally confused.
Answers
Answered by
Damon
radius = 3*10^-2 meters
calculate surface area 4 pi r^2
Intensity of visible = .05 * 75 / surface area
I = Emax Bmax /(2 uo) = (Emax)^2/(2 uo c)
I am sure this is in your Physics book!!! Look up intensity of electromagnetic wave and "Poynting vector".
calculate surface area 4 pi r^2
Intensity of visible = .05 * 75 / surface area
I = Emax Bmax /(2 uo) = (Emax)^2/(2 uo c)
I am sure this is in your Physics book!!! Look up intensity of electromagnetic wave and "Poynting vector".
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