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We can reasonaly model a 75 W incandescent lightbulb as a sphere 6 cm in diameter. Typically, only about 5% of the energy goes...Asked by Brittany
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to non-visible light 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?
I am really lost on how to solve this problem. The only thing I know for sure that I need to do is: (75)(.05)=3.75 W for the visible light. Can you please show your work?
(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?
I am really lost on how to solve this problem. The only thing I know for sure that I need to do is: (75)(.05)=3.75 W for the visible light. Can you please show your work?
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Answered by
Damon
find the surface area of the sphere (4 pi r^2) where r = .03 meters
then you have .05*75/ surface area
Your text has the relationship between intensity and E and B amplitudes.
then you have .05*75/ surface area
Your text has the relationship between intensity and E and B amplitudes.
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