Fall 2011 Homework 25 Solutions

Fall 2011 Homework 25 Solutions - Physics 21 Fall, 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-25 32-15 We can reasonably model a 60 W incandescent light- bulb as a sphere 5.2 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 at the surface of the bulb? b) What is the am- plitude of the electric field at this surface, for a sinusoidal wave with this intensity? c) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity? (a) The intensity of light is given by the power per unit area, so the visible light intensity I vis at the surface of the light bulb is the power due to visible light divided by the surface area of the light bulb: I vis = P vis A = . 05 P tot 4 ( d/ 2) 2 = (0 . 05)(60 . 0) 4 (0 . 026) 2 = 353 W / m 2 . (b) The intensity at the surface of the bulb calculated above is just the magnitude S of the time averaged Poynting vector, which is related to the electric field amplitude E by S = 1 2 cE 2 . Using this expression we can solve for the amplitude of the electric field: E = 2 S c = 2 I vis c = 516 V / m . (c) The amplitudes of the electric and magnetic fields are related by B = E c = 1 . 7 T . We could have found the magnetic field amplitude directly from S using an alternate expression for S : S = 1 2 c B 2 = B = 2 S c . 32-41 A small helium-neon laser emits red visible light with a power of 3.40 mW in a beam that has a diameter of 2.00 mm. (a) What is the amplitude of the electric field of the light? (b) What is the amplitude of the magnetic field of the light? (c) What is the average energy density associated with the electric field? (d) What is the averageassociated with the electric field?...
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Fall 2011 Homework 25 Solutions - Physics 21 Fall, 2011...

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