Example Problem 23.17 A 75 W incandescent light bulb acts as a point light source centered inside a 6.0 cm diameter spherical bulb. Only 5% of the total power goes to visible light – the rest is emitted in the thermal IR as heat. (a) What is the visible light intensity (in W/m²) at the surface of the bulb? (b) What are the amplitudes of the visible E and B fields at the surface of the bulb? Given: Power Ptot = 75 W, distance from point source r = ½(6.0 cm) = 3.0 cm, fraction of total intensity in visible = 5% Find: Total visible light intensity Visible light intensity = 1 = 0.05P/A = 0.0 5 P/4T (4/2)² = 380 w/m ² Amplitude of Electric field = u = 1/₂ {₁{² =540v/m 1/c = u Amplitude of Magnetic field = B = 2/C = 1.8μT =&= √³1/₂0 E=

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