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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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 Pot = 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 = 0.05 P/4T (4/2)² = 380 W/m²
Amplitude of Electric field = u = 1/₂ E₁ E²
= ε = √²1/c
=540v/m
1/c = u
Amplitude of Magnetic field = B = 2/C = 1.8μT
Transcribed Image Text: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 Pot = 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 = 0.05 P/4T (4/2)² = 380 W/m² Amplitude of Electric field = u = 1/₂ E₁ E² = ε = √²1/c =540v/m 1/c = u Amplitude of Magnetic field = B = 2/C = 1.8μT
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