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Chapter 12

Q. 12.7

100 W of radiant energy leave a frosted glass spherical light bulb 10 cm in diameter enclosed in a fixture having a flat glass plate as in Figure 12.2. If the glass is 2 cm thick and has a gray extinction coefficient of 0.05 cm^{−1}, find the intensity directly from a location on the bulb surface leaving the fixture at an angle of θ = 60° as shown in the figure. Neglect interface-interaction effects resulting from the difference in refractive index between the glass and surrounding air; these effects are included in Chapter 17.

100 W of radiant energy leave a frosted glass spherical light bulb 10 cm in diameter enclosed in a fixture having a flat glass plate as in Figure 12.2. If the glass is 2 cm thick and has a gray extinction coefficient of 0.05 cm−1, find the intensity directly from a location

Step-by-Step

Verified Solution

Integrating Equation 12.18 over λ and S results in the total intensity I(S,θ)=I(0,θ)e^{−(κ+σs )S(θ)}.To obtain I(0, θ), consider the bulb a diffuse sphere. The emissive power at the sphere surface is 100 W divided by the sphere area. The intensity is this diffuse emissive power divided by π,

I_\lambda (S)=I_\lambda (0)e^{-\int_{S^*=0}^{S}{\beta _\lambda }(S^*) dS^*}          (12.18)

I(0,θ) = 100 W/(π10^2cm^2×π \ sr)=0.101 W/(cm^2 ⋅ sr)

Then

I(S,θ) = 0.101e^{−0.05(2/cos 60°)}= 0.0827 ⋅ W/(cm^2.sr)