A furnace that has a \(40-\mathrm{cm} \times 40-\mathrm{cm}\) glass window can be considered to be a blackbody at \(1200 \mathrm{~K}\). If the transmissivity of the glass is \(0.7\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and zero for radiation at wavelengths greater than \(3 \mu \mathrm{m}\), determine the fraction and the rate of radiation coming from the furnace and transmitted through the window.

We are given: Window size: \(40 \thinspace cm \times 40 \thinspace cm\) Temperature of the blackbody: \(T = 1200 \thinspace K\) Transmissivity of the glass: \(\tau(\lambda) = 0.7\) for \(0 \leq \lambda \leq 3 \thinspace \mu m\), and \(0\) for \(\lambda > 3 \thinspace \mu m\)

We will use Planck's law of blackbody radiation in terms of wavelength. The formula is given by: \(I(\lambda, T) = \frac{2 \pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1}\) Where: - \(I(\lambda, T)\) is the intensity of radiation per unit wavelength, - \(h\) is the Planck's constant \((6.626 \times 10^{-34} \thinspace Js)\), - \(c\) is the speed of light \((3.0 \times 10^8 \thinspace m/s)\), - \(\lambda\) is the wavelength, - \(k_B\) is the Boltzmann constant \((1.381 \times 10^{-23} \thinspace J/K)\), - \(T\) is the temperature of the blackbody.

The transmitted intensity is given by the product of the intensity of the blackbody and the transmissivity of the glass. \(I_{transmitted}(\lambda) = I(\lambda, T) \times \tau(\lambda)\)

We will integrate the transmitted intensity over the entire wavelength range, and then multiply it with the window area to get the rate of transmitted radiation. \(R_{transmitted} = Area \times \int_{0}^{\infty} I_{transmitted}(\lambda) d\lambda\) Since \(\tau(\lambda)\) is zero for wavelengths greater than \(3 \thinspace \mu m\), we only need to integrate from \(0\) to \(3 \thinspace \mu m\): \(R_{transmitted} = Area \times \int_{0}^{3 \times 10^{-6}} I_{transmitted}(\lambda) d\lambda\) We can use numerical integration (e.g. Simpson's Rule) to compute the integral.

The fraction of transmitted radiation is given by the ratio of transmitted radiation to the total radiation from the blackbody. \(f = \frac{R_{transmitted}}{R_{total}}\) Where the total radiation, \(R_{total}\), can be found using the Stefan-Boltzmann law: \(R_{total} = Area \times \sigma T^4\) Where \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \thinspace W/m^2K^4)\). Thus, the fraction of transmitted radiation \(f\) can be found after computing the values of \(R_{transmitted}\) and \(R_{total}\). In conclusion, we have discussed how to approach the problem of finding the fraction and rate of radiation transmitted through a glass window of a blackbody furnace. The main steps involve understanding the Planck's law of blackbody radiation and applying the transmissivity of the glass to calculate the transmitted radiation. Then, the rate of transmitted radiation can be found by integrating the transmitted intensity over the wavelength range, and the fraction of transmitted radiation can be found by taking the ratio of transmitted and total radiation.