A small surface of area \(A_{1}=5 \mathrm{~cm}^{2}\) emits radiation as a blackbody at \(T_{1}=1000 \mathrm{~K}\). A radiation sensor of area \(A_{2}=\) \(3 \mathrm{~cm}^{2}\) is placed normal to the direction of viewing from surface \(A_{1}\) at a distance \(L\). An optical filter with the following spectral transmissivity is placed in front of the sensor: $$ \tau_{\lambda}= \begin{cases}\tau_{1}=0, & 0 \leq \lambda<2 \mu \mathrm{m} \\\ \tau_{2}=0.5, & 2 \mu \mathrm{m} \leq \lambda<\infty\end{cases} $$ If the distance between the radiation sensor and surface \(A_{1}\) is \(L=0.5 \mathrm{~m}\), determine the irradiation measured by the sensor.

To calculate the power emitted by the blackbody, we use the Stefan-Boltzmann law, which states that the power emitted per unit area of a blackbody is proportional to the fourth power of its temperature. The constant of proportionality is the Stefan-Boltzmann constant, denoted by \(\sigma\). \(P_{1}=\sigma T_{1}^{4} \cdot A_{1}\) Where, \(P_{1}\) is the power emitted, \(\sigma = 5.67\times10^{-8} W m^{-2}K^{-4}\) is the Stefan-Boltzmann constant, \(T_{1} = 1000K\) is the temperature of the blackbody, \(A_{1} = 5 cm^2 = 5\times10^{-4} m^2\) is the area of the blackbody. Now, let's calculate the power emitted by the blackbody: \(P_{1} = (5.67\times10^{-8} W m^{-2}K^{-4})(1000K)^4(5\times10^{-4} m^2) = 283.5W\)

Now, we will find the irradiation on the sensor before considering the optical filter. Irradiation on the sensor will be the power received per unit area of the sensor. \(I = \frac{P_{1}}{4\pi L^2}\) Where, \(I\) is the irradiation measured by the sensor, \(L = 0.5m\) is the distance between the sensor and the blackbody. Now, let's calculate the irradiation: \(I = \frac{283.5W}{4\pi (0.5m)^2} = 90.28 W m^{-2}\)

Now, we will consider the effect of the optical filter on the irradiation. The spectral transmissivity of the filter \(\tau_{\lambda}\) varies with wavelength. The problem states that for \(\lambda < 2\mu m\), the transmissivity is 0, and for \(\lambda \geq 2\mu m\), the transmissivity is 0.5. The Planck's law tells us that the spectral distribution of a blackbody's radiation is given by: \(B_{\lambda}(\lambda,T) = \frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_BT}}-1}\) Where, \(B_{\lambda}(\lambda,T)\) is the spectral radiance, \(h = 6.626\times10^{-34} J s\) is the Planck's constant, \(c = 3\times10^8 m s^{-1}\) is the speed of light, \(\lambda\) is the wavelength, \(k_B = 1.38\times10^{-23} J K^{-1}\) is the Boltzmann constant, \(T = 1000K\) is the blackbody temperature. Using this formula, we can express the irradiation measured by the sensor after the filter as an integral: \(I_{filtered} = \int_{0}^{\infty} \tau_{\lambda} B_{\lambda}(\lambda, T) d\lambda\) However, given the transmissivity of the filter, the integral simplifies to: \(I_{filtered} = 0.5 \cdot \int_{2\mu m}^{\infty} B_{\lambda}(\lambda, T) d\lambda\) This integral is a function of temperature and can be challenging to compute directly. An alternative approach is to find the fraction of power emitted by the blackbody with wavelengths greater than \(2\mu m\). We can use Wien's law, which states: \(\lambda_{max}T = b\) Where, \(\lambda_{max}\) is the peak wavelength, \(b = 2.898\times10^{-3} mK\) is the Wien's displacement constant, \(T = 1000K\) is the blackbody temperature. Let's calculate the peak wavelength first: \(\lambda_{max} = \frac{b}{T} = \frac{2.898\times10^{-3} mK}{1000K} = 2.898\mu m\) Since \(\lambda_{max} > 2\mu m\), we can assume that approximately 50% of the power emitted by the blackbody has wavelengths greater than \(2\mu m\). The irradiation measured by the sensor after the filter is: \(I_{filtered} = I \cdot 0.5 = 90.28 W m^{-2} \cdot 0.5 = 45.14 W m^{-2}\) The final answer for the irradiation measured by the sensor after the optical filter is \(45.14 W m^{-2}\).