Question

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: $${\tau }_{\lambda }=\{\begin{array}{ll}{\tau }_1=0,& 0\le \lambda <2\mu \mathrm{m}\\ {\tau }_2=0.5,& 2\mu \mathrm{m}\le \lambda <\mathrm{\infty }\end{array}$$ If the distance between the radiation sensor and surface $A_1$ is $L=0.5\mathrm{m}$, determine the irradiation measured by the sensor.

Short answer

Answer: The irradiation measured by the sensor after the optical filter is 45.14 W m^{-2}.

Step-by-step solution

Calculate the power emitted by the blackbody

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, $\sigma =5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}$ is the Stefan-Boltzmann constant, $T_1=1000K$ is the temperature of the blackbody, $A_1=5c{m}^2=5\times {10}^{-4}{m}^2$ is the area of the blackbody. Now, let's calculate the power emitted by the blackbody: $P_1=(5.67\times {10}^{-8}W{m}^{-2}{K}^{-4})(1000K{)}^4(5\times {10}^{-4}{m}^2)=283.5W$

Calculate irradiation before the filter

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: $I=\frac{283.5W}{4\pi (0.5m{)}^2}=90.28W{m}^{-2}$

Calculate irradiation after the filter

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 $\lambda \ge 2\mu m$, the transmissivity is 0.5. The Planck's law tells us that the spectral distribution of a blackbody's radiation is given by: $B_{\lambda }(\lambda ,T)=\frac{2\pi h{c}^2}{{\lambda }^5}\frac{1}{e^{\frac{hc}{\lambda k_{B}T}}-1}$ Where, $B_{\lambda }(\lambda ,T)$ is the spectral radiance, $h=6.626\times {10}^{-34}Js$ is the Planck's constant, $c=3\times {10}^{8}m{s}^{-1}$ is the speed of light, $\lambda$ is the wavelength, $k_B=1.38\times {10}^{-23}J{K}^{-1}$ 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: $I_{filtered}={\int }_0^{\mathrm{\infty }}{\tau }_{\lambda }{B}_{\lambda }(\lambda ,T)d\lambda$ However, given the transmissivity of the filter, the integral simplifies to: $I_{filtered}=0.5\cdot {\int }_{2\mu m}^{\mathrm{\infty }}{B}_{\lambda }(\lambda ,T)d\lambda$ 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, $b=2.898\times {10}^{-3}mK$ is the Wien's displacement constant, $T=1000K$ is the blackbody temperature. Let's calculate the peak wavelength first: ${\lambda }_{max}=\frac{b}{T}=\frac{2.898\times {10}^{-3}mK}{1000K}=2.898\mu m$ 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: $I_{filtered}=I\cdot 0.5=90.28W{m}^{-2}\cdot 0.5=45.14W{m}^{-2}$ The final answer for the irradiation measured by the sensor after the optical filter is $45.14W{m}^{-2}$.

Key concepts

Blackbody Radiation

Blackbody radiation is a fundamental concept in thermal physics that describes how objects emit light and heat. All objects with a temperature above absolute zero emit radiation. A blackbody is an idealized object that absorbs all incoming radiation and emits energy perfectly across all wavelengths, meaning it is a perfect emitter and absorber.
To better understand blackbody radiation, consider that the radiation emitted reaches a peak at a certain wavelength, which shifts depending on the temperature of the object. This emission is due to the object's temperature, and its energy output increases dramatically with temperature.
Some key aspects of blackbody radiation include:

• It is most fully described by the Planck's law, which gives the intensity of radiation at different wavelengths.
• This radiation can include visible light, infrared, ultraviolet, etc., depending on the object's temperature.
• In a practical sense, stars, including the sun, closely follow blackbody radiation characteristics.

Understanding blackbody radiation is crucial because it serves as a key reference in physics for thermal emission characteristics of real-world objects.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law bridges the gap between an object's temperature and the total energy radiated per unit surface area. It is a pivotal formula in thermodynamics, providing a way to calculate the power emitted by a blackbody.
According to this law, the power emitted per unit area of a blackbody is proportional to the fourth power of its absolute temperature. The mathematical expression is:

• $P=\sigma T^4\times A$

Here, $P$ is the power emitted, $\sigma$ is the Stefan-Boltzmann constant $5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}$, $T$ is the absolute temperature in Kelvin, and $A$ is the surface area of the blackbody.
This law indicates that even small increases in an object's temperature result in a significant boost in emitted radiation. Understanding this principle explains why stars, despite being far away, are visible: their enormous temperatures make them potent emitters of radiant energy.

Optical Filter

Optical filters are devices placed in the path of light or radiation to control which wavelengths pass through. They are widely used in scientific instruments, photography, and various technological applications.
In the context of the exercise, an optical filter with specified spectral transmissivity affects how much radiation reaches the sensor. This transmission property means the filter allows certain wavelengths to pass while blocking others. Typically, the filter's transmissivity is represented as:

• Complete absorption for some wavelengths (${\tau }_{\lambda }=0$).
• Partial transmission for others (${\tau }_{\lambda }=0.5$).

The sensor's received radiation is then computed by integrating the transmissivity across wavelengths. Filters play a crucial role in experiments and applications where specific wavelengths are of interest, like in the filtering of infrared or ultraviolet light.

Wien's Law

Wien's Law describes the relationship between the temperature of an ideal blackbody and the wavelength at which it emits radiation most intensely. Essentially, as the temperature of a blackbody increases, the peak wavelength of emitted radiation shifts to shorter wavelengths.
The mathematical expression of Wien’s law is given by:

• ${\lambda }_{max}=\frac{b}{T}$

where ${\lambda }_{max}$ is the peak wavelength, $T$ is the absolute temperature in Kelvin, and $b$ is Wien’s displacement constant $2.898\times {10}^{-3}mK$.
This law is impactful in understanding stellar emissions and thermal imaging technologies, as it helps us determine the temperature of stars and other celestial bodies based on their spectrum.
For example, if a star's peak emission is visible light in blues, it is hotter than one that peaks in reds or infrareds, illustrating why blue stars are hotter than red stars.