Question

A semi-transparent plate \(\left(A_{1}=2 \mathrm{~cm}^{2}\right)\) has an irradiation of \(500 \mathrm{~W} / \mathrm{m}^{2}\), where \(30 \%\) of the irradiation is reflected away from the plate and \(50 \%\) of the irradiation is transmitted through the plate. A radiometer is placed \(0.5 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. If the temperature of the plate is uniform at \(350 \mathrm{~K}\), determine the irradiation that the radiometer would detect.

Short answer

Answer: The total irradiation detected by the radiometer is approximately 181118.12 W/m².

Step-by-step solution

Calculate reflected and transmitted irradiation

Given that 30% of the irradiation is reflected away from the plate, we have: $$ I_{reflected} = 0.3 \times I_0 $$ And the 50% of the irradiation is transmitted through the plate, $$ I_{transmitted} = 0.5 \times I_0 $$ Now let's calculate these values: $$ I_{reflected} = 0.3 \times 500 W/m^2 = 150 W/m^2 $$ $$ I_{transmitted} = 0.5 \times 500 W/m^2 = 250 W/m^2 $$

Calculate emitted irradiation using Stefan-Boltzmann law

The Stefan-Boltzmann law states that the irradiation emitted by a black body at a temperature T is given by: $$ I_{emitted} = \sigma T^4 $$ Where \(\sigma\) is the Stefan-Boltzmann constant, approximately equal to \(5.67 \times 10^{-8} W/m^2K^4\). The plate's temperature is given as 350 K. Let's calculate emitted irradiation from the plate. $$ I_{emitted} = \sigma (350 K)^4 = 5.67 \times 10^{-8} W/m^2K^4 \times (350 K)^4 \approx 45242.03 W/m^2 $$

Calculate total irradiation detected by radiometer

To find the total irradiation detected by the radiometer, we need to sum the reflected irradiation and the emitted irradiation. However, since the radiometer is placed 0.5 m above the plate, some of the irradiation will spread out and decrease with distance. To account for this effect, we will divide the emitted irradiation by the square of the distance (0.5 m) between the plate and radiometer. $$ I_{total} = I_{reflected} + \frac{I_{emitted}}{distance^2} $$ Substitute with the values calculated before: $$ I_{total} = 150 W/m^2 + \frac{45242.03 W/m^2}{(0.5 m)^2} \approx 150 W/m^2 + 180968.12 W/m^2 = 181118.12 W/m^2 $$ Therefore, the radiometer would detect an irradiation of approximately \(181118.12 W/m^2\).

Key concepts

Understanding the Stefan-Boltzmann Law

The Stefan-Boltzmann law is pivotal when studying thermal radiation emitted by bodies, particularly perfect emitters known as black bodies. It defines the relationship between the temperature of a body and the energy it emits per unit area. According to this law, the total energy radiated per unit surface area of a black body across all wavelengths per unit time \( J \) is directly proportional to the fourth power of the black body's temperature \( T \) in kelvins. This is mathematically expressed as:

\[ J = \sigma T^{4} \]
where \( \sigma \) represents the Stefan-Boltzmann constant, equal to approximately \( 5.67 \times 10^{-8} W/m^{2}K^{4} \).

This law becomes very useful in a variety of applications, including understanding the radiant heat transfer in astrophysics, climate studies, and engineering. For instance, when dealing with problems similar to our exercise, it helps us to predict how much heat energy an object at a certain temperature would emit. In the solution provided, the law was utilized to calculate the emitted irradiation from a semi-transparent plate with a known temperature. Knowing this factor is crucial for determining the total irradiation a radiometer would detect, as demonstrated in the exercise solution.

The Role of a Semi-Transparent Plate in Irradiation

A semi-transparent plate is an important concept in the discussion of irradiation as it partially allows radiation to pass through while reflecting and absorbing the rest. In the context of our exercise, the plate reflects 30% and transmits 50% of the incident irradiation. The understanding of these percentages is critical to solving the problem accurately.

When radiation encounters a semi-transparent plate, the following possibilities occur:

• Some of the irradiation is reflected, which in this case is 30%.
• Transmission accounts for the portion that passes through the plate, 50% in this instance.
• The remaining percentage, if not stated as transmitted or reflected, is typically absorbed.

Standard calculations assume that any absorbed radiation may later be released through emission, as described by the Stefan-Boltzmann law. These properties are significant for researchers and engineers when designing systems for solar energy harnessing, climate control in buildings, or even in creating protective clothing. The solution showcases the practical use of these concepts in breaking down the contributions to the total irradiation detected by a radiometer.

Radiometer Measurements and Distance Effect

Radiometer measurements are utilized in various scientific fields to determine the quantity of radiant energy. A radiometer is an instrument that measures the intensity of electromagnetic radiation, most commonly used in the measurement of irradiation in terms of energy flux. In our exercise, a radiometer is placed above a semi-transparent plate to measure the irradiation.

However, it is essential to consider the inverse square law in these measurements. As radiated energy travels away from the source, it spreads out and its intensity diminishes with the square of the distance from the source. This effect was taken into account in the solution by dividing the emitted irradiation by the square of the distance to obtain accurate measurements of irradiation at the radiometer's location.

The radiometer's distance from the irradiation source can significantly impact the detected energy, especially when dealing with high-intensity sources or when precise measurements are required in scientific experiments, industrial processes, or even in the calibration of solar panels. The accurate calculation of this distance effect is fundamental for correctly interpreting the radiometer data, as seen in the calculation for the total irradiation detected in the provided exercise solution.