Suppose we are observing an infrared source that is 500 pc away. It radiates like a \(50 \mathrm{K}\) blackbody and is 1 pc in extent. (a) What is the total energy per second per square meter reaching the Earth from this source? How does that compare with the total amount of solar radiation reaching the Earth per second per square meter. (b) Suppose we observe this source using a satellite with a \(1 \mathrm{m}\) diameter mirror, and we observe at a wavelength of \(100 \mu \mathrm{m} .\) What is the energy/s/Hz striking the telescope? (c) Suppose the telescope radiates like a blackbody at \(300 \mathrm{K}\), but with an efficiency of \(1 \% .\) (That is, the spectrum looks like that of a blackbody but with an intensity reduced by a factor of \(100 .\). What is the energy/Hz/s given off by the telescope at this wavelength? How does your answer compare with that in (b). (d) Redo part (c), assuming that we can cool the mirror to \(30 \mathrm{K}\) (still with a \(1 \%\) emission efficiency

Step by step solution

01

Determine the source's flux reaching Earth

Use the inverse square law to determine the flux reaching Earth. The formula is: \[ F = \frac{L}{4 \pi d^2} \]where \(L\) is the luminosity and \(d\) is the distance.

02

Calculate luminosity of the blackbody source

Use the Stefan-Boltzmann law to calculate the luminosity:\[ L = 4 \pi R^2 \sigma T^4 \]where \( R = 0.5 \) pc (since the diameter is 1 pc), \( \sigma \) is the Stefan-Boltzmann constant, and \( T = 50 \) K.

03

Find flux from the source

Substitute the values from Step 2 into the inverse square law to calculate the flux at Earth. \( F = \frac{L}{4 \pi (500\, \text{pc})^2} \)

04

Compare with solar radiation

Compare this flux with the total solar radiation reaching the Earth, which is approximately \( F_\odot = 1361 \) W/m\(^2\).

05

Calculate energy/s/Hz striking the telescope

Use the equation: \[ E = F \cdot A \cdot \Delta u \] where \(A = \pi (0.5\,m)^2\) (mirror area) and \(\Delta u = \frac{c}{\lambda} \) (bandwidth), and \( \lambda = 100 \mu \text{m} \) to calculate the photon energy.

06

Compute telescope's blackbody radiation at 300 K with efficiency

Use the Planck's law and adjust for efficiency: \[ I(u, T) = \frac{2 h u^3}{c^2} \frac{1}{e^{hu/kT} - 1} \] where \(u = \frac{c}{\lambda}\) and \(T = 300\, \text{K}\). Then apply \(1\%\) efficiency.

07

Compare telescope's radiation with source at 300 K

Compare the energy calculated in Step 6 to the energy obtained in Step 5.

08

Redo calculations for 30 K

Repeat Step 6 calculation with \(T = 30\, \text{K}\).

09

Compare telescope's radiation with source at 30 K

Compare the energy calculated in Step 8 to the energy obtained in Step 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Blackbody Radiation

Blackbody radiation refers to the theoretical concept of an idealized object that absorbs and emits all frequencies of electromagnetic radiation flawlessly.
Basically, a blackbody absorbs all incident radiation and re-emits it according to its temperature.
This emission has a specific spectrum and intensity which only depends on the body's temperature and is described by Planck's law.

The Stefan-Boltzmann Law Explained

The Stefan-Boltzmann law links the total energy radiated per unit surface area of a blackbody to the fourth power of its temperature.
It is formulated as \[ L = 4 \pi R^2 \sigma T^4 \] where *L* is the luminosity, *R* is the radius, *\sigma* is the Stefan-Boltzmann constant, and *T* is the temperature in Kelvins.
This law shows that hotter objects radiate significantly more energy than cooler ones.

Inverse Square Law in Astronomy

The inverse square law states that the intensity of radiation decreases with the square of the distance from the source.
Mathematically, it is \[ F = \frac{L}{4 \pi d^2} \] where *F* is the flux, *L* is the luminosity, and *d* is the distance.
This law is crucial for calculating the energy we receive from distant celestial objects.

Deciphering Planck's Law

Planck's law describes the spectrum of blackbody radiation.
It tells us the amount of energy at different wavelengths for a given temperature and is given by \[ I(u, T) = \frac{2 h u^3}{c^2} \frac{1}{e^{hu/kT} - 1} \] where *I(u,T)* is the spectral radiance, *h* is Planck's constant, *u* is frequency, *c* is the speed of light, and *k* is the Boltzmann constant.
Essentially, it explains why objects of different temperatures emit different spectra.

Telescope Detection in Infrared Astronomy

Infrared telescopes detect radiation in the infrared part of the spectrum, which is invisible to the human eye.
They often observe celestial objects that are not visible in optical light due to dust obscuration or low temperatures.
The efficiency and capability of infrared telescopes rely on their ability to cool their mirrors and detectors, reducing their own thermal emissions and thereby improving detection sensitivity.