Assume that you have a telescope with an aperture of 1 meter. Compare the telescope's theoretical resolution when you are observing in the near-infrared region of the spectrum \((\lambda=1,000 \mathrm{nm})\) with that when you are observing in the violet region of the spectrum \((\lambda=400 \mathrm{nm})\).

Short Answer

Expert verified
Theoretical resolution is poorer in the near-infrared region compared to the violet region.

Step by step solution

01

- Understand Theoretical Resolution

The theoretical resolution of a telescope is determined by the Rayleigh criterion. The formula for the resolution \(\theta\) in radians is given by \(\theta = 1.22 \frac{\lambda}{D}\), where \(\lambda\) is the wavelength of the light, and \(D\) is the diameter of the telescope's aperture.
02

- Convert Wavelengths to Meters

First, convert the given wavelengths from nanometers to meters.For near-infrared, \(\lambda = 1,000 \mathrm{nm} = 1,000 \times 10^{-9} \mathrm{m} = 10^{-6} \mathrm{m}\).For violet, \(\lambda = 400 \mathrm{nm} = 400 \times 10^{-9} \mathrm{m} = 4 \times 10^{-7} \mathrm{m}\).
03

- Calculate Resolution for Near-Infrared

Substitute the near-infrared wavelength into the Rayleigh criterion formula: \(\theta_{\text{infrared}} = 1.22 \frac{\lambda_{\text{infrared}}}{D} = 1.22 \frac{10^{-6}}{1} = 1.22 \times 10^{-6} \mathrm{radians}\).
04

- Calculate Resolution for Violet

Substitute the violet wavelength into the Rayleigh criterion formula: \(\theta_{\text{violet}} = 1.22 \frac{\lambda_{\text{violet}}}{D} = 1.22 \frac{4 \times 10^{-7}}{1} = 4.88 \times 10^{-7} \mathrm{radians}\).
05

- Compare the Resolutions

Compare the calculated resolutions: \(\theta_{\text{infrared}} = 1.22 \times 10^{-6} \mathrm{radians}\) and \(\theta_{\text{violet}} = 4.88 \times 10^{-7} \mathrm{radians}\). The resolution in the near-infrared region is larger (poorer) than in the violet region.

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Key Concepts

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

Rayleigh criterion
The Rayleigh criterion is a key concept in optics that helps determine the resolution of a telescope. It defines the smallest angular separation between two point sources that can be resolved by the telescope. The formula for resolution \( \theta \) in radians is given by \( \theta = 1.22 \frac{\lambda}{D} \). Here, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's aperture. This criterion highlights that shorter wavelengths or larger apertures lead to better resolution, meaning we can distinguish finer details in the observed objects.

By calculating the resolution using the Rayleigh criterion, astronomers can understand the limits of what their telescopes can observe. This is crucial for tasks such as distinguishing between closely situated stars or other celestial objects in the sky.
wavelength conversion
Wavelength conversion is the process of changing the units of wavelength to make calculations more manageable. In the given exercise, wavelengths are converted from nanometers (nm) to meters (m). This step simplifies the use of wavelengths in formulas. For instance, 1 nanometer is equal to \( 1 \times 10^{-9} \) meters.

Let's look at the conversions used in our example:
  • For near-infrared light, \( \lambda = 1,000 \text{nm} \rightarrow 1,000 \times 10^{-9} \text{m} = 10^{-6} \text{m} \).
  • For violet light, \( \lambda = 400 \text{nm} \rightarrow 400 \times 10^{-9} \text{m} = 4 \times 10^{-7} \text{m} \).

These conversions are essential because they allow us to plug the values directly into the Rayleigh criterion formula.
near-infrared observing
Near-infrared observing refers to observing light in the near-infrared region of the electromagnetic spectrum. This region typically covers wavelengths from about 700 nm to 1,000 nm. Due to longer wavelengths, infrared light can penetrate dust clouds that visible light cannot, allowing astronomers to study regions of space that are otherwise obscured.

In the given problem, the wavelength used for near-infrared is \( 1,000 \text{nm} \). When we calculate the resolution for near-infrared using the Rayleigh criterion:
\( \theta_{\text{infrared}} = 1.22 \frac{10^{-6}}{1} = 1.22 \times 10^{-6} \text{radians} \).

This value indicates the theoretical resolution for a 1-meter aperture telescope observing near-infrared light. Since the wavelength is larger, the resolution is larger (poorer) compared to shorter wavelengths.
violet light observing
Violet light observing involves analyzing light in the violet region of the spectrum, which typically covers wavelengths between 380 nm and 450 nm. Violet light has shorter wavelengths compared to infrared, leading to a better resolution according to the Rayleigh criterion.

In our example, we use a wavelength of \( 400 \text{nm} \). When plugged into the Rayleigh criterion formula:
\( \theta_{\text{violet}} = 1.22 \frac{4 \times 10^{-7}}{1} = 4.88 \times 10^{-7} \text{radians} \).

This smaller angle (better resolution) means that the telescope can distinguish finer details when observing violet light compared to near-infrared light. This superior resolution is essential for capturing detailed images of celestial objects.

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Most popular questions from this chapter

The VLBA employs an array of radio telescopes ranging across \(8,000 \mathrm{km}\) of Earth's surface from the Virgin Islands to Hawaii. a. Calculate the angular resolution of the array when radio astronomers are observing interstellar water molecules at a microwave wavelength of \(1.35 \mathrm{cm}\) b. How does this resolution compare with the angular resolution of two large optical telescopes separated by 100 meters and operating as an interferometer at a visible wavelength of \(550 \mathrm{nm}\) ?

You are shopping for telescopes online. You find two in your price range. One of these has an aperture of \(20 \mathrm{cm},\) and the other has an aperture of \(30 \mathrm{cm} .\) If aperture size is the only difference, which should you choose, and why? a. The \(20 \mathrm{cm}\), because the light-gathering power will be better. b. The \(20 \mathrm{cm}\), because the image size will be larger. c. The \(30 \mathrm{cm}\), because the light-gathering power will be better. d. The \(30 \mathrm{cm}\), because the image size will be larger.

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