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Two stars are \(3.7 \times 10^{11} \mathrm{~m}\) apart and are equally distant from the earth. A telescope has an objective lens with a diameter of \(1.02 \mathrm{~m}\) and just detects these stars as separate objects. Assume that light of wavelength \(550 \mathrm{~nm}\) is being observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the telescope. Find the maximum distance that these stars could be from the earth.

Short Answer

Expert verified
The maximum distance is approximately \( 5.63 \times 10^{17} \mathrm{~m} \).

Step by step solution

01

Identify the Formula

To solve this problem, we need to determine the maximum distance the stars could be from Earth using the resolving power of the telescope. The formula we will use is the Rayleigh criterion for resolution, given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution (in radians), \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's objective lens.
02

Calculate Angular Resolution

Given \( \lambda = 550 \mathrm{~nm} = 550 \times 10^{-9} \mathrm{~m} \) and \( D = 1.02 \mathrm{~m} \), calculate \( \theta \) using the formula \( \theta = 1.22 \frac{550 \times 10^{-9}}{1.02} \). Solve this to find \( \theta \).
03

Convert Angular Resolution to Radians

Calculate \( \theta \) as \( \theta = 1.22 \times \frac{550 \times 10^{-9}}{1.02} \approx 6.57 \times 10^{-7} \mathrm{~radians}\). This is the minimum angular separation that can be resolved by the telescope.
04

Use Small Angle Approximation

Since the stars are very far away, we can use the small angle approximation that says \( \theta \approx \frac{s}{L} \), where \( s \) is the separation of the stars \( (3.7 \times 10^{11} \mathrm{~m}) \) and \( L \) is the distance to the stars (from the Earth).
05

Calculate Maximum Distance

Rearrange the small angle approximation \( \theta = \frac{s}{L} \) to solve for \( L \): \( L = \frac{s}{\theta} \). Substitute \( s = 3.7 \times 10^{11} \mathrm{~m} \) and \( \theta = 6.57 \times 10^{-7} \mathrm{~radians} \) to find \( L = \frac{3.7 \times 10^{11}}{6.57 \times 10^{-7}} \approx 5.63 \times 10^{17} \mathrm{~m} \).

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Key Concepts

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

Angular Resolution
Angular resolution is a crucial concept in optical systems like telescopes, as it dictates the system's ability to distinguish between two closely spaced objects. In simple terms, it refers to how close two objects can be before they appear as one blurred object. For telescopes, higher angular resolution means being able to separate finer details when observing celestial objects.
When light from these objects reaches the telescope, they form a diffraction pattern due to the wave nature of light. This is where the Rayleigh criterion comes in, which we use to determine the smallest angle between them that can still be resolved. Mathematically, this resolution is given by:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
Here, \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of the light, and \( D \) is the diameter of the telescope's lens. Therefore, larger telescopes or shorter wavelengths result in better angular resolution.
Diffraction Limit
The diffraction limit is a fundamental limitation in optical systems, defined by the wave nature of light. It is the smallest detail that a telescope or other optical device can resolve, determined by the interference of light waves diffracting around the edges of the aperture.
For instance, even with perfect lenses, the light waves that form an image spread out into a pattern consisting of a central bright spot surrounded by rings, called an Airy disk. The Rayleigh criterion specifies that two objects can be resolved as separate if their images are closer than the peak of one Airy disk falls on the first minimum of another. - Limitations set by diffraction directly impact the effectiveness of observing faint, distant objects in space. - The diffraction limit is influenced by the wavelength of light being observed; shorter wavelengths can help achieve a higher resolution.
Telescope Optics
Telescope optics is the science of designing optical systems to gather and focus light from distant objects in space. The main goal is to capture faint light and reveal details using lenses or mirrors. Optical telescopes enhance stargazing by providing a view of stars, planets, and galaxies. - **Objective Lens/Mirror**: This is the primary component of a telescope that collects light. A larger objective allows more light to be captured, improving brightness and detail. - **Resolving Power**: Determined by the Rayleigh criterion, which predicts how two points of light will appear as separate or a single blur. - **Focal Length**: Longer focal lengths provide higher magnifications resulting in more detailed observations of small objects at vast distances. Advancements in telescope optics have vastly improved our understanding of the universe. They're not only limited by their design but also by atmospheric conditions; however, space telescopes bypass this limitation.
Wavelength of Light
Wavelength is a critical trait of light, representing the distance between two consecutive peaks of a wave. It's fundamental to understanding many optical phenomena. Wavelengths are typically measured in nanometers (nm) when dealing with visible light, like the 550 nm often used in astronomical observations. - **Visible Spectrum**: Ranges from about 400 nm (violet) to 700 nm (red). The light emitted or reflected by stars often falls within this spectrum. - **Influence on Resolution**: Shorter wavelengths allow better resolution. This is why telescopes like the Hubble Space Telescope, with the capability to capture ultraviolet light, yield higher resolution images. - **Role in Diffraction**: The wavelength affects when light diffresses while passing through lenses or around edges, contributing to the diffraction limit. Understanding the wavelength of light helps in tailoring telescopes for optimal performance, such as deciding the best filter or sensor to capture light from distant celestial objects.

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Most popular questions from this chapter

A mix of red light \(\left(\lambda_{\text {vacuum }}=661 \mathrm{nm}\right)\) and green light \(\left(\lambda_{\text {vacuum }}=551 \mathrm{nm}\right)\) is directed perpendicularly onto a soap film \((n=1.33)\) that has air on either side. What is the minimum nonzero thickness of the film, so that destructive interference causes it to look red in reflected light?

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{~m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{~m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{~nm}\), what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\)

A hunter who is a bit of a braggart claims that, from a distance of \(1.6 \mathrm{~km}\), he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{~nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{~mm}\), the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{~nm}\).

Light shines through a single slit whose width is \(5.6 \times 10^{-4} \mathrm{~m}\). A diffraction pattern is formed on a flat screen located \(4.0 \mathrm{~m}\) away. The distance between the middle of the central bright fringe and the first dark fringe is \(3.5 \mathrm{~mm}\). What is the wave length of the light?

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