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A telescope having an objective lens with a diameter of 10.0 \(\mathrm{cm}\) will be used to view two equally bright small sources of \(550-\mathrm{nm}\) light. (a) Determine the angular separation of the sources if they are just resolvable. Use Rayleigh's criterion. (b) How far apart can they be at a distance of \(1000 \mathrm{km} ?\)

Short Answer

Expert verified
(a) Angular separation is \(6.71 \times 10^{-6}\) radians. (b) The sources can be \(6.71\) meters apart.

Step by step solution

01

Understand Rayleigh's Criterion

According to Rayleigh's criterion, two points are just resolvable when the central maximum of the diffraction pattern of one image coincides with the first minimum of the diffraction pattern of the other. The formula is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \(\lambda\) is the wavelength of the light and \(D\) is the diameter of the objective lens.
02

Calculate Angular Separation

First, convert the given wavelength \(\lambda\) from nm to meters: \( \lambda = 550 \times 10^{-9} \text{ m} \). Now use Rayleigh's criterion: \( \theta = 1.22 \times \frac{550 \times 10^{-9}}{0.10} \). This simplifies to \( \theta = 6.71 \times 10^{-6} \text{ radians } \).
03

Determine the Linear Separation

Given the angle \(\theta\) and the distance \(d = 1000 \times 10^3 \text{ m} \) from the sources, use the small angle approximation: \( s = d \cdot \theta \). Plug in the values: \( s = 1000 \times 10^3 \times 6.71 \times 10^{-6} \). Simplifying, we get \( s \approx 6.71 \text{ meters} \).

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Key Concepts

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

Angular Separation
The concept of angular separation is crucial when discussing the ability of optical systems, like telescopes, to resolve two closely placed point sources. Angular separation is essentially the angle between the lines extending from a lens or mirror to two points that are barely discernible. In the case of Rayleigh's criterion, determining the minimum angular separation is essential to understand when two light sources can be considered as separate entities.

To achieve this distinction, we use the Rayleigh criterion, which offers a mathematical rule to specify this angular limit. It involves diffraction, a fundamental optical effect occurring when light passes through an aperture, like a lens. Using the formula given by Rayleigh, \[ \theta = 1.22 \frac{\lambda}{D} \]we can find the smallest angular separation (\(\theta\)). Here, \(\lambda\) represents the wavelength of light, and \(D\) indicates the diameter of the objective lens. The factor 1.22 is introduced owing to the diffraction pattern's characteristics of circular apertures.
Diffraction Pattern
A diffraction pattern is a series of light and dark bands, known as fringes, that occurs when light waves bend around obstacles, such as the edges of the lens aperture, and interfere with each other. This interference can be understood as the overlapping of light waves that overlap constructively or destructively at different points. In a telescope or any similar optical system, the central maximum of a diffraction pattern is the brightest spot. It is surrounded by a series of concentric circles of diminishing intensity, known as diffraction rings. The first minimum around the central maximum is particularly significant in resolving images. For Rayleigh's criterion, two point sources are considered 'just resolvable' if the central maximum of one image coincides exactly with the first minimum of another. This establishes the foundation of calculating the angular separation threshold for an optical system. Understanding this principle provides insights into the optical resolution limits of telescopes and similar instruments.
Objective Lens Diameter
The objective lens diameter in optical devices like telescopes is critical in determining resolution. Larger diameters allow for more light to be gathered, and significantly influence the resolving power of the device.

When considering Rayleigh's criterion, the diameter of the objective lens (\(D\)) directly impacts the angular separation formula: \[ \theta = 1.22 \frac{\lambda}{D} \]This implies that a larger objective lens results in a smaller minimum angular separation, allowing it to distinguish between two close light sources. Hence, increasing the lens diameter enhances an optical system's ability to resolve fine details, which is why telescopes with larger lenses can see more distant and detailed celestial objects.

When astronomers design and choose telescopes, the lens diameter is a key factor determining its observational strength and capability.
Wavelength of Light
The wavelength of light is a crucial factor in defining the characteristics of the light that our eyes and optical instruments observe. Essentially, it is the distance over which the wave's shape repeats, and it's typically measured in nanometers (nm) for visible light.In the context of Rayleigh's criterion, the wavelength (\(\lambda\)) is part of the equation determining the resolving power of an optical system: \[ \theta = 1.22 \frac{\lambda}{D} \]Since the formula states that \(\lambda\) is in the numerator, a larger wavelength will result in a larger minimum angular separation, implying poorer resolution. This means that shorter wavelengths, like blue light, provide better resolving capabilities than longer wavelengths, like red light.Therefore, astronomers often consider the wavelength when adjusting their telescopes for observing celestial objects, tailoring the observation to the part of the spectrum that best suits their needs. Understanding wavelength helps in grasping how light interacts with optical equipment.

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Most popular questions from this chapter

A narrow single slit (in air) in an opaque screen is illuminated by infrared from a He-Ne laser at \(1152.2 \mathrm{nm}\), and it is found that the center of the tenth dark band in the Fraunhofer pattern lies at an angle of \(6.2^{\circ}\) off the central axis. Determine the width of the slit. At what angle will the tenth minimum appear if the entire arrangement is immersed in water \(\left(n_{w}=1.33\right)\) rather than air \(\left(n_{a}=1.00029\right) ?\)

Collimated light from a krypton ion laser at \(568.19 \mathrm{nm}\) impinges normally on a circular aperture. When viewed axially from a distance of \(1.00 \mathrm{m}\), the hole uncovers the first half-period Fresnel zone. Determine its diameter.

Consider a Fresnel zone plate having a transparent circular disk at its center. This is the \(m=1\) region, and the tenth transparent region has a diameter of \(6.00 \mathrm{mm}\). Determine the plate's principal focal length when \(\lambda_{0}=600 \mathrm{nm}\).

consider the single-slit Fraunhofer diffraction pattern formed on a screen by a lens of focal length \(f .\) Show that the peak of the first subsidiary bright band is a distance \(Y\) (measured from the central axis) on the viewing screen, given by $$\approx 1.4303 \frac{\lambda f}{b}$$

Consider the Fraunhofer diffraction pattern for eight very narrow parallel slits under monochromatic illumination. (a) Sketch the resulting irradiance distribution. (b) Explain why the first minimum occurs, from a phasor perspective. (c) Why is the electric field zero midway between principal peaks? (d) What does the phasor diagram of the field amplitude look like for the second minimum (measured from the zeroth principal maximum)? (e) What are the angles between successive phasors at each minimum considered above?

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