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Suppose that two points are separated by \(2.0 \mathrm{~cm} .\) If they are viewed by an eye with a pupil opening of \(5.0 \mathrm{~mm}\), what distance from the viewer puts them at the Rayleigh limit of resolution? Assume a light wavelength of \(500 \mathrm{~nm}\).

Short Answer

Expert verified
The points are at the Rayleigh limit at approximately 164 meters from the viewer.

Step by step solution

01

Understand Rayleigh's Criterion

Rayleigh's criterion for resolution describes the minimum angular separation (θ) at which two bright points can be distinctly resolved. It is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the pupil.
02

Calculate Angular Resolution

Convert the given measurements to consistent units: wavelength \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} \) and pupil diameter \( D = 5.0 \text{ mm} = 0.005 \, \text{m} \). Substituting into Rayleigh's criterion: \( \theta = 1.22 \times \frac{500 \times 10^{-9}}{0.005} = 1.22 \times 10^{-4} \, \text{radians}. \)
03

Find Distance Using Small Angle Approximation

Use the small angle approximation: \( \theta \approx \frac{s}{L} \), where \( s \) is the separation between the points (2.0 cm = 0.02 m) and \( L \) is the distance from the viewer. Set \( \theta = 1.22 \times 10^{-4} \, \text{radians} \): \[ 1.22 \times 10^{-4} = \frac{0.02}{L} \].
04

Solve for L

Re-arrange the equation to solve for \( L \): \( L = \frac{0.02}{1.22 \times 10^{-4}} = 163.934 \). Round this to three significant figures, we get \( L = 164 \, \text{m} \).

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Key Concepts

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

Understanding Angular Resolution
Angular resolution is a crucial concept in optics that determines the ability to distinguish between two closely spaced objects. In practical terms, it's the smallest angle through which two point sources (like stars) can be seen as distinct from each other. This concept is vital in determining how clear or sharp an image can be.

The Rayleigh criterion is used to quantify angular resolution. According to this criterion, the minimum angular separation (\( \theta \)) that allows us to distinguish two points is given as:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
where:
  • \( \lambda \) is the wavelength of the light being used, and
  • \( D \) is the diameter of the aperture (like a telescope or a camera lens).
This equation reveals that angular resolution improves with larger aperture sizes, allowing us to resolve finer details. Therefore, more substantial apertures in optical devices lead to better clarity under the same conditions.
The Role of Wavelength of Light
Wavelength, often denoted by \( \lambda \), is the distance between successive peaks of a wave. In optics, it plays a pivotal role in determining the resolution limits due to its influence on diffraction patterns. When light passes through an aperture or around an edge, it spreads out, creating a pattern of constructive and destructive interference called a diffraction pattern.

The diffraction pattern is responsible for the limits on how finely details can be resolved. A shorter wavelength (like blue light) will have smaller diffraction patterns, allowing for better resolution compared to longer wavelengths (like red light). This is why optical instruments performing observations at shorter wavelengths can distinguish finer details.

Thus, in the context of Rayleigh's Criterion, a smaller value of \( \lambda \) reduces the minimum resolvable angle \( \theta \), allowing observers to see more detail. Light with a wavelength of \( 500\ nm \) (nanometers) typically represents green light and is a standard approximation when discussing average resolution limits in the visible spectrum.
The Small Angle Approximation Simplified
The small angle approximation is a mathematical tool often used in physics to simplify equations when dealing with very small angles, typically measured in radians. This approximation stems from the fact that when angles are small, their sine, tangent, and the angle itself (in radians) are nearly equal.

For practical applications, the small angle approximation assumes:
  • \( \tan(\theta) \approx \theta \)
  • \( \sin(\theta) \approx \theta \)
In angular resolution problems, this approximation allows the equation \( \theta \approx \frac{s}{L} \) to be used, where:
  • \( s \) is the separation between points or objects, and
  • \( L \) is the distance from the viewer or the point of observation.
Applying the small angle approximation simplifies many complex trigonometric calculations, especially in the context of optics. It streamlines the process of determining distances or resolving powers without delving into more intricate math. In this exercise, it provided a straightforward way to find the necessary distance \( L \) for objects at the Rayleigh limit of resolution, translating angular measures into practical distances.

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Most popular questions from this chapter

Estimate the linear separation of two objects on Mars that can just be resolved under ideal conditions by an observer on Earth (a) using the naked eye and (b) using the 200 in. \((=5.1 \mathrm{~m})\) Mount Palomar telescope. Use the following data: distance to Mars \(=8.0 \times 10^{7} \mathrm{~km}\), diameter of pupil \(=5.0 \mathrm{~mm}\), wavelength of light \(=550 \mathrm{~nm}\).

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm} .\) The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) By what factor is the beam intensity reduced in transit to the target? (The laser is fired from space, so that atmospheric absorption can be ignored.)

A diffraction grating is made up of slits of width \(300 \mathrm{~nm}\) with separation \(900 \mathrm{~nm}\). The grating is illuminated by monochromatic plane waves of wavelength \(\lambda=600 \mathrm{~nm}\) at normal incidence. (a) How many maxima are there in the full diffraction pattern? (b) What is the angular width of a spectral line observed in the first order if the grating has 1000 slits?

With light from a gaseous discharge tube incident normally on a grating with slit separation \(1.73 \mu \mathrm{m}\), sharp maxima of green light are experimentally found at angles \(\theta=\pm 17.6^{\circ}, 37.3^{\circ},-37.1^{\circ}\), \(65.2^{\circ}\), and \(-65.0^{\circ} .\) Compute the wavelength of the green light that best fits these data.

If Superman really had x-ray vision at \(0.10 \mathrm{~nm}\) wavelength and a \(4.0 \mathrm{~mm}\) pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by \(5.0 \mathrm{~cm}\) to do this?

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