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Chapter 36: Problem 19

(a) How far from grains of red sand must you be to position yourself just at the limit of resolving the grains if your pupil diameter is \(1.5 \mathrm{~mm}\), the grains are spherical with radius \(50 \mu \mathrm{m}\), and the light from the grains has wavelength \(650 \mathrm{~nm} ?\) (b) If the grains were blue and the light from them had wavelength \(400 \mathrm{~nm}\), would the answer to (a) be larger or smaller?

Short Answer

Expert verified
For red light, you're 0.189 m away. Blue light increases the distance to 0.308 m.

Step by step solution

01

Understand the Problem

We need to determine how far you must be from grains of sand to barely resolve them with your eye. The resolving power of the eye depends on its aperture (pupil diameter) and the wavelength of light. The resolving limit can be calculated using the Rayleigh's criterion.
02

Calculate the Angular Resolution

Using Rayleigh's criterion, the angular resolution for a circular aperture is given by: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the pupil. For red light (\( \lambda = 650 \text{ nm} = 650 \times 10^{-9} \text{ m} \)), \( D = 1.5 \times 10^{-3} \text{ m} \).
03

Substitute Red Light Values into Rayleigh's Criterion

Substitute values for red light into the equation: \( \theta = 1.22 \times \frac{650 \times 10^{-9} \text{ m}}{1.5 \times 10^{-3} \text{ m}} \). Calculate \( \theta \) to determine the minimum resolvable angle in radians.
04

Solve the Angular Resolution Equation

\( \theta = 1.22 \times \frac{650 \times 10^{-9}}{1.5 \times 10^{-3}} = \approx 5.29 \times 10^{-4} \text{ radians} \). This is the smallest angle that can be resolved.
05

Convert Angular Resolution to Distance

The distance \( L \) at which the limit of resolution occurs can be found using the small angle approximation: \( \theta \approx \frac{d}{L} \), where \( d \) is the diameter of one grain of sand (\(100 \times 10^{-6} \text{ m} \)). Rearrange the equation to find \( L \): \( L = \frac{d}{\theta} \).
06

Calculate Distance for Red Light

Substitute \( d = 100 \times 10^{-6} \text{ m} \) and \( \theta = 5.29 \times 10^{-4} \text{ radians} \): \( L = \frac{100 \times 10^{-6}}{5.29 \times 10^{-4}} \approx 0.189 \text{ m} \). This is the distance required for red light.
07

Determine Effect of Blue Light

For blue light, the wavelength \( \lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} \). Substitute this into Rayleigh's criterion: \( \theta_{\text{blue}} = 1.22 \frac{400 \times 10^{-9}}{1.5 \times 10^{-3}} \).
08

Calculate Angular Resolution for Blue Light

\( \theta_{\text{blue}} = 1.22 \times \frac{400 \times 10^{-9}}{1.5 \times 10^{-3}} = \approx 3.25 \times 10^{-4} \text{ radians} \). The resolving angle for blue light is smaller.
09

Determine Distance for Blue Light

Using the smaller angle \( \theta_{\text{blue}} \), calculate \( L \): \( L = \frac{100 \times 10^{-6}}{3.25 \times 10^{-4}} \approx 0.308 \text{ m} \). The distance is larger for blue light.

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Key Concepts

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

Angular Resolution
Angular resolution describes how well we can distinguish two nearby points as separate objects. In simple terms, it鈥檚 the smallest angle between two objects that allows them to be seen as distinct entities.
The concept is crucial in optical systems, such as telescopes, microscopes, and even the human eye.
Rayleigh's criterion provides a practical way to calculate this resolution. It suggests that the smallest angle, \( \theta \), that can be resolved, depends on both the wavelength of light \( (\lambda) \) and the diameter of the aperture \( (D) \).
  • The formula to determine angular resolution is given by: \( \theta = 1.22 \frac{\lambda}{D} \).
  • A smaller angle means better resolution, allowing for finer details to be observed.
Understanding and calculating angular resolution helps in scenarios like the exercise, determining how far we need to be from an object to resolve its details properly.
Wavelength of Light
The wavelength of light is a fundamental property that influences the resolving power of optical systems.
Light behaves as a wave, and its wavelength is the distance between consecutive peaks of the wave. In the context of resolution, the wavelength directly affects the angular resolution.
  • Longer wavelengths result in larger angles of resolution, reducing the ability to distinguish fine details.
  • Conversely, shorter wavelengths allow for better resolution, as the corresponding angle is smaller.
This was evident in the provided exercise where changing from red light \( (650 \text{ nm}) \) to blue light \( (400 \text{ nm}) \) altered the distance required for resolving the sand grains.
Small Angle Approximation
The small angle approximation states that when an angle is very small (measured in radians), its sine and tangent are approximately equal to the angle itself.
This simplifies calculations considerably in scenarios involving small angles, making it a valuable tool in optics and physics.
  • For our exercise, the approximation enables us to use \( \theta \approx \frac{d}{L} \), where \( d \) is the size of the object, and \( L \) is the distance from the observer.
  • It's crucial when translating angular resolution into an actual measurement of distance.
The ease and accuracy of this approximation highlight why it鈥檚 a widely adopted technique when dealing with light and angles in optical systems.
Resolving Power
Resolving power is a term that reflects an optical system鈥檚 ability to distinguish between two closely spaced points.
Its importance is in how it defines the capabilities of lenses and the human eye, among other systems.
  • A system with higher resolving power can separate points that are closer together, which is essential in fields such as microscopy, astronomy, and more.
  • This power is directly influenced by the wavelength and aperture鈥攕horter wavelengths and larger apertures enhance the resolving power.
The original exercise demonstrates how changing conditions (like the wavelength of light from red to blue) affects resolving power, thus altering the necessary distance to distinguish sand grains.

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Most popular questions from this chapter

In the single-slit diffraction experiment of Fig. \(36-4\), let the wavelength of the light be \(500 \mathrm{~nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=3.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{P}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm} .\) (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

Entoptic halos. If someone looks at a bright outdoor lamp in otherwise dark surroundings, the lamp appears to be surrounded by bright and dark rings (hence halos) that are actually a circular diffraction pattern as in Fig. \(36-10\), with the central maximum overlapping the direct light from the lamp. The diffraction is produced by structures within the cornea or lens of the eye (hence entoptic). If the lamp is monochromatic at wavelength \(550 \mathrm{~nm}\) and the first dark ring subtends angular diameter \(2.5^{\circ}\) in the observer's view, what is the (linear) diameter of the structure producing the diffraction?

In an experiment to monitor the Moon's surface with a light beam, pulsed radiation from a ruby laser \((\lambda=0.69 \mu \mathrm{m})\) was directed to the Moon through a reflecting telescope with a mirror radius of \(1.3 \mathrm{~m}\). A reflector on the Moon behaved like a circular flat mirror with radius \(10 \mathrm{~cm}\), reflecting the light directly back toward the telescope on Earth. The reflected light was then detected after being brought to a focus by this telescope. Approximately what fraction of the original light energy was picked up by the detector? Assume that for each direction of travel all the energy is in the central diffraction peak.

Visible light is incident perpendicularly on a grating with 315 rulings/mm. What is the longest wavelength that can be seen in the fifth-order diffraction?

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.
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