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Chapter 36: Problem 39

(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 \(1.8 \mathrm{~mm}\), 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
Red light: 0.228 m; Blue light: larger distance (0.369 m).

Step by step solution

01

Understand the Parameters

In this problem, we are trying to find out how far away we can be from grains of sand and still resolve them, given specific parameters: pupil diameter, grain size, and light wavelength.
02

Use Rayleigh's Criterion for Resolution

According to Rayleigh's criterion, the minimum angular resolution (胃) is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture (the pupil, in this case).
03

Calculate Angular Resolution for Red Sand

For red light, \( \lambda = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \) and \( D = 1.8 \, \text{mm} = 1.8 \times 10^{-3} \, \text{m} \). Substitute these into Rayleigh's criterion: \( \theta = 1.22 \times \frac{650 \times 10^{-9}}{1.8 \times 10^{-3}} \). This yields \( \theta \approxeq 4.39 \times 10^{-4} \, \text{radians} \).
04

Relate Angular Resolution to Physical Distance

The angular resolution can also be expressed as the angle subtended by the grain when viewed from a distance \( R \). This is \( \theta = \frac{d}{R} \), where \( d = 100 \times 10^{-6} \, \text{m} \) (diameter of the grain: 2 times the radius).
05

Solve for Distance R (Red Light)

Rearrange \( \frac{d}{R} = 4.39 \times 10^{-4} \) to find \( R \). Substituting \( d = 100 \times 10^{-6} \, \text{m} \), we have \( R = \frac{100 \times 10^{-6}}{4.39 \times 10^{-4}} \approx 0.228 \, \text{m} \).
06

Calculate Resolution for Blue Light

For blue light, \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \). Compute \( \theta = 1.22 \times \frac{400 \times 10^{-9}}{1.8 \times 10^{-3}} \approx 2.71 \times 10^{-4} \, \text{radians} \).
07

Compare Distances for Red and Blue Light

Use the same process as Step 5 to find \( R \) for blue light: \( R = \frac{100 \times 10^{-6}}{2.71 \times 10^{-4}} \approx 0.369 \text{m} \). 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 is a key concept when it comes to the ability to distinguish between two close objects. Imagine looking at two nearby grains of sand through a camera lens or your eyes. Angular resolution determines how clear this distinction can be.
Rayleigh's criterion is often used to explain angular resolution. According to this criterion, the angular resolution \( \theta \) is given by:
  • \[ \theta = 1.22 \frac{\lambda}{D} \]
Here, \( \lambda \) is the wavelength of the light, and \( D \) is the aperture diameter. A smaller angular resolution means a clearer distinction between two objects viewed at a distance.
In practical terms, if the angular resolution is high (meaning \( \theta \) is small), you can separate fine details and view them clearly. If it's low (\( \theta \) is large), the details blend together and may appear as one. This concept is vital in fields like astronomy, photography, and eye health, where resolving details clearly is crucial.
Wavelength of Light
The wavelength of light plays an essential role in determining how well you can resolve fine details. Different colors of light have different wavelengths. For instance, red light has a longer wavelength compared to blue light.
  • Red light: \( \lambda = 650 \) nm
  • Blue light: \( \lambda = 400 \) nm
The formula for angular resolution shows that a longer wavelength results in a larger angular resolution, translating to less clarity. Therefore, using shorter wavelengths, like blue light, allows for finer details to be seen clearly compared to longer wavelengths like red light. In the exercise, for red light, the resulting distance for resolving grains is smaller compared to blue light because a shorter wavelength like blue light allows you to see details from further away. Hence, different light wavelengths affect clarity and resolving power differently.
Aperture Diameter
Aperture diameter signifies the size of the opening through which light enters a system, such as your eye or a camera lens. A crucial aspect in determining how much light can be gathered. A larger aperture can gather more light, which helps in achieving better image clarity.
In Rayleigh's criterion, the aperture diameter \( D \) is vital:
  • The formula \( \theta = 1.22 \frac{\lambda}{D} \) implies that a larger \( D \) results in a smaller angular resolution, which means greater clarity.
In the exercise example, the pupil diameter of 1.8 mm is the aperture diameter. A larger eye or lens would be able to resolve the sand grains from a further distance due to a smaller value of \( \theta \). In practical scenarios, devices like telescopes and cameras leverage large apertures to enhance resolution.

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Most popular questions from this chapter

Assume that the limits of the visible spectrum are arbitrarily chosen as \(430 \mathrm{~nm}\) and \(680 \mathrm{~nm}\). Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of \(12.0^{\circ}\).

Nuclear-pumped \(x\)-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(1500 \mathrm{~km}\). 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 \(0.400 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

Light of wavelength \(420 \mathrm{~nm}\) is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by \(\sin \theta=0.2\) and \(\sin \theta=0.3\). The fourth-order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width, what are the (c) largest, (d) second largest, and (e) third largest values of the order number \(m\) of the maxima produced by the grating?

The telescopes on some commercial surveillance satellites can resolve objects on the ground as small as \(85 \mathrm{~cm}\) across (see Google Earth), and the telescopes on military surveillance satellites reportedly can resolve objects as small as \(10 \mathrm{~cm}\) across. Assume first that object resolution is determined entirely by Rayleigh's criterion and is not degraded by turbulence in the atmosphere. Also assume that the satellites are at a typical altitude of \(420 \mathrm{~km}\) and that the wavelength of visible light is \(550 \mathrm{~nm}\). What would be the required diameter of the telescope aperture for (a) \(85 \mathrm{~cm}\) resolution and (b) \(10 \mathrm{~cm}\) resolution? (c) Now, considering that turbulence is certain to degrade resolution and that the aperture diameter of the Hubble Space Telescope is \(2.4 \mathrm{~m}\), what can you say about the answer to (b) and about how the military surveillance resolutions are accomplished?

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(650 \mathrm{~nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first and third diffraction minima on the same side of the central diffraction maximum?
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