/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 It is claimed that some professi... [FREE SOLUTION] | 91Ó°ÊÓ

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It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward home plate. One way to judge this claim is to estimate the distance at which a batter can first hope to resolve two points on opposite sides of a baseball, which has a diameter of \(0.0738 \mathrm{m}\). (a) Estimate this distance, assuming that the pupil of the eye has a diameter of \(2.0 \mathrm{mm}\) and the wavelength of the light is \(550 \mathrm{nm}\) in vacuum. (b) Considering that the distance between the pitcher's mound and home plate is \(18.4 \mathrm{m},\) can you rule out the claim based on your answer to part (a)?

Short Answer

Expert verified
No, you can't rule out the claim based on the calculations.

Step by step solution

01

Understand the Problem

We need to determine the distance at which the human eye can resolve two points on a baseball rotating on the way to home plate. We'll use the concept of angular resolution and Rayleigh's criterion for the eye.
02

Use Rayleigh's Criterion

The angular resolution \( \theta \) for the human eye is given by Rayleigh's criterion: \[ \theta = 1.22 \frac{\lambda}{D} \] where \( \lambda = 550 \times 10^{-9} \mathrm{m} \) is the wavelength of light and \( D = 2.0 \times 10^{-3} \mathrm{m} \) is the diameter of the pupil.
03

Calculate the Angular Resolution

Substitute the given values into Rayleigh's criterion: \[ \theta = 1.22 \times \frac{550 \times 10^{-9}}{2.0 \times 10^{-3}} = 3.355 \times 10^{-4} \text{ radians} \].
04

Understand the Geometry

The distance \( d \) at which two points \( L \) meters apart can be resolved is given by \( \theta = \frac{L}{d} \). Here, \( L = 0.0738 \, \mathrm{m} \) is the diameter of the baseball, representing the resolved points.
05

Solve for the Distance

Rearrange the formula to find \( d \): \[ d = \frac{L}{\theta} = \frac{0.0738}{3.355 \times 10^{-4}} \approx 220 \mathrm{m} \].
06

Evaluate the Result Against Claims

The calculated resolving distance (\( 220 \mathrm{m} \)) is much greater than the distance from the pitcher's mound to home plate (\( 18.4 \mathrm{m} \)). This suggests that players can potentially resolve the baseball's rotation at the distance they are positioned.

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Key Concepts

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

Rayleigh Criterion
The Rayleigh criterion is a fundamental principle in optics that helps us determine when two points of light can be distinctly seen as separate. This is particularly interesting when applied to the human eye and its ability to discern fine details. According to the Rayleigh criterion, the angular resolution \( \theta \) of an optical system, like the human eye, is calculated using the formula: \[ \theta = 1.22 \frac{\lambda}{D} \] - Here, \( \lambda \) represents the wavelength of light, and \( D \) is the diameter of the aperture through which the light enters, such as the eye's pupil.- The constant 1.22 is derived from diffraction properties and accounts for the circular shape of apertures and the overlapping diffraction patterns that arise. This principle helps explain why certain objects might appear blurry or indistinguishable unless viewed from a close enough range or with sufficient magnification.
Angular Resolution
Angular resolution refers to the smallest angular separation at which two points can be resolved as distinct entities. In simpler terms, it determines the clarity or detail we can discern in the objects we are viewing. For the human eye, this resolution is linked to factors like:
  • The wavelength of light being observed
  • The size of the optical elements — primarily the diameter of the eye's pupil
Using the previously mentioned Rayleigh criterion, we find that smaller wavelengths and larger apertures generally enhance our ability to resolve smaller angles. This principle not only applies to human vision but also to cameras, telescopes, and other optical devices, guiding us in designing these systems for better precision and detail.
Wavelength of Light
The wavelength of light is an important physical property that influences how we perceive and resolve objects. In our context, we're using light with a wavelength of \( 550 \, \text{nm} \), which is in the visible range — often associated with a greenish-yellow color. The wavelength affects the diffraction pattern, crucial in applications like:
  • Determining the Rayleigh criterion for resolving power
  • Influencing color perception
  • Affecting the efficiency of optical instruments
Light behaves like a wave, and different wavelengths bend or diffract in varying degrees when passing through small apertures like the human pupil. This diffraction can impact the resolution, causing light from separate points to spread out and potentially overlap.
Human Eye Anatomy
Understanding the human eye's ability to resolve detail involves appreciating its anatomy. The human eye functions much like a sophisticated optical instrument, with components designed to focus light and produce clear images: - **Cornea and lens:** These elements work together to refract incoming light, adjusting the eye's focus. - **Pupil:** The aperture of the eye, regulating the amount of light entering the eye. - **Retina:** The surface at the back of the eye where light-sensitive cells convert light into electrical signals sent to the brain. - **Rods and cones:** Specialized cells that detect light intensity and color, respectively. The pupil's size can change depending on ambient light levels, which directly influences the eye's resolution capability. A wider pupil allows more light and better resolution, which is particularly beneficial in low-light conditions. These anatomical features combined allow humans to detect fine details in our environment, giving rise to our naturally impressive, albeit limited, ability to discern objects like a spinning baseball in motion.

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Most popular questions from this chapter

Two gratings A and B have slit separations \(d_{\lambda}\) and \(d_{\mathrm{B}},\) respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating \(\mathrm{B},\) it is observed that the first-order maximum of \(A\) is exactly replaced by the second-order maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

Light of wavelength \(410 \mathrm{nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(1.2 \times 10^{-5} \mathrm{m} .\) The distance between the grating and the viewing screen is \(0.15 \mathrm{m}\). A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation \(\tan \theta=\sin \theta\) can be used. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right),\) what is the distance \(y ?\)

A soap film \((n=1.33)\) is \(375 \mathrm{nm}\) thick and coats a flat piece of glass \((n=1.52) .\) Thus, air is on one side of the film and glass is on the other side, as the figure illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through air and strikes the film nearly perpendicularly. Concepts: (i) What, if any, phase change occurs when light, traveling in air, reflects from the air-film interface? (ii) What, if any, phase change occurs when light, traveling in the film, reflects from the film-glass interface? (iii) Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum? Calculations: For what wavelengths in the range of 380 to 750 nm does constructive interference cause the film to look bright in reflected light?

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm}),\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths. The ability to exhibit interference effects is a fundamental characteristic of any kind of wave. Our understanding of these effects depends on the principle of linear superposition, which we first encountered in Chapter 17\. Only by means of this principle can we understand the constructive and destructive interference of light waves that lie at the heart of every topic in this chapter. Problem 67 serves as a review of the essence of this principle. Problem 68 deals with thin-film interference and reviews the factors that must be considered in such cases.

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and \(1.40 .\) Light that has a wavelength of \(625 \mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

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