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Chapter 27: Problem 34

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 m. (a) Estimate this distance, assuming that the pupil of the eye has a diameter of 2.0 mm and the wavelength of the light is 550 nm in vacuum. (b) Considering that the distance between the pitcher鈥檚 mound and home plate is 18.4 m, can you rule out the claim based on your answer to part (a)?

Short Answer

Expert verified
Seeing the spin between mound and plate is unlikely given resolution limit.

Step by step solution

01

Understand the Formula for Angular Resolution

The resolution of the eye is determined by the Rayleigh criterion. This criterion gives the minimum angular separation \( \theta \) that can be resolved, which is \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture (in this case, the pupil of the eye).
02

Convert Given Values to Consistent Units

We need to convert all given values into consistent units for calculation:- Diameter of the baseball: \( 0.0738 \) m - Diameter of the pupil: \( 2.0 \) mm = \( 0.002 \) m - Wavelength of light: \( 550 \) nm = \( 550 \times 10^{-9} \) m
03

Calculate Minimum Angular Resolution

Substitute the given values into the Rayleigh criterion formula: \( \theta = 1.22 \frac{\lambda}{D} \) = 1.22 \( \frac{550 \times 10^{-9}}{0.002} \) rad. This works out to: \[ \theta \approx 3.355 \times 10^{-4} \text{ rad} \]
04

Calculate the Minimum Distance for Resolution

Using the small angle approximation, the minimum distance \( s \) at which the two points can be resolved is related to \( \theta \) and the diameter of the baseball \( d \) by: \[ \theta = \frac{d}{s} \]Rearranging, we find: \[ s = \frac{d}{\theta} \]Substitute \( d = 0.0738 \) m and \( \theta = 3.355 \times 10^{-4} \): \[ s \approx \frac{0.0738}{3.355 \times 10^{-4}} \approx 220 \text{ m} \]
05

Compare with The Pitcher's Mound to Home Plate Distance

Now, compare the calculated resolution distance of \( 220 \) meters with the actual distance from the pitcher's mound to home plate, which is \( 18.4 \) meters. As \( 220 \) meters is significantly larger than \( 18.4 \) meters, the eye cannot resolve the spin of the ball at 18.4 meters.

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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 key principle in understanding how well we can distinguish two separate points of light. It defines the minimum angular separation, \( \theta \), that can be resolved by an optical system. The formula for the Rayleigh criterion is given by:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
  • Where \( \lambda \) is the wavelength of light.
  • And \( D \) is the diameter of the aperture, which in this case is the pupil of the eye.
This criterion is crucial as it helps determine if we can actually see two separate objects or if they will blur into one. It's especially interesting in contexts like sports, where players might try to see small details at a distance.
wavelength of light
The wavelength of light is a fundamental concept in physics, representing the distance between successive peaks of a wave. In the context of the Rayleigh criterion, it plays a significant role in determining angular resolution. The wavelength is often measured in nanometers (nm) when dealing with visible light:
  • In this scenario, the wavelength used is \( 550 \) nm, which is a common approximation for sunlight.
  • This value translates to \( 550 \times 10^{-9} \) meters when converted to standard units.
Understanding wavelength is vital because it directly affects how we perceive light and resolve different objects. Shorter wavelengths can potentially offer higher resolution than longer ones, given the same aperture diameter.
diameter of pupil
The diameter of the pupil is essential in optics because it acts as the aperture for the eye, affecting how much light enters and how well it can resolve detail.
  • In this exercise, the pupil's diameter is given as \( 2.0 \) mm, which converts to \( 0.002 \) meters.
  • A larger pupil allows more light to enter, which can enhance visual acuity under certain conditions.
The size of the pupil constantly changes with the amount of ambient light鈥攄ilating in dim conditions and constricting in bright environments. In resolution calculations, the pupil's diameter is crucial since it directly influences the minimum angular resolution according to the Rayleigh criterion.
small angle approximation
The small angle approximation is a mathematical simplification used when dealing with tiny angles, such as those found in optical resolution problems. The idea is that for small angles (measured in radians), the sine of the angle is approximately equal to the angle itself:
  • In formulas: \( \sin \theta \approx \theta \).
  • This simplification is valid when \( \theta \) is much less than 1 radian.
In the context of this exercise, the small angle approximation is applied when calculating the minimum distance \( s \) at which the baseball's sides can be told apart:
  • It aids in rearranging \( \theta = \frac{d}{s} \) to solve for \( s \).
  • This approximation simplifies complex trigonometric equations into linear ones, making it easier to find practical solutions in resolving tiny angular separations.
This tool is useful in physics for reducing computational complexity without significant loss of accuracy, especially in conditions where angular measurements are minimal.

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Most popular questions from this chapter

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{nm},\) what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

The distance between adjacent slits of a certain diffraction grating is \(1.250 \times 10^{-5} \mathrm{m}\) . The grating is illuminated by monochromatic light with a wavelength of \(656.0 \mathrm{nm},\) and is then heated so that its temperature increases by 100.0 \(\mathrm{C}^{\circ} .\) Determine the change in the angle of the seventh order principal maximum that occurs as a result of thermal expansion of the grating. The cofficient of linear expansion for the diffaction grating is \(1.30 \times 10^{-4}\left(\mathrm{C}^{9}\right)^{-1} .\) Be sure to include the proper algebraic sign with your answer: \(+\) if the angle increases, - if the angle decreases.

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure 27.6\(a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm} )\) of the monochromatic light?

In a single-slit diffraction pattern on a flat screen, the central bright fringe is 1.2 \(\mathrm{cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{m} .\) When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of 1.9 \(\mathrm{cm} .\) What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\)

Light of wavelength 410 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 \(\tan \theta \approx \sin \theta\) can be used.) (b) If the entire apparatus is submerged in water \(\left(n_{\text { water }}=1.33\right),\) what is the distance \(y ?\)
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