/*! 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 62 The maximum resolution of the ey... [FREE SOLUTION] | 91Ó°ÊÓ

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The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Short Answer

Expert verified
(a) No, Rayleigh criterion does not resolve 50-μm; (b) Shortest is about 34 μm; (c) Object subtends about 8.24 min, greater than 1 min; (d) Retinal cell size limits.

Step by step solution

01

Understanding Rayleigh's Criterion

Rayleigh's criterion for resolution limits is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture. Here, \( \lambda = 550 \) nm and \( D = 2.0 \) mm.
02

Calculate Angular Resolution

Substitute the given values into Rayleigh's criterion formula: \( \theta = 1.22 \times \frac{550 \times 10^{-9} \, \text{m}}{2.0 \times 10^{-3} \, \text{m}} \). Compute \( \theta \).
03

Converting Angular Resolution to Object Size

Convert \( \theta \) to the linear resolution at the 25 cm near point using \( h = \theta \times d \), where \( h \) is the height of the object (50 \( \mu \)m) and \( d = 0.25 \) m.
04

Resolving 50-\( \mu \)m Object

Comparing the computed height from Step 3 with 50 \( \mu \)m to determine if the Rayleigh criterion allows resolution of an object this size.
05

Shortest Resolvable Object Calculation

Given the angular resolution \( \theta \), calculate the smallest object size \( h_{min} \) that can be resolved at 25 cm: \( h_{min} = \theta \times 0.25 \).
06

Calculate Subtended Angle

For the object size found in Step 5, calculate its angular size in radians as \( \theta_{object} = \frac{h_{min}}{0.25} \) and convert to minutes: multiply by \( \frac{180 \times 60}{\pi} \).
07

Compare and Conclude on Resolution Limitations

Compare the obtained angular size with the experimental value (1 minute), and discuss whether diffraction or retinal cell size more significantly limits resolution.

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Key Concepts

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

Diffraction Effects
Diffraction refers to the way waves, such as light, spread out when they pass through an opening or around an obstacle. This spreading causes overlapping and limits the ability to see fine details.
In the human eye, diffraction occurs at the pupil. The smaller the pupil, the more significant the diffraction, which limits the eye's maximum resolution capability.
Light waves entering the eye through the pupil bend, creating a blur that limits the detail you can see. Diffraction effectively sets a fundamental boundary on how small details can be resolved.
The Rayleigh criterion provides a formula to determine when two points of light can be considered separate based on angular resolution. This criterion, defined as \( \theta = 1.22 \frac{\lambda}{D} \), gives the minimum angular separation between two sources that the eye can distinguish. Here, \( \theta \) is the angular resolution, \( \lambda \) is the wavelength of light (550 nm), and \( D \) is the pupil diameter (2.0 mm).
By calculating \( \theta \), you can determine how diffraction limits the resolution of objects viewed by the eye.
Retinal Cell Size
Retinal cells are minute structures in the eye responsible for detecting light and forming images.
Each retinal cell collects light, and their size sets physical limits on how small details can be recorded and converted into visual information.
In the exercise problem, each retinal cell is about 5.0 µm in diameter, restricting the eye’s capability to discern small objects or fine details without blurring.
For instance, at a standard close viewing distance of 25 cm, this retinal cell size limits objects to a minimum height of about 50 µm to be resolved clearly.
If an object is smaller than this threshold, it will generally blur into surrounding details. Therefore, both retinal cell size and diffraction work together to set the ultimate boundaries on visual acuity.
This interplay between the physical structure of the eye and light behavior emphasizes the complexity and limitation of human vision.
Angular Resolution
Angular resolution is a critical concept that helps describe how well the eye can distinguish two separate points. This separation is expressed in terms of angles. The smaller this angle, the finer the details that can be resolved.
The Rayleigh criterion mentioned earlier helps quantify this by providing the minimal resolvable angle based on light wavelength and pupil diameter.
When the angular resolution is calculated, it can be further converted into a linear object size. This allows us to determine the smallest visible object size at a given distance (here, 25 cm), using \( h = \theta \times d \), where \( h \) is the height and \( d \) is the distance.
For this problem, the angular resolution can be converted into an angle expressed in minutes, revealing how closely it aligns with experimental observations of about 1 arcminute.
Angular resolution illustrates the intrinsic interaction of optical physics with biological structures, dictating how sharply and clearly we perceive the world.

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Most popular questions from this chapter

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm\)61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (380-750 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm. (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

Diffraction occurs for all types of waves, including sound waves. High- frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point \(O\) in Fig. 36.5a. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point \(O\). At what distances from \(O\) will the intensity detected by the microphone be zero?

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