/*! 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 49 \(\bullet$$\bullet\) Resolution ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) Resolution of the eye, I. Even if the lenses of our eyes functioned perfectly, our vision would still be limited due to diffraction of light at the pupil. Using Rayleigh's criterion, what is the smallest object a person can see clearly at his near point of 25.0 \(\mathrm{cm}\) with a pupil 2.00 \(\mathrm{mm}\) in diameter and light of wavelength 550 \(\mathrm{nm}\) ? (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.) Based upon your answer, does it seem that diffraction plays a significant role in limiting our visual acuity?

Short Answer

Expert verified
The smallest object visible is approximately 83.87 micrometers. Yes, diffraction significantly limits our visual acuity.

Step by step solution

01

Understanding Rayleigh's Criterion

Rayleigh's criterion is an optical criterion used to determine the minimum angular separation at which two point sources can be distinguished. It is given by the formula \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture, or in this case, the pupil.
02

Substitute Values into Rayleigh's Criterion

Use the provided values: the wavelength \( \lambda = 550 \mathrm{nm} = 550 \times 10^{-9} \mathrm{m} \), and the pupil diameter \( D = 2.00 \mathrm{mm} = 2.00 \times 10^{-3} \mathrm{m} \). Substitute these into the formula to find \( \theta \). \[ \theta = 1.22 \frac{550 \times 10^{-9} \mathrm{m}}{2.00 \times 10^{-3} \mathrm{m}} \] Calculate \( \theta \) to find the angular resolution in radians.
03

Calculate the Angular Resolution

Perform the calculation for \( \theta \): \[ \theta = 1.22 \times \frac{550 \times 10^{-9}}{2.00 \times 10^{-3}} = 3.355 \times 10^{-4} \text{ radians} \].This gives us the minimum angular separation that can be resolved by the eye at the given conditions.
04

Determine the Smallest Object Visible

Convert the angular resolution to a linear separation at the near point of 25 cm using the small angle approximation, where \( s = r \theta \) with \( r = 0.25 \mathrm{m} \).\[ s = 0.25 \mathrm{m} \times 3.355 \times 10^{-4} = 8.387 \times 10^{-5} \mathrm{m} = 83.87 \mathrm{\mu m} \].Thus, the smallest object a person can see clearly at the near point is approximately 83.87 micrometers.
05

Evaluate the Role of Diffraction

Considering that the diameter of the smallest objects, like a human hair, is typically around 70 to 100 micrometers, diffraction does impose a limit to the eye's ability to resolve fine detail. Based on the calculated resolution limit of approximately 84 micrometers, it indeed plays a significant role in visual acuity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rayleigh's criterion
Rayleigh's criterion is a principle used to define the minimum distance at which two points of light can be distinguished from one another. This concept is vital in understanding how lenses and optical systems work, including the human eye. According to the criterion, the minimum angular resolution, denoted by \( \theta \), can be determined using the formula: \( \theta = 1.22 \frac{\lambda}{D} \). Here, \( \lambda \) represents the wavelength of light used, and \( D \) is the diameter of the aperture through which the light passes (such as the pupil of the eye).

For example, if you have a pupil diameter of 2 mm and a light wavelength of 550 nm, you can insert these values into the formula to calculate the smallest angular separation the eye can resolve. This helps to understand how well the eye can distinguish between two closely spaced objects.
diffraction of light
Diffraction of light occurs when light waves encounter an obstacle or slit that is comparable in size to its wavelength. This phenomenon causes the light waves to bend around the edges of the obstacle, resulting in a pattern of dark and light bands known as a diffraction pattern. In the context of the eye, the pupil acts as the opening through which light enters, causing diffraction.

Diffraction limits the resolving power of the eye. Even if the eye's lenses are flawless, the effects of diffraction mean that there will be a fundamental limit to how closely two points of light can be situated before the eye can no longer distinguish them. This is why, despite having clear and healthy optic structures, people still encounter a natural limit to vision due to light waves spreading and interacting with each other.
angular resolution
Angular resolution refers to the ability of an optical instrument, like the human eye, to distinguish two points as separate objects rather than a single blurry one. It is measured in terms of the angle subtended by the two points. With better angular resolution, two points close together can still be seen as distinct.

The formula for angular resolution, \( \theta = 1.22 \frac{\lambda}{D} \), tells us how the wavelength of light and the diameter of the aperture (eye's pupil) affect the resolution. As \( \theta \) gets smaller, the resolution improves, allowing for finer details to be observed. For example, changes in pupil size or the environment's light conditions can affect how well someone can see detailed structures or distinguish between close objects. This is critical in applications ranging from astronomy to daily human vision.
visual acuity
Visual acuity is the clarity or sharpness of vision, heavily influenced by the angular resolution and diffraction of light. It describes how well a person can detect the fine details and distinguish small objects at different distances.

Determining visual acuity involves understanding components like the diffraction limit, which sets the smallest detail the eye can resolve. For instance, if we consider the diffraction effect and use Rayleigh's criterion, we calculate that the smallest object discernable at a standard near-viewing distance (like 25 cm) depends on the diameter of the pupil and the wavelength of light. If diffraction limits result in the minimum resolvable size resembling everyday small items like a human hair, it underscores how significant diffraction is in determining how well we see fine details.
  • This becomes evident in various tasks, from reading tiny text to observing small distant objects.
  • It highlights why optical devices like glasses and microscopes are used to counteract these natural limitations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet$$\bullet\) Thickness of human hair. Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength 632.8 \(\mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.25 \(\mathrm{m}\) away, the first dark fringes on either side of the central bright spot were 5.22 \(\mathrm{cm}\) apart. How thick was this strand of hair?

\(\bullet\) A laser beam of wavelength 600.0 \(\mathrm{nm}\) is incident normally on a transmission grating having 400.0 lines/mm. Find the angles of deviation in the first, second, and third orders of bright spots.

\(\bullet\) Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit with a width of 0.437 \(\mathrm{mm} .\) What is the distance from the central maximum to the first minimum on a screen 1.75 \(\mathrm{m}\) away from the slit?

\(\bullet$$\bullet\) A glass sheet measuring 10.0 \(\mathrm{cm} \times 25.0 \mathrm{cm}\) is covered by a very thin opaque coat- ing. In the middle of this sheet is a thin, straight scratch 0.00125 \(\mathrm{mm}\) thick, as shown in Figure \(26.50 .\) The sheet is totally immersed beneath the surface of a liquid having an index of refraction of \(1.45 .\) Monochromatic light strikes the sheet perpendicular to its surface and passes through the scratch. A screen is placed under water a distance 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on this screen are 22.4 \(\mathrm{cm}\) apart. What is the wavelength of the light in air?

\(\bullet$$\bullet\) Sensitive eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 \(\mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.