/*! 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 40 If you can read the bottom row o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If you can read the bottom row of your doctor’s eye chart, your eye has a resolving power of 1 arcminute, equal to \(\frac{1}{60}\) degree. If this resolving power is diffraction limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume \(\lambda=550 \mathrm{nm}\).

Short Answer

Expert verified
The effective diameter of the eye's optical system corresponding to the given resolving power is approximately \(2.94 \times 10^{-3}\) m or \(2.94\) mm.

Step by step solution

01

Convert units

The given angle in arcminute should be converted to degrees and then to radians since in physics, angles are usually in radians. We know that 1 arcminute = \(\frac{1}{60}\) degree and 1 degree = \(\frac{\pi}{180}\) radians. The given resolution \(1\) arcminute becomes \(\frac{1}{60} \times \(\frac{\pi}{180}\) = \(\frac{\pi}{10800}\) radian.
02

Calculate the diameter using Rayleigh's criterion

Using Rayleigh's criterion formula for diffraction limited system, we can find the diameter (D) of the eye's optical system. The equation goes as follows: \(\theta = 1.22 \times \(\frac{\lambda}{D}\)\), where \(\theta\) is the resolution angle in radians and \(\lambda\) is the wavelength of light. Substituting the known values and solving for D: D = \(1.22 \times \frac{\lambda}{\theta}\) = \(1.22 \times \frac{550 \times 10^-9 m}{\frac{\pi}{10800} rad}\)
03

Evaluate the diameter

By inputting the above values and solving for D, we find the effective diameter of the eye's optical system

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
Understanding Rayleigh's criterion is crucial for comprehending why there's a limit to how closely we can distinguish two light sources, such as stars in the night sky or letters on an eye chart. According to Lord Rayleigh, two light sources can be considered resolved—meaning, distinct from one another—if the principal diffraction maximum of one image coincides with the first minimum of the other.

This criterion is the standard for minimum resolvable detail and is the basis for calculating the resolving power of optical systems like telescopes, microscopes, and your own eyes. When an optical system is 'diffraction-limited', it means that the resolution of the system is primarily influenced by the wave nature of light, rather than imperfections or aberrations in its optics.
Resolving Power
The 'resolving power' of an optical instrument is a measure of its ability to distinguish small details of an object. For an optical system with a circular aperture, such as a human eye, telescope, or camera lens, the resolving power is affected by both the wavelength of the light being observed, usually denoted by \( \lambda \) (lambda), and the diameter of the aperture (or opening), expressed as \( D \).

When discussing vision or imaging, higher resolving power allows for better clarity and more detailed images. As per Rayleigh's criterion, the angle \( \theta \) under which an optical system can barely resolve two points is given by \( \theta = 1.22 \times \frac{\lambda}{D} \), where \( \theta \) is expressed in radians. This relationship is essential in determining the capabilities of imaging systems and underpins many design considerations for optical instruments.
Optical System Diameter
In optical terminology, the 'diameter' of an optical system refers to the size of its aperture—the part that allows light to enter. For example, in human eyes, this would correspond to the pupil opening. The diameter plays a pivotal role in determining the system's resolving power: Larger apertures can gather more light and generally provide better resolution.

The effective diameter lets us predict how fine the details we can resolve through that system will be, based on the relationship established in Rayleigh's criterion. It helps in comparing the performance of different optical devices and understanding the limitations imposed by physical properties such as diffraction.
Radian Conversion
Radians are the standard unit of angular measure used in physics and mathematics. Unlike degrees, which are based on dividing a circle into 360 arbitrary units, radians are derived from the properties of a circle itself. One radian is the angle created when the arc length is equal to the radius of the circle.

Conversion between degrees and radians is essential in calculations involving angular measurements. As seen in the provided exercise, to apply Rayleigh's criterion, the angular resolution stated in arcminutes must first be converted to degrees and then to radians using the conversion factor \( \frac{\pi}{180} \) radians per degree. There are \( 60 \) arcminutes in a degree, so \( 1 \) arcminute is \( \frac{1}{60} \) of a degree, which in turn can be converted to radians for the application of formulas in wave optics.

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

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen \(2.50 \mathrm{~m}\) from the slit. If the width of the central maximum is \(6.00 \mathrm{~mm},\) what is the slit width \(a\) if the wavelength is (a) \(500 \mathrm{nm}\)(visible light); (b) \(50.0 \mu \mathrm{m}\) (infrared radiation); (c) \(0.500 \mathrm{nm}\) (x rays)?

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

In the 1920s Clinton Davisson and Lester Germer accidentally observed diffraction when electrons with \(54 \mathrm{eV}\) of energy were scattered off crystalline nickel. The diffraction peak occurred when the angle between the incident beam and the scattered beam was \(50^{\circ}\). (a) What is the corresponding angle \(\theta\) relevant for Eq. (36.16)\(?\) (b) The planes in crystalline nickel are separated by \(0.091 \mathrm{nm}\), as determined by x-ray scattering experiments. According to the Bragg condition, what wavelength do the electrons in these experiments have? (c) Given the mass of an electron as \(9.11 \times 10^{-31} \mathrm{~kg},\) what is the corresponding classical speed \(v_{\mathrm{cl}}\) of the diffracted electrons? (d) Assuming the electrons correspond to a wave with speed \(v_{\mathrm{cl}}\) and wavelength \(\lambda,\) what is the frequency \(f\) of the diffracted waves? (e) Quantum mechanics postulates that the energy \(E\) and the frequency \(f\) of a particle are related by \(E=h f,\) where \(h\) is known as Planck's constant. Estimate \(h\) from these observations. (f) Our analysis has a small flaw: The relevant wave velocity, known as a quantum phase velocity, is half the classical particle velocity, for reasons explained by deeper aspects of quantum physics. Re-estimate the value of \(h\) using this modification. (g) The established value of Planck's constant is \(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\). Does this agree with your estimate?

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?

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.