/*! 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 13 A converging lens \(7.20 \mathrm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 13

A converging lens \(7.20 \mathrm{~cm}\) in diameter has a focal length of \(300 \mathrm{~mm}\). If the resolution is diffraction limited, how far away can an object be if points on it \(4.00 \mathrm{~mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{~nm}\).

Short Answer

Expert verified
The object can be resolved up to approximately 430 m away.

Step by step solution

01

Understanding the Problem

We need to find the maximum distance at which two points, 4.00 mm apart, can be resolved by a lens with a given diameter, focal length, and light wavelength. This requires applying the Rayleigh criterion for optical resolution.
02

Apply Rayleigh's Criterion

According to Rayleigh's criterion, the minimum angular resolution \( \theta_{min} \) is given by \( \theta_{min} = 1.22 \times \frac{\lambda}{D} \), where \( D \) is the diameter of the lens and \( \lambda \) is the wavelength. Give the values: \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m} \) and \( D = 7.20 \text{ cm} = 0.072 \text{ m} \). Substitute these values into the formula.
03

Calculate Minimum Angular Resolution

Substitute the values: \[ \theta_{min} = 1.22 \times \frac{550 \times 10^{-9}}{0.072} = 9.305 \times 10^{-6} \text{ radians} \]. This is the smallest angle at which two points can be resolved by this lens.
04

Relate Angular Resolution to Linear Resolution

The linear resolution \( R \) at a distance \( L \) is given by \( R = L \times \theta_{min} \). We need to resolve points 4.00 mm apart, which means \( R = 4.00 \text{ mm} = 0.004 \text{ m} \). Set up the equation: \[ \theta_{min} = \frac{R}{L} \].
05

Solve for Distance L

Rearrange the equation \( \theta_{min} = \frac{R}{L} \) to solve for \( L \): \[ L = \frac{R}{\theta_{min}} = \frac{0.004}{9.305 \times 10^{-6}} \approx 430.0 \text{ m} \]. This is the maximum distance at which the two points can be resolved.

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 91影视!

Key Concepts

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

Angular Resolution
Angular resolution refers to the smallest angle between two objects that can be distinguished as separate. In optics, it is essential for determining the clarity and detail an optical instrument can provide.
For example, if two stars are close together in the sky, angular resolution determines if we see them as distinct points or just a blur.
The Rayleigh criterion helps us find this angle, known as the minimum angular resolution, \(\theta_{min}\), using the formula:
  • \(\theta_{min} = 1.22 \times \frac{\lambda}{D}\)
where \(\lambda\) is the wavelength of light used, and \(D\) is the diameter of the aperture.
A smaller \(\theta_{min}\) means better resolution and more detail.
Diffraction Limited
An optical system is said to be diffraction limited when its performance is only hindered by diffraction effects, rather than imperfections in the lens or mirror.
Diffraction occurs when light waves encounter an object or aperture, causing them to spread out.
  • This limits how finely details can be resolved in an image.
  • Even a perfect lens cannot resolve details smaller than this limit.
For example, in the given exercise, the lens's resolution is diffraction limited, setting the sharpness boundaries for the objects it can view. This means when using components like lenses or telescopes, fine structure might be blurred at certain boundary levels due to this inherent nature.
Focal Length
The focal length of a lens is the distance over which parallel rays of light converge to a single point.
It's a critical parameter that determines the lens's ability to magnify or reduce the size of an image.
  • Shorter focal lengths provide a wider field of view but less magnification.
  • Longer focal lengths offer higher magnification with a narrower view.
In the exercise, the focal length is given as 300 mm, which helps dictate the resolving power of the lens in combination with angular resolution and diameter. Knowing the focal length is key to calculating how far an object can be for points on it to be viewed as separate.
Optical Resolution
Optical resolution is the ability of an imaging system to distinguish between closely spaced objects. It is determined largely by the aperture's diameter and the wavelength of light used.
  • Higher optical resolution means greater detail and clarity in the image.
  • The resolution can be limited by factors such as lens quality and alignment.
Using Rayleigh's criterion, the exercise examines how a lens can distinguish points that are 4.00 mm apart. Optical resolution ensures that the finest details in the object being observed are clear and distinct, with the exercise focusing on achieving resolution under diffraction limited conditions.

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

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about \(8000 \mathrm{~km}\). When this radio telescope is focusing radio waves of wavelength \(2.0 \mathrm{~cm}\), what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength \(550 \mathrm{~nm}\) so that the visible-light telescope has the same resolution as the radio telescope?

Two satellites at an altitude of \(1400 \mathrm{~km}\) are separated by 28 \(\mathrm{km}\). If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

A glass sheet \(12 \times 10^{-3} \mathrm{~mm}\) thick is placed in the path of one of the interfering beams in a Young's double slit interference arrangement using monochromatic light of wavelength \(6000 \AA\). If the central bright fringe shifts a distance equal to width of 10 bands, find the refractive index of glass. What is the thickness of the sheet of diamond of refractive index \(2.5\) that has to be introduced in the path of second beam to bring the central bright fringe to original position?

Laser light of wavelength \(632.8 \mathrm{~nm}\) falls normally on a slit that is \(0.0250 \mathrm{~mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is \(8.50 \mathrm{~W} / \mathrm{m}^{2}\). Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all.

Two radio transmitter each transmitting a radio of frequency \(12 \mathrm{MHz}\) in phase with each other are placed at a distance \(l=100 \mathrm{~m}\) on either side of the runway on the towers as shown in the figure. An aircraft with a ground speed of \(V\) is flying towards the airport such that its velocity makes an angle \(\theta\) with the runway. The aircraft is heading directly for the mid-point of the towers. (a) At the aircraft's current position, the intensity of the signal from each tower separately would be \(I_{0}\). Find the intensity of the combined signal from both towers received by the aircraft for heading of \(\theta=0\) and \(\theta=\pi / 2 .\) (b) For what heading would the aircraft receive no signal. Assuming distance of plane much greater than \(l\). (c) If the aircraft is heading from large distance directly towards one of the transmitter in the direction perpendicular to 'l' find the position of the aircraft from the nearest tower when the signal dropped to a minimum first time.
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Electromagnetism

Read Explanation

Force

Read Explanation

Geometrical and Physical Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Energy Physics

Read Explanation
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.