/*! 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 37 A slit \(1.00 \mathrm{~mm}\) wid... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 37

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(650 \mathrm{~nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first and third diffraction minima on the same side of the central diffraction maximum?

Short Answer

Expert verified
The distance is approximately \(3.90 \, \mathrm{mm}\).

Step by step solution

01

Understand the Diffraction Concept

In single-slit diffraction, a pattern of dark and bright fringes appears on a screen when light passes through a narrow slit. The dark bands, or minima, occur where the path difference between light from different parts of the slit cause destructive interference. These minima are located at specific angles which can be found using the formula for minima in single-slit diffraction.
02

Identify the Formula for Minima Location

The angles for the diffraction minima in a single-slit diffraction pattern are found where the condition \( a \sin(\theta) = m\lambda \) is satisfied. Here, \(a\) is the slit width, \(m\) is the order of the minimum (1 for first, 2 for second, etc.), and \(\lambda\) is the wavelength of the light.
03

Calculate the Angles for the First and Third Minima

For the first minimum (\(m=1\)), using \(a = 1.00 \, \mathrm{mm} = 1.00 \times 10^{-3} \, \mathrm{m}\), the angle is determined by \( \sin(\theta_1) = \frac{1 \times 650 \times 10^{-9}}{1 \times 10^{-3}} \). Solving, we find \( \sin(\theta_1) = 6.5 \times 10^{-4} \). For the third minimum (\(m=3\)), \( \sin(\theta_3) = \frac{3 \times 650 \times 10^{-9}}{1 \times 10^{-3}} \), so \( \sin(\theta_3) = 1.95 \times 10^{-3} \).
04

Calculate Linear Distance on the Screen

The linear distance \( y \) from the central maximum to a minimum on a screen is given by \( y = L \tan(\theta) \), where \( L \) is the distance from the slit to the screen, 3.00 m here. Approximating \( \tan(\theta) \approx \sin(\theta) \) for small angles, for the first minimum, \( y_1 = 3.00 \times 6.5 \times 10^{-4} \), and for the third minimum, \( y_3 = 3.00 \times 1.95 \times 10^{-3} \).
05

Calculate the Distance Between Minima

Now find the linear distance between the first and third minima, which is \( y_3 - y_1 = (3.00 \times 1.95 \times 10^{-3}) - (3.00 \times 6.5 \times 10^{-4}) \). Performing the calculation gives the distance.

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.

Single-slit diffraction
Single-slit diffraction occurs when light passes through a narrow slit, creating a pattern of alternating bright and dark regions on a screen. This phenomenon arises because light behaves as a wave, and as it encounters the edges of the slit, it spreads out, or diffracts. The central bright band is the most intense, flanked by gradually diminishing maxima.
- **Diffraction Patterns**: These are the distinct arrangements of light and dark bands observed. - **Light as Waves**: The wave nature of light causes overlapping and interference, leading to diffraction.
Understanding single-slit diffraction helps in realizing how light bends when passing through obstacles, providing insights into wave optics.
Diffraction minima
Diffraction minima occur when light waves destructively interfere, creating dark bands on the screen. In single-slit diffraction, these minima appear at angles where the wave path differences equal half-wavelengths.
The formula for determining the angles of these dark bands in a single-slit setup is given by:\[ a \sin(\theta) = m\lambda\]where:
  • \(a\) is the slit width,
  • \(\theta\) is the angle from the central axis,
  • \(\lambda\) is the light wavelength,
  • m is the order of the minima (1 for first, 2 for second, and so on).
This formula is crucial in predicting where the dark bands will appear and how they relate to the wavelength and width of the slit.
Interference pattern
An interference pattern is the result of constructive and destructive interference between overlapping light waves. In the context of single-slit diffraction, it's what creates the series of bright and dark fringes on the screen.
Whenever light from different parts of the slit meets:
  • Constructive interference occurs if their crests and troughs align, forming bright bands.
  • Destructive interference happens if the crests of one meet troughs of another, creating dark bands.
This alternating pattern demonstrates the wave nature of light, as particles do not show such intricate interference.
Optics
Optics is the study of light and its interactions with different mediums. Understanding diffraction and interference patterns is a significant part of optics.
- **Geometrical Optics**: Focuses on light as rays, useful in understanding lenses and mirrors.
- **Physical Optics**: Considers wave characteristics, crucial for explaining phenomena like diffraction.
Diffraction studies reveal much about the behavior of light beyond straight-line propagation, bridging the gap between theoretical concepts and practical applications in lenses, lasers, and more.
Wave optics
Wave optics, or physical optics, deals with the study of light as a wave, in contrast to ray optics, which treats light as rays. This understanding is pivotal in explaining how light interacts with matter on a scale where its wave characteristics dominate.
- **Wave Nature**: explains phenomena such as diffraction, interference, and polarization. - **Applications**: Essential for modern technologies like fiber optics and lasers, where wave properties are crucial.
Single-slit diffraction and other similar experiments underpin this broader field, emphasizing the need to see light as more than simple rays. Wave optics helps comprehend the nuances of how light behaves under various circumstances.

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

A source containing a mixture of hydrogen and deuterium atoms emits red light at two wavelengths whose mean is \(656.3 \mathrm{~nm}\) and whose separation is \(0.180 \mathrm{~nm}\). Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the second order.

\mathrm{X}\( rays of wavelength \)0.10 \mathrm{~nm}\( are found to undergo secondorder reflection at a Bragg angle of \)23^{\circ}$ from a lithium fluoride crystal. What is the interplanar spacing of the reflecting planes in the crustal?

Estimate the linear separation of two objects on Mars that can just be resolved under ideal conditions by an observer on Earth (a) using the naked eye and (b) using the \(200 \mathrm{in} .(=5.1 \mathrm{~m})\) Mount Palomar telescope. Use the following data: distance to Mars \(=8.0 \times 10^{7} \mathrm{~km}\), diameter of pupil \(=5.0 \mathrm{~mm}\), wavelength of light \(=550 \mathrm{~nm}\).

The wall of a large room is covered with acoustic tile in which small holes are drilled \(6.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm}\), and the wavelength of the room light to be \(550 \mathrm{~nm}\) ?

A beam of light consisting of wavelengths from \(460.0 \mathrm{~nm}\) to \(640.0 \mathrm{~nm}\) is directed perpendicularly onto a diffraction grating with 150 lines/mm. (a) What is the lowest order that is overlapped by another order? (b) What is the highest order for which the complete wavelength range of the beam is present? In that highest order, at what angle does the light at wavelength (c) \(460.0 \mathrm{~nm}\) and (d) \(640.0 \mathrm{~nm}\) appear? (e) What is the greatest angle at which the light at wavelength \(460.0 \mathrm{~nm}\) appears?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Turning Points in Physics

Read Explanation

Thermodynamics

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation

Geometrical and Physical Optics

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.