/*! 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 21 Light that has a wavelength of \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light that has a wavelength of \(668 \mathrm{~nm}\) passes through a slit \(6.73 \times 10^{-6} \mathrm{~m}\) wide and falls on a screen that is \(1.85 \mathrm{~m}\) away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either side?

Short Answer

Expert verified
The distance to the third dark fringe is approximately 549 mm.

Step by step solution

01

Understanding Diffraction and the Formula

This problem involves single-slit diffraction. For a single slit, the position of dark fringes is given by the formula for minima: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength, \( \theta \) is the angle, and \( m \) is the order of the dark fringe (1, 2, 3, ...). In this case, we need the third dark fringe, so \( m = 3 \).
02

Calculation of the Angle \( \theta \) for the Third Dark Fringe

Using the formula \( a \sin \theta = m \lambda \), substitute \( a = 6.73 \times 10^{-6} \) m, \( \lambda = 668 \times 10^{-9} \) m, and \( m = 3 \). Calculate \( \sin \theta \):\[ \sin \theta = \frac{3 \times 668 \times 10^{-9}}{6.73 \times 10^{-6}} \approx 0.297 \].Since \( \sin \theta \) is relatively small, \( \theta \approx \sin \theta \) in radians.
03

Determining the Distance from the Central Maximum

The distance to the fringe on the screen \( y \) can be found using the small angle approximation \( y = L \tan \theta \approx L \sin \theta \). Substitute \( L = 1.85 \) m and \( \sin \theta \approx 0.297 \):\[ y = 1.85 \times 0.297 = 0.54945 \text{ m} \approx 549 \text{ mm}. \]

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.

Understanding Diffraction Minima
In the realm of single-slit diffraction, the term "diffraction minima" is crucial. When light waves pass through a narrow slit, they spread out and create patterns of dark and bright areas called fringes. The dark fringes, where the light intensity is minimal, are known as diffraction minima. This occurs because of destructive interference, where certain parts of the wave overlap and cancel each other out.
To calculate the position of these minima on a screen, we use the formula:
  • \(a \sin \theta = m \lambda\)
The variables include:
  • \(a\): Slit width
  • \(\theta\): Angle of diffraction
  • \(m\): Order of the fringe
  • \(\lambda\): Wavelength of light
For the third dark fringe \( m = 3 \). Remembering this formula is essential for understanding why light and dark patterns form when waves go through small openings.
Exploring the Wavelength of Light
The wavelength of light, denoted as \( \lambda \), is a major player in diffraction scenarios. It represents the distance between successive peaks of a wave. In this exercise, the light's wavelength is given as 668 nanometers (nm), which is a microscopic scale more commonly used in optics.
The wavelength directly influences where the diffraction minima occur, as evident in the formula \( a \sin \theta = m \lambda \). A longer wavelength leads to more spread out fringes, while a shorter wavelength results in fringes that are closely packed.
Understanding the wavelength is vital not only for solving problems but also for interpreting how light behaves when it interacts with different materials.
The Significance of Slit Width in Diffraction Patterns
Slit width, symbolized as \( a \), is a measure of how wide the opening is through which light passes. In our problem, the slit width is \(6.73 \times 10^{-6} \text{ m}\).
The width Plays a pivotal role in determining the diffraction pattern because it affects how significantly the waves spread out after passing through the slit.
A smaller slit width causes broader diffraction patterns, meaning the dark and bright fringes appear farther apart. Conversely, a wider slit results in closely spaced fringes.
The slit width appears in the diffraction minima formula, showing its direct impact on where the dark fringes will occur on a screen.
Calculating Distance to the Screen and its Role
Distance to the screen, often denoted as \( L \), is the distance from the slit to the observing surface where diffraction patterns are displayed. In this scenario, the screen is 1.85 meters away from the slit.
This distance is critical when calculating the actual position of the diffraction minima on the screen. Using the approximated angle \( \theta \) from earlier calculations, the physical distance \( y \) to the specific fringe can be determined using:
  • \( y = L \sin \theta \)
A larger distance to the screen increases the distance between fringes, making the diffraction pattern appear larger and easier to observe. Conversely, a shorter distance results in a more compact pattern.

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

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{~m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{~m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{~nm}\), what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

(a) What, if any, phase change occurs when light, traveling in air, reflects from the interface between the air and a soap film \((n=1.33) ?\) (b) What, if any, phase change occurs when light, traveling in a soap film, reflects from the interface between the soap film and a glass plate \((n=1.52) ?(\mathrm{c})\) Is the wavelength of the light in a soap film greater than, smaller than, or equal to the wavelength in a vacuum? A soap film \((n=1.33)\) is \(465 \mathrm{~nm}\) thick and lies on a glass plate \((n=1.52)\). Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{~nm}\), travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

Orange light \(\left(\lambda_{\text {vacuum }}=611 \mathrm{nm}\right.\) ) shines on a soap film \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is \(675 \mathrm{~nm}\). Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{~m}\) and (b) \(1.8 \times 10^{-6} \mathrm{~m}\).

At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{~m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{~m}\), and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.

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.