/*! 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 4 Light of wavelength 633 \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light of wavelength 633 \(\mathrm{nm}\) from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide, and the resulting diffraction pattern is observed on a screen 3.50 \(\mathrm{m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

Short Answer

Expert verified
The distance between the dark fringes is approximately 5.90 mm.

Step by step solution

01

Understand the Question

We want to find the distance between two dark fringes on either side of the central bright fringe in the diffraction pattern formed when light passes through a slit. The first step is to understand that these dark fringes arise at specific angles where destructive interference occurs.
02

Use the Single-Slit Diffraction Formula

The position of dark fringes in a single-slit diffraction pattern is given by the formula \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the dark fringe from the central line, \( m \) is the order of the dark fringe (1, 2, 3,...), and \( \lambda \) is the wavelength of the light.
03

Calculate for First Order Dark Fringes

For the first order dark fringes (\( m = 1 \)), we have the formula: \( 0.750 \times 10^{-3} \sin \theta = 1 \times 633 \times 10^{-9} \). Solve for \( \sin \theta \):\[ \sin \theta = \frac{633 \times 10^{-9}}{0.750 \times 10^{-3}} \]Upon calculation, \( \sin \theta \approx 8.44 \times 10^{-4} \).
04

Calculate Angle \( \theta \)

Since \( \sin \theta \) is very small, we can approximate \( \theta \approx \sin \theta \). Thus, \( \theta \approx 8.44 \times 10^{-4} \) radians. This angle represents the direction to the first dark fringe on one side of the central peak.
05

Find the Distance on the Screen

Convert the angle \( \theta \) into a distance on the screen. The distance \( y \) from the central maximum to the first dark fringe is given by \( y = L \tan \theta \). With small angles, \( \tan \theta \approx \sin \theta \), so:\[ y = 3.50 \ imes 10^{-4} \]Calculate to find \( y \approx 2.95 \ ext{mm} \).
06

Calculate the Distance Between Two First Order Dark Fringes

The distance between the first order dark fringes on either side of the central peak is twice the distance \( y \), so:\[ ext{Distance} = 2 imes 2.95 \approx 5.90 \ \text{mm} \].

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.

Single-Slit Diffraction
Single-slit diffraction is a fundamental concept in wave physics. It occurs when a wave, such as light or sound, passes through a narrow opening, known as a slit. This phenomenon leads to the spreading out of the wave as it exits the slit, creating a pattern of fringes on a screen where the intensity of the light varies. This pattern is known as a diffraction pattern.

Key aspects of single-slit diffraction include:
  • **Central Maximum:** The most intense light spot right in the center of the pattern, which is larger and brighter compared to others.
  • **Dark Fringes:** Occur at specific angles where the light waves interfere destructively, effectively canceling out to produce darkness between the bright fringes.

This phenomenon can be mathematically described using the condition for dark fringes, where the path difference between overlapping waves leads to destructive interference. Understanding the pattern formed by single-slit diffraction is pivotal in learning how light behaves under different conditions.
Destructive Interference
Destructive interference is a crucial principle when discussing diffraction and wave behavior. It happens when two waves meet in such a way that their crests and troughs are out of sync. In this situation, they cancel each other out, resulting in a reduction in amplitude.

In relation to single-slit diffraction, destructive interference is what creates the dark fringes in the diffraction pattern. The light waves passing through the different parts of the slit have slightly different path lengths. When these path lengths match conditions for destructive interference, they cancel out.

For dark fringes, the condition is defined by the equation:
  • \[ a \sin \theta = m \lambda \]
where:
  • \(a\) is the width of the slit,
  • \(\theta\) is the angle at which the fringe occurs,
  • \(m\) is the order of the fringe (1, 2, 3,...), and
  • \(\lambda\) is the wavelength of the light.

Understanding this concept helps us in calculating not just the angles and positions of the dark fringes but also the overall behavior of waves in different media.
Wavelength
Wavelength is a fundamental property of waves that describes the distance between two consecutive points that are in phase, such as crest to crest or trough to trough. It is usually denoted by the Greek letter \(\lambda\) and is typically measured in units like meters or nanometers (nm).

Wavelength plays a significant role in diffraction and interference patterns because it determines the angles and spacing of the resulting fringes. In the context of single-slit diffraction, the diffraction pattern is highly dependent on the wavelength of the light used. Shorter wavelengths tend to produce closely spaced fringes, while longer wavelengths will have more widely spaced fringes.

In practical terms, knowing the wavelength of the light source helps in predicting and calculating the positions of various interference patterns.
  • This becomes particularly important in experiments and applications such as optics, where precise knowledge of the wave properties is crucial.
Understanding wavelength is essential for anyone studying the behavior of waves, as it affects how waves interact with the world around us.

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

Photography. A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0,\) what would be the width of the smallest resolvable feature on the bear?

Measuring Refractive Index. A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm 38.2^{\circ}\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm 21.6^{\circ} .\) Find the refractive index of the liquid.

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?

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 \(\lambda=550 \mathrm{nm} .\)

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits/cm? (b) What is the second order deviation of this wavelength? Assume normal incidence.

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.