/*! 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 Red light of wavelength 633 \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Red light of wavelength 633 \(\mathrm{nm}\) from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

Short Answer

Expert verified
(a) 10.851 mm, (b) 5.426 mm

Step by step solution

01

Understanding Diffraction Minima Formula

Diffraction minima occur at angles \( \theta \) where \( a \sin \theta = m \lambda \), with \( a \) being the slit width, \( \lambda \) the wavelength, and \( m \) the order of the minimum. Here, we need \( m = 1 \) for the first minimum because the central maximum lies between the first minima.
02

Calculating Sin θ for Central Minima

For the central bright fringe, set \( m = 1 \). Thus, \( a \sin \theta = \lambda \). Plug in \( a = 0.350 \times 10^{-3} \text{ m} \) and \( \lambda = 633 \times 10^{-9} \text{ m} \) to find \( \sin \theta \): \[ \sin \theta = \frac{633 \times 10^{-9}}{0.350 \times 10^{-3}} \approx 0.0018085 \].
03

Calculating Distance to Minima on Screen

With small angles, \( \sin \theta \approx \tan \theta = \frac{y}{L} \), where \( y \) is the distance from the central maximum to a minimum on the screen, and \( L = 3.00 \text{ m} \) is the distance to the screen. Calculate \( y \) for the first minimum: \[ y_1 = L \times \sin \theta \approx 3.00 \times 0.0018085 \approx 0.0054255 \text{ m} \].
04

Finding Width of Central Bright Fringe

The central bright fringe extends from the first minimum on the left to the first minimum on the right, so its width is \( 2y_1 \). Thus, \( 2 \times 0.0054255 \approx 0.010851 \text{ m} \) or \( 10.851 \text{ mm} \).
05

Calculating Width of First Bright Fringe (One Side)

For the first bright fringe on one side of the center, use \( m = 2 \) for the second minimum: \[ \sin \theta \approx \frac{2 \times 633 \times 10^{-9}}{0.350 \times 10^{-3}} \approx 0.003617 \].
06

Calculate Distance to Second Minima on Screen

Find the second minimum distance: \[ y_2 = L \times 0.003617 \approx 3.00 \times 0.003617 \approx 0.010851 \text{ m} \].
07

Finding Width of the First Bright Fringe

The first bright fringe extends from \( y_1 \) to \( y_2 \). Calculate its width: \( y_2 - y_1 = 0.010851 - 0.0054255 = 0.0054255 \text{ m} \) or \( 5.4255 \text{ mm} \).

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.

Diffraction pattern
When light encounters an obstacle or a gap that is comparable in size to its wavelength, it tends to bend around the edges. This phenomenon is known as diffraction. As a result, light forms a pattern consisting of bright and dark regions on a screen. This arrangement, often referred to as the diffraction pattern, showcases a central bright spot flanked by alternating dark and bright bands. These bands, or fringes, are a direct consequence of the light waves interfering with each other either constructively or destructively.
The understanding of the diffraction pattern provides valuable insights into the wave nature of light. The exact positions and intensities of these patterns are determined by variables such as wavelength, slit width, and the distance to the screen. Observing these patterns allows us to measure key optical properties like slit width and light wavelength.
Central bright fringe
The central bright fringe is the dominant feature in a diffraction pattern. It is the widest and the brightest, positioned prominently in the center. This central maximum is where constructive interference is strongest, resulting in maximum brightness. The light waves arrive in phase and reinforce each other in this region.
To determine the width of the central bright fringe, one must identify the first minima on either side. This width represents the distance between these points. Specifically, the distance is twice the position of the first minimum. In the context provided, the width of the central bright fringe was calculated to be approximately 10.851 mm. This is derived from the slit width, wavelength, and distance to the observation screen.
First bright fringe
On either side of the central maximum, the first bright fringe can be observed. This fringe results from secondary constructive interference conditions, though not as intense as the central one. Each bright fringe occurs between minima and corresponds to a higher order of interference.
To find the position and width of this fringe, look for the second minima that mark its boundaries. From the calculations provided, we determine the width of the first bright fringe as the difference between the positions of the first and second minima. In the problem given, this first bright fringe was estimated to have a width of about 5.4255 mm. Understanding the separation and distribution of these fringes is essential for exploring the properties of light and wave behaviors.
Wavelength of light
The wavelength of a wave denotes the distance between consecutive peaks or troughs in a wave. For light, it's a crucial factor that influences the diffraction pattern. In this context, a wavelength of 633 nm signifies that the light possesses a red hue, similar to what is emitted by a helium-neon laser.
This specific wavelength plays a pivotal role in determining both the angle and the position of the minima and maxima in the diffraction pattern. When calculating diffraction effects, it's crucial to know the wavelength because it dictates interference conditions. Longer wavelengths result in broader diffraction patterns, while shorter ones create narrower bands. Accurate measurements of wavelength enable precise predictions and analyses of optical phenomena like diffraction.
Slit width
The slit width is a significant parameter in determining the nature of the diffraction pattern. It represents the physical size of the gap through which light passes, and in this problem, it measures 0.350 mm. The slit width influences the degree of bending and spreading of light waves around the slit.
A narrow slit compared to the wavelength results in a broader diffraction pattern, while a wider slit narrows the fringes. In diffraction calculations, the slit width is essential to determine the angle to the minima, given by the formula: \(a \sin \theta = m \lambda\). As the slit width changes, so does the entire diffraction setup, affecting the spacing and intensity of the bright and dark fringes on a screen.

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 light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is \(\frac{1}{7}\) the center-to-center distance between the slits. (a) Which interference maxima are missing in the pattem on the screen? (b) Does the answer to part (a) depend on the wavelength of the light used? Does the location of the missing maxima depend on the wavelength?

Monochromatic light with wavelength 620 \(\mathrm{nm}\) passes through a circular aperture with diameter 7.4\(\mu \mathrm{m}\) . The resulting diffraction pattern is observed on a screen that is 4.5 \(\mathrm{m}\) from the aperture. What is the diameter of the Airy disk on the screen?

Phased-Array Radar. In one common type of radar installation, a rotating antenna sweeps a radio beam around the sky. But in a phased-array radar system, the antennas remain stationary and the beam is swept electronically. To see how this is done, consider an array of \(N\) antennas that are arranged along the horizontal \(x\) -axis at \(x=0, \pm d, \pm 2 d, \ldots, \pm(N-1) d / 2 .\) (The number \(N\) is odd.) Each antenna emits radiation uniformly in all directions in the horizontal \(x y\) -plane. The antennas all emit radiation coherently, with the same amplitude \(E_{0}\) and the same wavelength \(\lambda\) . The relative phase \(\delta\) of the emission from adjacent antennas can be varied, however. If the antenna at \(x=0\) emits a signal that is given by \(E_{0} \cos \omega t,\) as measured at a point next to the antenna, the antenna at \(x=d\) emits a signal given by \(E_{0} \cos (\omega t+\delta),\) as measured at a point next to that antenna. The corresponding quantity for the antenna at \(x=-d\) is \(E_{0} \cos (\omega t-\delta) ;\) for the antennas at \(x=\pm 2 d,\) it is \(E_{0} \cos (\omega t \pm 2 \delta) ;\) and so on. \((a)\) If \(\delta=0\) , the interference pattern at a distance from the antennas is large compared to \(d\) and has a principal maximum at \(\theta=0\) (that is, in the angular range \(-90^{\circ}<\theta<90^{\circ} .\) Hence this principal maximum describes a beam emitted in the direction \(\theta=0\) . As described in Section \(36.4,\) if \(N\) is large, the beam will have a large intensity and be quite narrow. (b) If \(\delta \neq 0\) , show that the principal intensity maximum described in part (a) is located at $$ \boldsymbol{\theta}=\arcsin \left(\frac{\delta \boldsymbol{\lambda}}{2 \pi d}\right) $$ where \(\delta\) is measured in radians. Thus, by varying \(\delta\) from positive to negative values and back again, which can easily be done electronically, the bearn can be made to sweep back and forth around \(\boldsymbol{\theta}=\mathbf{0} .\) (c) A weather radar unit to be installed on an airplane emits radio waves at 8800 \(\mathrm{MHz}\) . The unit uses 15 antennas in an array 28.0 \(\mathrm{cm}\) long (from the antenna at one end of the array to the antenna at the other end). What must the maximum and minimum values of \(\delta\) be (that is, the most positive and most negative values) if the radar beam is to sweep \(45^{\circ}\) to the left or right of the air- plane's direction of flight? Give your answer in radians.

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)?

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm} )\) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

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.