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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 two dark fringes is 0.005908 m.

Step by step solution

01

Understand the Diffraction Formula

The angle \( \theta \) for a dark fringe in a single slit diffraction pattern can be determined using the formula \( a \sin \theta = m \lambda \). Here, \( a \) is the slit width, \( \lambda \) is the wavelength, and \( m \) is the order of the dark fringe (\( m = \pm 1, \pm 2, .... \)). For the first dark fringe, \( m = \pm 1 \).
02

Calculate Angle for First Dark Fringe

Rearrange the formula to find \( \theta \):\[\sin \theta = \frac{m \lambda}{a}.\]For \( m = 1 \), \( \lambda = 633 \times 10^{-9} \) m and \( a = 0.750 \times 10^{-3} \) m:\[\sin \theta = \frac{1 \times 633 \times 10^{-9}}{0.750 \times 10^{-3}} = 0.000844.\]
03

Calculate the Angle \( \theta \)

Find \( \theta \) by taking the arcsin of \( 0.000844 \):\[\theta = \arcsin(0.000844) \approx 0.000844 \, \text{radians}\] (since \( \sin \theta \) is very small, \( \theta \approx \sin \theta \) in radians).
04

Calculate the Distance to the Dark Fringe

The position \( y \) of the dark fringe on the screen is given by \( y = L \tan \theta \), where \( L \) is the distance to the screen.Since \( \theta \) is very small, \( \tan \theta \approx \theta \):\[y = 3.50 \times 0.000844 = 0.002954 \, \text{m}.\]
05

Determine Fringes on Both Sides

The distance between two dark fringes on either side of the central fringe is twice the value found for one side:\[\text{Total distance} = 2 \times 0.002954 = 0.005908 \, \text{m}.\]

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Key Concepts

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

Diffraction Pattern
When light passes through a single slit, it doesn't simply project a single line on a screen. Instead, it spreads out in a pattern due to diffraction. This is known as a diffraction pattern. The diffraction pattern consists of a central bright fringe, which is the brightest and widest part of the pattern, and several alternating dark and bright fringes.

These alternating fringes happen because different parts of the wavefront emerging from the slit interfere with each other. Constructive interference leads to bright fringes, while destructive interference results in dark fringes. The entire pattern provides insight into the wave nature of light.
  • Central bright fringe: the primary and most intense part of the diffraction pattern.
  • Dark fringes: regions where destructive interference causes a reduction in light intensity.
  • Side bright fringes: less intense than the central bright fringe, resulting from constructive interference.
Dark Fringe
Dark fringes in a diffraction pattern are regions where the intensity of light is minimized due to destructive interference. When light waves pass through the slit and spread out, they encounter each other and, depending on their phase, can cancel out. This results in areas known as dark fringes.

To find the position of these dark fringes, we use the following formula: \( a \sin \theta = m \lambda \). Here, \(a\) represents the width of the slit, \( \lambda \) the wavelength of the light, and \( m \) is the order of the fringe (an integer that indicates which dark fringe it is, such as the first or second). For the first dark fringe, \( m = \pm 1 \). The formula helps to calculate the angle, \( \theta \), at which these dark fringes occur.
  • Destructive interference: causes dark fringes.
  • The angle \( \theta \) is key to finding the dark fringe's position.
  • The order \( m \) signifies the sequence number of the fringe.
Wavelength
The wavelength of light, denoted by \( \lambda \), is the distance between consecutive peaks of the light wave. It is a fundamental property of the wave, often measured in nanometers (nm) for visible light. In single slit diffraction, the wavelength is crucial because it determines how strongly light waves interfere with each other after passing through the slit.

The formula \( a \sin \theta = m \lambda \) shows this relationship clearly, highlighting how wavelength affects the pattern of light on the screen. In our given problem, the wavelength \( \, \lambda = 633 \mathrm{nm} \). This specific wavelength was used to observe how the diffraction pattern materialized, especially the spacing between the dark fringes around the central bright fringe.
  • Determines interference patterns.
  • Critical for calculating positions of bright and dark fringes.
  • Directly influences the entire diffraction pattern viewable on the screen.
Central Bright Fringe
The central bright fringe is the brightest part of a single slit diffraction pattern. It appears at the center of the pattern on the screen and is due to the constructive interference of light waves. All the light waves coming through the slit travel similar paths and merge to reinforce each other.

This central bright fringe is wider and more intense than the other fringes. Any understanding of diffraction patterns starts with this prominent feature. Notably, the central bright fringe is symmetrical, with equal numbers of dark and bright fringes extending outward on either side. The problem involves finding the distance from this central bright fringe to the dark fringes on both sides, emphasizing its pivotal role in the pattern.
  • Result from constructive interference of waves.
  • Wider and more intense than other fringes.
  • Acts as a reference point for measuring distances to other features in the pattern.

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Most popular questions from this chapter

A thin glass slide \((n=1.53)\) that is \(0.485 \mu \mathrm{m}\) thick and surrounded by air is illuminated by a monochromatic electromagnetic wave. The wave is incident along the normal to the slide. What is the lowest frequency of the wave that will produce (a) an intensified reflected wave and (b) a canceled reflected wave?

Young's experiment is performed with light of wavelength \(502 \mathrm{nm}\) from excited helium atoms. Fringes are measured carefully on a screen \(1.20 \mathrm{~m}\) away from the double slit, and the center of the \(20 \mathrm{th}\) fringe (not counting the central bright fringe) is found to be \(10.6 \mathrm{~mm}\) from the center of the central bright fringe. What is the separation of the two slits?

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Parallel rays of green mercury light with a wavelength of \(546 \mathrm{nm}\) pass through a slit with a width of \(0.437 \mathrm{~mm}\). What is the distance from the central maximum to the first minimum on a screen \(1.75 \mathrm{~m}\) away from the slit?

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a thin film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2},\) which has an index of refraction of \(2.62,\) as the coating, what is the minimum film thickness that will cancel light of wavelength \(505 \mathrm{nm} ?\) (b) If this coating is too thin to stand up to wear, what other thicknesses would also work? Find only the three thinnest ones.

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