/*! 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 30 The central bright fringe in a s... [FREE SOLUTION] | 91Ó°ÊÓ

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The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

Short Answer

Expert verified
\( \frac{\lambda}{W} \approx 1 \)

Step by step solution

01

Understanding the Problem

In a single-slit diffraction pattern, the central bright fringe is twice the width of the distance between the first minima on either side. The width of the central bright fringe is given to be equal to the distance between the screen and the slit.
02

Use the Condition for Minima

The position of the minima in a single-slit diffraction is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the diffraction angle, \( m \) is the order of the minimum, and \( \lambda \) is the wavelength of the light.
03

Relate Width of Central Fringe to Distance

For the first minimum (\( m = \pm 1 \)), the angle \( \theta \) relates the slit and screen distance \( L \) as \( a \sin \theta \approx L \). This gives us the condition \( \sin \theta \approx \frac{\lambda}{a} = \frac{W}{a} \), where \( W = L \).
04

Substitute and Solve for Ratio

The given condition implies \( \frac{\lambda}{a} = \frac{W}{a} \). Solving, we find the ratio to be \( \frac{\lambda}{W} = \sin \theta \approx 1 \) under the small angle approximation used here.

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Key Concepts

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

Central Bright Fringe
In single-slit diffraction, the central bright fringe is a key feature. When light passes through a narrow slit, it spreads out and creates a diffraction pattern with a bright center. This center section is called the central bright fringe. It is the most intense part of the pattern.

The width of the central bright fringe can be measured from the middle of the first dark fringe on one side to the first dark fringe on the other side. In the given problem, the central bright fringe's width is the same as the distance between the screen and the slit. This is a special condition that simplifies how we can relate different quantities in the diffraction pattern.
Wavelength
The wavelength \(\lambda\) is a fundamental property of waves, including light. It's the distance between successive crests of a wave. In diffraction patterns, the wavelength is crucial because it determines how the wave spreads as it passes through the slit.

In our example, the relationship between the wavelength and the physical size of the slit and screen setup is examined. Specifically, the exercise involves finding the ratio \(\frac{\lambda}{W}\), which tells us how the characteristics of the light wave (its wavelength) compare to the size of the slit (represented by \(W\) in this problem). This ratio is significant because it influences the overall shape and size of the diffraction pattern.
Diffraction Minima
Diffraction minima are the dark regions in a diffraction pattern where destructive interference occurs. For single-slit diffraction, these minima occur at certain angles where the waves from the slit cancel each other out.

The positions of these minima are found using the formula \(a \sin \theta = m \lambda\), where \(a\) is the width of the slit, \(\theta\) is the angle to the minima, and \(m\) denotes the order of the minimum (e.g., first, second). In simple terms, if the light's waves are slightly out of phase, they can interfere destructively and create these minima.
  • For the first minimum, the angle occurs at \(m = \pm 1\).
  • This condition helps define the boundaries of the central bright fringe, since that fringe is exactly between these first minima.
Diffraction Angle
The diffraction angle \(\theta\) is the angle at which light is bent around the slit edges. It's a vital part of understanding how the diffraction pattern forms. In the formula \(a \sin \theta = m \lambda\), \(\theta\) tells us how much the light wave spreads and at which angles the minima occur.

The small angle approximation is often used, particularly when \(\theta\) is small. This approximation simplifies calculations because it assumes \(\sin \theta \approx \theta\) when \(\theta\) is measured in radians. In the problem at hand, this approximation helps simplify the expression for how \(\lambda\) and \(W\) (slit width) relate as \(\sin \theta \approx \frac{W}{a}\). This relationship is particularly useful for finding the desired ratio, as it shows that the light spreads out in a manner dependent on both \(\lambda\) and \(W\).
  • The diffraction angle is vital for analyzing and predicting where light will constructively and destructively interfere, thus forming the characteristic light and dark fringes of the diffraction pattern.

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Most popular questions from this chapter

Light waves with two different wavelengths, \(632 \mathrm{nm}\) and \(474 \mathrm{nm}\), pass simultaneously through a single slit whose width is \(7.15 \times 10^{-5} \mathrm{m}\) and strike a screen \(1.20 \mathrm{m}\) from the slit. Two diffraction patterns are formed on the screen. What is the distance (in \(\mathrm{cm}\) ) between the common center of the diffraction patterns and the first occurrence of a dark fringe from one pattern falling on top of a dark fringe from the other pattern?

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is \(296 \mathrm{nm} .\) What is the vacuum wavelength of the light in \(\mathrm{nm} ?\)

A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ), and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\)

In a single-slit diffraction pattern on a flat screen, the central bright fringe is \(1.2 \mathrm{cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{m} .\) When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of \(1.9 \mathrm{cm} .\) What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\)

Two gratings A and B have slit separations \(d_{A}\) and \(d_{\mathrm{B}},\) respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of \(\mathrm{A}\) is exactly replaced by the second-order maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

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