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Chapter 36: Problem 90

A single-slit diffraction experiment is set up with light of wavelength \(420 \mathrm{nm}\), incident perpendicularly on a slit of width \(5.10 \mu \mathrm{m} .\) The viewing screen is \(3.20 \mathrm{~m}\) distant. On the screen, what is the distance between the center of the diffraction pattern and the second diffraction minimum?

Short Answer

Expert verified
The distance is approximately 0.527 m.

Step by step solution

01

Understanding Diffraction

In a single-slit diffraction experiment, light waves spread as they pass through a narrow slit, creating a pattern of bright and dark regions on the screen. The dark regions are known as diffraction minima.
02

Formula for Minima Position

The position of the diffraction minima is given by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength of light, \( \theta \) is the angle relative to the central axis, and \( m \) is the order of the minimum (with \( m = 1, 2, \ldots \) ).
03

Calculate the Angle

To find the angle for the second minimum \( (m = 2) \), rearrange the formula to \( \sin \theta = \frac{m \lambda}{a} \). Substituting \( m = 2 \), \( \lambda = 420 \times 10^{-9} \mathrm{m} \), and \( a = 5.10 \times 10^{-6} \mathrm{m} \), we get \( \sin \theta = \frac{2 \times 420 \times 10^{-9}}{5.10 \times 10^{-6}} \approx 0.164706 \).
04

Calculate the Angle (Approximation)

Since \( \theta \) is small in practical diffraction, \( \sin \theta \approx \theta \) (in radians). Thus, \( \theta \approx 0.164706 \mathrm{radians} \).
05

Calculate the Distance on Screen

The linear distance \( y \) on the screen from the central maximum to the second minimum is given by \( y = L \tan \theta \approx L \theta \) (as \( \tan \theta \approx \theta \) for small angles). Using \( L = 3.20 \mathrm{~m} \) and \( \theta = 0.164706 \), we get \( y = 3.20 \times 0.164706 \approx 0.527 \mathrm{~m} \).
06

Conclusion

Thus, the distance between the center of the diffraction pattern and the second diffraction minimum is approximately \( 0.527 \mathrm{~m} \).

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Key Concepts

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

Diffraction Pattern
A diffraction pattern emerges when light waves encounter an obstacle or a slit that is comparable in size to their wavelength. As the light passes through a narrow slit, it bends and spreads out, creating a series of light and dark fringes on a screen. These alternating patterns are a result of the constructive and destructive interference of the light waves.
  • Bright fringes occur where the light waves reinforce each other, leading to maximum intensity.
  • Dark fringes, or minima, occur where the waves cancel each other out.
The central bright fringe, also known as the central maximum, is the brightest and widest. As you move away from the center, the intensity and width of the fringes decrease. Understanding this pattern is crucial as it allows us to analyze the structure of light and the behavior of waves in different mediums.
Wavelength of Light
The wavelength of light, often denoted by the symbol \( \lambda \), is a critical factor in determining how light behaves in diffraction experiments. It is the distance between two consecutive points of similar phase, such as crests or troughs, in a wave. The wavelength affects the diffraction pattern by influencing the spacing between the fringes.
In the case of a single-slit diffraction experiment, the wavelength of the incident light helps to establish the positions of the minima and maxima. Longer wavelengths result in wider spacing between the fringes, while shorter wavelengths lead to closer fringes. Given the wavelength \( \lambda = 420\, \mathrm{nm} \) in the current exercise, its impact can be calculated using the formula for minima positions: \( a \sin \theta = m \lambda \). As seen, the wavelength plays a key role in determination of angles for the diffraction minima.
Diffraction Minima
In a diffraction pattern, diffraction minima are the dark stripes on the screen, where the light intensity is at its lowest. These are points of destructive interference, where waves from different parts of the slit arrive out of phase and cancel each other. The minima occur at angles where the path difference between waves equals integral multiples of the wavelength.
The position of these minima can be calculated using the formula \( a \sin \theta = m \lambda \), where:
  • \( a \) is the slit width.
  • \( \theta \) is the angle from the central maximum.
  • \( m \) is the order of the minimum.
  • \( \lambda \) is the wavelength of light.
For each integer \( m \) (excluding zero), a minimum is formed. In practical applications, determining the position of these minima helps in measuring the light's properties and understanding wave behavior in physical systems.

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Most popular questions from this chapter

An \(x\) -ray beam of wavelength \(A\) undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is \(23^{\circ}\), and an x-ray beam of wavelength 97 pm undergoes third-order reflection when its angle of incidence to that face is \(60^{\circ} .\) Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength \(A\).

Monochromatic light of wavelength \(441 \mathrm{nm}\) is incident on a narrow slit. On a screen \(2.00 \mathrm{~m}\) away, the distance between the second diffraction minimum and the central maximum is \(1.50 \mathrm{~cm}\). (a) Calculate the angle of diffraction \(\theta\) of the second minimum. (b) Find the width of the slit.

The \(D\) line in the spectrum of sodium is a doublet with wavelengths 589.0 and \(589.6 \mathrm{nm}\). Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum.

In a double-slit experiment, the slit separation \(d\) is 2.00 times the slit width \(w\). How many bright interference fringes are in the central diffraction envelope?

A beam of light consisting of wavelengths from \(460.0 \mathrm{nm}\) to \(640.0 \mathrm{nm}\) is directed perpendicularly onto a diffraction grating with 160 lines/mm. (a) What is the lowest order that is overlapped by another order? (b) What is the highest order for which the complete wavelength range of the beam is present? In that highest order, at what angle does the light at wavelength (c) \(460.0 \mathrm{nm}\) and (d) \(640.0 \mathrm{nm}\) appear? (e) What is the greatest angle at which the light at wavelength \(460.0 \mathrm{nm}\) appears?
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