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Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem \(without\) calculating all the angles! (\(Hint\): What is the largest that sin \theta can be? What does this tell you is the largest that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Short Answer

Expert verified
226 completely dark fringes occur. The most distant one is at approximately 82.8°.

Step by step solution

01

Understand the Problem

We are given the wavelength of light \( \lambda = 585 \ nm \) and the width of the slit \( a = 0.0666 \ mm \). Our task is to find the number of dark fringes on both sides of the central bright spot using the diffraction condition for single slits.
02

Recall the Condition for Dark Fringes

For a single-slit diffraction pattern, dark fringes (minimum intensity) occur at angles where \( a \sin \theta = m \lambda \), where \( m \) is an integer. Specifically, \( m = \pm 1, \pm 2, \ldots \). The maximum value for \( \sin \theta \) is 1.
03

Calculate Maximum Possible m

Set \( \sin \theta = 1 \) to find the maximum \( m \) using the formula for dark fringes:\[ a \cdot 1 = m \lambda \]\[ m = \frac{a}{\lambda} \]Convert \( a = 0.0666 \ mm \) to \( nm \) for consistency with \( \lambda \):\[ a = 0.0666 \ mm = 66600 \ nm \]Now compute \( m \):\[ m = \frac{66600 \ nm}{585 \ nm} \approx 113.85 \]Since \( m \) must be an integer, the maximum \( m \) is 113.
04

Determine Total Number of Dark Fringes

The total number of dark fringes includes both positive and negative values of \( m \), from \(-113\) to \(+113\). Don't forget to include \( m = 0 \) once (central bright spot), then:\[ 113 + 113 = 226 \text{ (sides)} \] adding the central point gives a total of 227 positions, so subtract the central spot:\[ 113 + 113 = 226 \text{ dark fringes} \]
05

Calculate Angle for Most Distant Dark Fringe

Using \( m = 113 \), calculate \( \theta \) for the most distant dark fringe:\[ a \sin \theta = 113 \lambda \]\[ \sin \theta = \frac{113 \times 585 \ nm}{66600 \ nm} \approx 0.992 \]Hence, \( \theta = \arcsin(0.992) \approx 82.8^\circ \).

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Key Concepts

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

Diffraction Pattern
In the realm of light and optics, a diffraction pattern is an intriguing phenomenon. It occurs when light bends around the edges of an obstacle, such as a slit, resulting in a unique arrangement of bright and dark bands on a screen. Think of it like the shadow effect of objects not blocking light completely, but rather bending it all around, creating a pattern. The central bright spot in this pattern is the largest and most intense, while the surrounding dark and light regions, known as fringes, vary in intensity. The arrangement is defined by the constructive and destructive interference of light waves. Constructive interference happens when light waves amplify each other, forming bright fringes. In contrast, destructive interference leads to dark fringes where light waves cancel each other out. These patterns, known for their symmetry, help us understand the wave nature of light.
Dark Fringes
Dark fringes are the lines on the screen where light and dark areas alternate in the diffraction pattern. In a single-slit diffraction experiment, dark fringes signify points of minimum intensity due to destructive interference. These fringes form where the path difference between light waves results in a complete wave cancellation. Mathematically, the condition for dark fringes is expressed by the formula:\[ a \sin \theta = m \lambda \]Where:
  • \( a \) is the slit width.
  • \( \lambda \) is the wavelength of light.
  • \( m \) is the order number, an integer representing the fringe sequence.
The value of \( m \) determines the exact position of each dark fringe on the screen. Intriguingly, the number of visible dark fringes depends on the maximum possible value of \( m \), calculated from the physical constraints of the setup like slit width and wavelength.
Wavelength of Light
The wavelength of light is a crucial element in studying diffraction. It defines the color and type of light wave in the electromagnetic spectrum, typically measured in nanometers (nm).In this task, the given wavelength is 585 nm. The wavelength is vital because it directly influences the diffraction pattern produced. This is because it determines how far light bends around slits. A longer wavelength would result in more widely spaced fringes. The relationship between wavelength and diffraction can be visualized with the formula \( m \lambda = a \sin \theta \), linking wavelength, order number \( m \), and angle \( \theta \) for each fringe. Understanding wavelength and its role in diffraction helps grasp why different wavelengths scatter light in unique ways.
Angle of Diffraction
The angle of diffraction is the angle at which light deviates from its initial path due to passing through a narrow opening, like a slit. It is crucial in determining the positions of the dark and bright fringes in a diffraction pattern.For dark fringes, the angle is calculated using the formula:\[ a \sin \theta = m \lambda \]Here:
  • \( \theta \) is the diffraction angle.
  • \( a \) is the slit width.
  • \( \lambda \) is the wavelength.
  • \( m \) is the order of the fringe.
The maximum value of \( \sin \theta \) is 1, determining the highest possible angle. In this case, the most distant dark fringe occurs at an angle calculated using the highest integer \( m \) possible, which is where \( \sin \theta \approx 0.992 \), leading to an angle of about \( 82.8^\circ \). Recognizing these angles allows us to predict where light will vanish (dark fringes) on a screen.

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Most popular questions from this chapter

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

Visible light passes through a diffraction grating that has 900 slits/cm, and the interference pattern is observed on a screen that is 2.50 m from the grating. (a) Is the angular position of the first-order spectrum small enough for sin \(\theta \approx \theta\) to be a good approximation? (b) In the first- order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 mm. What is the difference in these wavelengths?

A slit \(0.240 \mathrm{~mm}\) wide is illuminated by parallel light rays of wavelength \(540 \mathrm{nm} .\) The diffraction pattern is observed on a screen that is \(3.00 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

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