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Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of wavelength 585 nm. (a) On a very large distant screen, what is the \(total\) number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (\(Hint\): What is the largest that sin u can be? What does this tell you is the largest value of \(m\)?) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Short Answer

Expert verified
There are 39 bright fringes. The angle for the most distant fringe is approximately 73.5°.

Step by step solution

01

Understand the Concept

This problem is about the double-slit experiment, which demonstrates interference patterns. The formula for bright fringes is given by \( d \sin \theta = m\lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the fringe order (an integer), and \( \lambda \) is the wavelength of the light.
02

Determine the Maximum Value of \( m \)

Since \( \sin \theta \) cannot exceed 1, the maximum value of \( m \) for which a fringe can occur is when \( m\lambda \leq d \). Thus, the maximum \( m \) is determined as \( m = \frac{d}{\lambda} \). Substituting given values: \( d = 0.0116 \) mm = \( 11.6 \times 10^{-6} \) m and \( \lambda = 585 \) nm = \( 585 \times 10^{-9} \) m, which calculates to \( m = \frac{11.6 \times 10^{-6}}{585 \times 10^{-9}} \approx 19.82 \). This means \( m \) can be at most 19 each side.
03

Calculate Total Number of Bright Fringes

For \( m = 0 \), we have the central bright fringe. For \( m = -19 \) to \( m = 19 \), we include fringes on both sides. Therefore, the number of bright fringes is \( 2 \times 19 + 1 = 39 \).
04

Determine the Angle for the Most Distant Fringe

For the most distant fringe (\( m = 19 \)), we use \( \sin \theta = \frac{m\lambda}{d} \). Substituting \( m = 19 \), \( \lambda = 585 \times 10^{-9} \) m, and \( d = 11.6 \times 10^{-6} \) m into the formula gives \( \sin \theta \approx \frac{19 \times 585 \times 10^{-9}}{11.6 \times 10^{-6}} \approx 0.957 \). Thus, \( \theta = \arcsin(0.957) \approx 73.5^\circ \).

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Key Concepts

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

Constructive Interference
In the double-slit experiment, constructive interference happens when waves from two slits meet in phase. This means that the crests and troughs of the waves align perfectly, amplifying the light and creating a bright spot on the screen. This bright spot is known as a bright fringe. Constructive interference occurs at specific angles, where the path difference between waves from each slit equals an integer multiple of the wavelength.
In our problem, the condition for constructive interference is given by the formula: \[ d \sin \theta = m\lambda \] Here, \(d\) is the distance between the slits, \(\theta\) is the angle at which the constructive interference occurs, \(m\) is the order of the bright fringe, and \(\lambda\) is the wavelength of the light. The order \(m\) can be positive, negative, or zero, corresponding to different bright fringes on the screen.
Understanding this concept is key to predicting where and how light intensifies in patterns.
Interference Patterns
Interference patterns are a series of light and dark bands on a screen resulting from the superposition of light waves. The double-slit experiment creates these patterns when coherent light, like from a laser, passes through two closely spaced slits. On the screen, you'll notice alternating bright and dark stripes. Bright stripes occur where waves constructively interfere, while dark stripes are regions of destructive interference, where waves cancel each other out.
The spacing and intensity of these patterns depend on factors such as:
  • Distance between the slits
  • Wavelength of the light used
  • Distance from the slits to the screen
By calculating the angles for various orders of interference, our problem demonstrates how to predict the positions of bright fringes in the interference pattern.
Wavelength of Light
The wavelength of light, denoted by \(\lambda\), is a crucial factor in determining interference patterns in the double-slit experiment. It is the distance between successive peaks of a wave. In our exercise, \(\lambda\) is given as 585 nm, where a nanometer (nm) is one-billionth of a meter, a common unit for measuring wavelengths of light.
Shorter wavelengths, like blue light, require more precise slits to create extensive interference patterns. Longer wavelengths, like red light, typically result in more spread-out patterns. The light's wavelength determines how waves interfere with each other and effectively dictates the position and number of bright and dark fringes observed.
Understanding how wavelength affects interference helps explain why the formula \(d \sin \theta = m\lambda\) is vital in calculating where interference maxima will appear.
Angle of Diffraction
In the context of interference patterns, the angle of diffraction \(\theta\) relates directly to where bright fringes appear on the projection screen. The angle \(\theta\) is measured from the normal—or perpendicular line—to the screen or central fringe. Depending on the conditions and parameters of the experiment, such as light wavelength and distance between slits, this angle dictates the exact positioning of bright spots.The relationship is defined by the equation \[d \sin \theta = m\lambda\] which allows the calculation of \(\theta\) at different fringe orders \(m\). For example, in our problem, by plugging known values into this equation, we determine that the most distant fringe creates an angle of about 73.5° relative to the original light direction.Understanding how to derive this angle without complex calculations is essential, as it helps visualize and predict how light behaves as it encounters obstacles.

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Most popular questions from this chapter

Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Two speakers \(A\) and \(B\) are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\). For points far from the speakers, find all the angles relative to the centerline (Fig. P35.44) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

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