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Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 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 approximately 5.91 mm.

Step by step solution

01

Identify the Known Variables

The problem provides the wavelength of light, \( \lambda = 633 \) nm, which is \( 633 \times 10^{-9} \) m. The slit width is \( a = 0.750 \) mm or \( 0.750 \times 10^{-3} \) m. The screen distance is \( L = 3.50 \) m. We need to find the distance between the first dark fringes on either side of the central maximum.
02

Use the Single-Slit Diffraction Formula

The position of the dark fringes in a single-slit diffraction pattern is given by the formula: \( a \sin \theta = m \lambda \), where \( m \) is the order of the fringe and \( m eq 0 \). For the first dark fringe, \( m = 1 \). Thus, \( 0.750 \times 10^{-3} \sin \theta = 1 \times 633 \times 10^{-9} \).
03

Solve for \( \sin \theta \)

Rearrange the equation to find \( \sin \theta \): \( \sin \theta = \frac{633 \times 10^{-9}}{0.750 \times 10^{-3}} \). Perform the calculation: \( \sin \theta \approx 0.000844 \).
04

Approximate \( \theta \) for Small Angles

For small angles, \( \sin \theta \approx \tan \theta \approx \theta \), so \( \theta \approx 0.000844 \) radians. This approximation is valid here because the angle is small.
05

Calculate the Linear Distance to the Dark Fringe

The linear distance from the central maximum to the first dark fringe is \( y = L \tan \theta \approx L \theta \). With \( L = 3.50 \) m, \( y = 3.50 \times 0.000844 \approx 0.002954 \) m.
06

Find Total Distance Between Two Dark Fringes

The distance between the two first dark fringes on either side of the central maximum is \( 2y \). Thus, \( 2y = 2 \times 0.002954 \approx 0.005908 \) m.

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Key Concepts

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

Wavelength
Wavelength is a crucial factor in understanding the behavior of waves, including light waves. It is the distance between consecutive crests of a wave, which greatly influences how the wave will interact with obstacles, such as a slit. In the original exercise, light has a wavelength of 633 nm, equivalent to 633 nanometers or 633 x 10-9 meters. This short length means it is in the visible spectrum, which means our eye can see it as a specific color. Wavelength is directly related to the energy of the light; shorter wavelengths have higher energy and tend to interact more strongly with materials, causing different patterns. In single-slit diffraction, the wavelength determines the spacing and position of the diffraction fringes. The variation in wavelength changes how the light bends and where the dark and bright fringes are located.
Central Maximum
In single-slit diffraction, the central maximum is the brightest part of the diffraction pattern observed on a screen. It is located directly opposite the light source's entry point into the slit. The central maximum stands out due to its intensity compared to adjacent maxima, making it the most noticeable feature. The central maximum occurs because light wavefronts from different points across the slit aperture converge in-phase at this particular point on the screen. As a result, their amplitudes add constructively, leading to significant light intensity. The broadness of the central maximum depends on factors like the light's wavelength and the slit width. A smaller slit or a longer wavelength results in a wider central maximum. Understanding the central maximum is essential when calculating distances to dark fringes, as it serves as the reference point.
Screen Distance
Screen distance, represented by the variable \( L \) in our calculations, is the straight-line distance from the slit to the screen where the diffraction pattern is observed. In this particular exercise, the screen distance is 3.50 meters. This value plays a critical role in determining the scale of the diffraction pattern.A larger screen distance generally results in more widely spaced fringes, making the pattern easier to measure and observe. When dealing with small angles, the relation \( y = L \theta \), where \( y \) is the linear distance from the central maximum to a fringe, allows us to convert angular movement into linear distance on the screen. This makes calculations more practical, as we often measure results like the fringe spacing in meters or centimeters.
Dark Fringes
Dark fringes are the darker spots on a diffraction pattern where light waves cancel each other out, resulting in minimal or no light intensity. These fringes occur due to destructive interference, where wavefronts from different parts of the slit arrive out of phase.The position of dark fringes is determined using the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the diffraction angle, \( m \) is the fringe order (not equal to zero), and \( \lambda \) is the wavelength. For the first dark fringe, \( m = \pm 1 \).In practice, we calculate the angle \( \theta \) for the first order \( m \) and use the small-angle approximation to find the distance from the central maximum to the dark fringe, \( y = L \theta \). This process demonstrates the predictable nature of wave diffraction patterns, ensuring precise calculation of fringe positions.

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Most popular questions from this chapter

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

A laser beam of wavelength \(\lambda\) = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 \(\mu\)m apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 \(\mu\)m apart. Repeat the calculation of part (a) for the DVD.

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?

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?

A loudspeaker with a diaphragm that vibrates at 960 Hz is traveling at 80.0 m/s directly toward a pair of holes in a very large wall. The speed of sound in the region is 344 m/s. Far from the wall, you observe that the sound coming through the openings first cancels at \(\pm11.4^\circ\) with respect to the direction in which the speaker is moving. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

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