Chapter 36: Problem 4
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.