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35.8. Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the center of the 20 \(\mathrm{th}\) fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

Short Answer

Expert verified
The slit separation is approximately 1.137 mm.

Step by step solution

01

Identify the Formula for the Double-Slit Experiment

In Young's double-slit experiment, the position of the bright fringes is given by the formula \( y_m = \frac{m \cdot \lambda \cdot L}{d} \), where \( y_m \) is the distance from the central maximum, \( m \) is the fringe order, \( \lambda \) is the wavelength of the light used, \( L \) is the distance from the slits to the screen, and \( d \) is the separation between the slits.
02

Substitute Known Values

The values given are \( \lambda = 502 \text{ nm} = 502 \times 10^{-9} \text{ m} \), \( L = 1.20 \text{ m} \), \( y_m = 10.6 \times 10^{-3} \text{ m} \) (converted from mm), and \( m = 20 \) (since the 20th fringe is measured). Substitute these into the formula: \[ 10.6 \times 10^{-3} = \frac{20 \cdot 502 \times 10^{-9} \cdot 1.20}{d} \]
03

Solve for Slit Separation \( d \)

Re-arrange the formula to solve for \( d \): \[ d = \frac{20 \cdot 502 \times 10^{-9} \cdot 1.20}{10.6 \times 10^{-3}} \] Calculate the value:\[ d = \frac{20 \times 502 \times 10^{-9} \times 1.20}{10.6 \times 10^{-3}} \approx 0.001137 \, \text{m} \] So, the slit separation is approximately \( 1.137 \times 10^{-3} \text{ m} \) or \( 1.137 \text{ mm} \).

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Key Concepts

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

Fringe Order
In Young's double-slit experiment, the concept of fringe order helps us understand the pattern of alternating light and dark bands. Fringe order is denoted by the symbol \( m \) and refers to the sequence number of bright or dark fringes in the interference pattern. - The central bright fringe, also known as the central maximum, is considered the zero-order fringe (\( m=0 \)).- The first bright fringe on either side of the central maximum is the first-order fringe (\( m=1 \)), the second is \( m=2 \), and so on. In the given exercise, we are interested in the 20th bright fringe, which means the fringe order is \( m=20 \). The order tells us how many wavelengths fit into the path difference between the light arriving at the screen from the two slits. This concept is crucial because it allows us to use the fringe order in mathematical formulas to predict the location of specific fringes on the screen. The fringe order provides a systematic way to analyze and predict the positions of these light and dark bands.
Wavelength of Light
The wavelength of light, represented by \( \lambda \), is a fundamental factor in the formation of interference patterns in Young's double-slit experiment. It describes the distance between consecutive crests or troughs in a wave. Different wavelengths correspond to different colors of light; for example, the helium light used here has a wavelength of 502 nanometers (nm), which emits a specific shade of blue-green light. - In the fringe formula \( y_m = \frac{m \cdot \lambda \cdot L}{d} \), \( \lambda \) is crucial because variations in wavelength alter the spacing and position of interference fringes. - Longer wavelengths result in fringes that are spaced further apart, while shorter wavelengths create fringes that are closer together. In the context of the exercise, knowing the wavelength allows us to use the mathematics of the experiment to calculate other properties, such as the separation between the slits. Understanding the wavelength is also critical because it directly influences how much diffraction occurs when light passes through the slits.
Slit Separation
Slit separation, denoted by \( d \), is an important variable in understanding the interference pattern generated in Young's double-slit experiment. It represents the distance between the two slits through which light passes, and it plays a crucial role in determining the position and spacing of the interference fringes on the screen. - The formula \( y_m = \frac{m \cdot \lambda \cdot L}{d} \) shows that slit separation \( d \) is inversely proportional to the fringe spacing. - As the slit separation increases, the fringes become closer together; conversely, decreasing the slit separation causes the fringes to move further apart. In the provided exercise, calculating \( d \) helps us find out the precise configuration needed to produce a specific interference pattern. It is a pivotal aspect of the experiment since it helps control how the fringes manifest on the screen.
Interference Pattern
The interference pattern generated in Young's double-slit experiment is a series of alternating light (bright) and dark bands that appear on a screen when light passes through two closely spaced slits. This intriguing phenomenon is a result of the principle of superposition, where two waves that overlap combine to form a new wave pattern. - Constructive interference occurs when the peaks of two waves align, resulting in a bright fringe. This happens when the path difference is a whole number multiple of the wavelength \( m \lambda \). - Destructive interference occurs when the peaks of one wave coincide with the troughs of another, leading to a dark fringe, which occurs when the path difference is a half-number multiple of the wavelength \((m + 0.5) \lambda\). The regular spacing and pattern of these fringes depend on several factors, including the wavelength of light, the distance between the slits and the screen \( L \), and the slit separation \( d \). Understanding interference patterns is key to understanding wave behavior and is fundamental in many applications ranging from spectroscopy to the design of optical instruments. By carefully analyzing and measuring these patterns, important properties about the light and the system can be discovered and quantified.

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Most popular questions from this chapter

35.20. Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 \(\mathrm{cm}\) . Point \(P\) is 4.86 \(\mathrm{m}\) from \(A\) and 5.24 \(\mathrm{m}\) from \(B\) . What is the phase difference at \(P\) between these two waves?

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?

35.16. Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

35.15. Coherent light with wavelength 600 \(\mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slits. The first-order bright fringe is at 4.84 \(\mathrm{mm}\) 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?

35.14. Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) ( red) and 470 \(\mathrm{nm}\) (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

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