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In the double-slit setup using monochromatic illumination, what is the value of the path-length difference for the first bright bands above and below the central band? Explain your answer.

Short Answer

Expert verified
Path-length difference for the first bright band is \( \lambda \).

Step by step solution

01

Understanding the Double-Slit Experiment

The double-slit experiment involves light waves passing through two slits and interfering with each other to form a pattern of bright and dark bands (fringes) on a screen. These bands result from the constructive and destructive interference of the light waves.
02

Identifying Path-Length Difference for Bright Bands

The path-length difference is determined by the condition for constructive interference, where bright bands are formed. The path-length difference for constructive interference is given by the formula \( \Delta L = m \lambda \), where \( m \) is the order of the fringe (an integer) and \( \lambda \) is the wavelength of the light.
03

Finding the First Bright Band Order

The first bright band above or below the central band is the first-order bright fringe, which corresponds to \( m = 1 \). The central band (central maximum) is \( m = 0 \). Therefore, the first order is the first bright fringe.
04

Calculating Path-Length Difference for the First Bright Band

Using the order \( m = 1 \) for the first bright band, we apply the formula \( \Delta L = m \lambda \) to find the path-length difference: \( \Delta L = 1 \cdot \lambda = \lambda \). This indicates that the path-length difference for the first bright band is equal to one wavelength of the light used.

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Key Concepts

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

Constructive Interference
When two waves meet, they can interact in a way that enhances their combined effect. This is known as constructive interference. In the context of the double-slit experiment, constructive interference occurs when the crests of two light waves coincide, leading to increased light intensity. This phenomenon is responsible for creating bright bands on the observation screen, which are referred to as bright fringes.

Constructive interference happens under specific conditions. The key condition is that the path-length difference between the waves should be an integral multiple of their wavelength. This can be mathematically expressed as:
  • \( \Delta L = m \lambda \)
Here, \( \Delta L \) represents the path-length difference, \( m \) is an integer (indicating the order of the fringe), and \( \lambda \) is the wavelength of light. When these conditions are met, the resultant light wave has a heightened amplitude, which makes the fringe appear bright.
Path-Length Difference
The concept of path-length difference is fundamental to understanding interference patterns, such as those seen in the double-slit experiment. This is the difference in distance that two waves travel before meeting at a point. For light waves passing through the two slits, each wave will travel a slightly different distance to reach any given point on the screen.

The path-length difference determines whether the waves will interfere constructively or destructively. If the path-length difference corresponds to a whole number multiple of the wavelength, constructive interference occurs, resulting in a bright fringe. Conversely, if the path-length difference corresponds to a half-number multiple, destructive interference happens, resulting in a dark fringe. The formula to calculate path-length difference for constructive interference is:
  • \( \Delta L = m \lambda \)
Hence, understanding path-length difference is crucial to predicting where the bright and dark fringes will occur on the screen.
First-Order Fringes
First-order fringes refer to the bright bands that appear immediately next to the central bright band, known as the central maximum, in the double-slit experiment. The central maximum corresponds to the zeroth-order fringe, characterized by a path-length difference of zero. The first-order fringes, however, are the first set of bright fringes on either side of this central peak.

To find the path-length difference that identifies the first-order fringes, the order of the fringe \( m \) is set to 1 in the equation \( \Delta L = m \lambda \). Thus, for first-order fringes, the path-length difference is:
  • \( \Delta L = 1 \times \lambda = \lambda \)
This reveals that the path-length difference for these fringes is exactly one wavelength of the light used. Understanding first-order fringes is essential as it provides a reference point for measuring and analyzing other interference fringes.
Wavelength of Light
The wavelength of light is a fundamental property that describes the distance between consecutive peaks of the light wave. It is denoted by \( \lambda \) and measured in meters (often in nanometers for visible light due to its small scale). In the double-slit experiment, the wavelength plays a critical role in determining the pattern of interference fringes observed on the screen.

The wavelength is central to calculating the path-length difference required for constructive interference and thus the formation of bright fringes. The equation \( \Delta L = m \lambda \) directly ties the wavelength to the order of the fringe and the path-length difference. When the necessary conditions of this equation are met, a bright fringe occurs at the point of interference.

The wavelength affects the spacing and position of the fringes. A shorter wavelength results in more closely spaced fringes, while a longer wavelength leads to wider spacing. Understanding wavelength is essential for interpreting and predicting patterns in optical experiments like the double-slit setup.

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Most popular questions from this chapter

A double-slit experiment is done in the usual way with \(480-\mathrm{nm}\) light and narrow slits that are \(0.050 \mathrm{~cm}\) apart. At what angle to the central axis will one observe \((a)\) the third-order bright spot and \((b)\) the second minimum from the central maximum?

In Young's experiment using monochromatic light at a vacuum wavelength of \(589.3 \mathrm{~nm}\), the two narrow slits are separated center to center by \(2.40 \mathrm{~mm}\). Determine the spacing between successive bright bands on a screen \(1.00 \mathrm{~m}\) away. Explain your answer.

When one leg of a Michelson interferometer is lengthened slightly, 150 dark fringes sweep through the field of view. If the light used has a wavelength of \(\lambda=480 \mathrm{~nm}\), how far was the mirror in that leg moved? Darkness is observed when the light beams from the two legs are \(180^{\circ}\) out-of- phase. As the length of one leg is increased by \(1-2 \lambda\) the path length (down and back) increases by \(\lambda\) and the field of view changes from dark to bright to dark. When 150 fringes pass, the leg is lengthened by an amount $$ (150)\left(\frac{1}{2} \lambda\right)=(150)(240 \mathrm{~nm})=36000 \mathrm{~nm}=0.0360 \mathrm{~mm} $$

A single slit of width \(0.140 \mathrm{~mm}\) is illuminated by monochromatic light, and diffraction bands are observed on a screen \(2.00 \mathrm{~m}\) away. If the second dark band is \(16.0 \mathrm{~mm}\) from the central bright band, what is the wavelength of the light?

A narrow beam of yellow light of wavelength \(600 \mathrm{~nm}\) is incident normally on a diffraction grating ruled 2000 lines \(/ \mathrm{cm}\), and images are formed on a screen parallel to the grating and \(1.00 \mathrm{~m}\) distant. Compute the distance along the screen from the central bright line to the first-order lines.

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