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Chapter 35: Problem 21

In a double-slit experiment, the distance between slits is \(5.0 \mathrm{~mm}\) and the slits are \(1.0 \mathrm{~m}\) from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength \(480 \mathrm{~nm}\), and the other due to light of wavelength \(600 \mathrm{~nm}\). What is the separation on the screen between the third-order \((m=3)\) bright fringes of the two interference patterns?

Short Answer

Expert verified
The separation is 0.072 mm.

Step by step solution

01

Identify the Formula

To find the separation between the third-order bright fringes, we use the formula for fringe separation: \( y_m = \frac{m \cdot \lambda \cdot L}{d} \), where \( y_m \) is the distance from the central maximum to the \( m \)-th bright fringe, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, \( d \) is the separation between the slits, and \( m \) is the order of the fringe.
02

Calculate the Third-order Fringe for 480 nm

Plug in the values for the first set of conditions: \( d = 5.0 \times 10^{-3} \mathrm{~m} \), \( L = 1.0 \mathrm{~m} \), \( \lambda = 480 \times 10^{-9} \mathrm{~m} \), and \( m = 3 \). The formula becomes \( y_3 = \frac{3 \cdot 480 \times 10^{-9} \cdot 1.0}{5.0 \times 10^{-3}} \). Calculating this gives \( y_3 = 0.288 \times 10^{-3} \mathrm{~m} = 0.288 \mathrm{~mm} \).
03

Calculate the Third-order Fringe for 600 nm

Now, use the same formula for the second wavelength: \( \lambda = 600 \times 10^{-9} \mathrm{~m} \). Substitute into the formula: \( y_3 = \frac{3 \cdot 600 \times 10^{-9} \cdot 1.0}{5.0 \times 10^{-3}} \). This gives us \( y_3 = 0.36 \times 10^{-3} \mathrm{~m} = 0.36 \mathrm{~mm} \).
04

Find the Separation Between the Fringes

The separation between the third-order bright fringes for the two wavelengths is the difference between the two fringe distances. Subtract the two values: \( y_{3,600} - y_{3,480} = 0.36 \mathrm{~mm} - 0.288 \mathrm{~mm} = 0.072 \mathrm{~mm} \).

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Key Concepts

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

Interference Pattern
The double-slit experiment is a classic demonstration of the wave-like behavior of light. It shows how coherent light passing through two closely spaced slits can create an alternating pattern of dark and bright bands on a screen. This is known as an interference pattern. Each bright band, or fringe, results from constructive interference where the light waves from each slit meet in phase. Dark bands occur where the waves meet out of phase, causing destructive interference. In essence, the interference pattern is a visual representation of the light wave properties adjusting as they overlap. As the waves meet with varying phases at different points on the screen, they add up or cancel out, creating areas of brightness or darkness.
  • Bright fringes: Areas where waves meet in phase.
  • Dark fringes: Areas where waves meet out of phase.
This alternating pattern underscores the fundamental concept of wave interference and provides insight into the wave nature of light.
Fringe Separation Formula
The fringe separation formula is essential to determining the position of bright fringes in an interference pattern on a screen. The formula is expressed as: \[ y_m = \frac{m \cdot \lambda \cdot L}{d} \] This equation relates several important variables:
  • \( y_m \): The distance from the central maximum to the \( m \)-th bright fringe.
  • \( m \): The order of the bright fringe.
  • \( \lambda \): The wavelength of the light used.
  • \( L \): The distance from the slits to the screen.
  • \( d \): The separation between the slits.
By plugging the known values into the equation, the position of specific bright fringes can be calculated with absolute precision. This formula is crucial in experimental physics, helping scientists predict and verify the locations of interference patterns created by different wavelengths of light.
Wavelength
Wavelength is a critical factor in understanding the behavior of light waves in the interference patterns of a double-slit experiment. It is defined as the distance between successive crests of a wave. In the context of the experiment, wavelengths determine how the light waves overlap when they emerge from the slits and subsequently interact with each other. The longer the wavelength, the farther apart the bright fringes are in the interference pattern. In contrast, shorter wavelengths result in fringes that are closer together. In our original problem, two wavelengths are analyzed: 480 nm and 600 nm. These two values illustrate how different wavelengths can lead to distinct interference patterns on the screen. Understanding wavelength and its impact is key to predicting how a light wave will behave when it encounters an obstacle or opening, such as the slits in the double-slit experiment. This knowledge enables us to not only comprehend light's wave nature but also accurately calculate the resulting interference patterns._FADE

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Most popular questions from this chapter

Three electromagnetic waves travel through a certain point \(P\) along an \(x\) axis. They are polarized parallel to a \(y\) axis, with the following variations in their amplitudes. Find their resultant at \(P\). $$ \begin{aligned} &E_{1}=(10.0 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(2.0 \times 10^{14} \mathrm{rad} / \mathrm{s}\right) t\right] \\ &E_{2}=(5.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(2.0 \times 10^{14} \mathrm{rad} / \mathrm{s}\right) t+45.0^{\circ}\right] \\ &E_{3}=(5.00 \mu \mathrm{V} / \mathrm{m}) \sin \left[\left(2.0 \times 10^{14} \mathrm{rad} / \mathrm{s}\right) t-45.0^{\circ}\right] \end{aligned} $$

A disabled tanker leaks kerosene \((n=1.20)\) into the Persian Gulf, creating a large slick on top of the water \((n=1.30) .\) (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is \(460 \mathrm{~nm}\), for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are. scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at \(600 \mathrm{~nm}\) and a minimum at \(450 \mathrm{~nm}\), with no minimum in between. If \(n=1.33\) for the film, what is the film thickness, assumed uniform?

Suppose that Young's experiment is performed with blue-green light of wavelength \(500 \mathrm{~nm}\). The slits are \(1.20 \mathrm{~mm}\) apart, and the viewing screen is \(5.40 \mathrm{~m}\) from the slits. How far apart are the bright fringes near the center of the interference pattern?

A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of 500 and \(700 \mathrm{~nm}\) and for no wavelengths in between. If the index of refraction of the oil is \(1.30\) and that of the glass is \(1.50\), find the thickness of the oil film.
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