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In a Young's double-slit experiment, the wavelength of the light used is \(520 \mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{~m}\). Determine the angle that locates (a) the dark fringe for which \(m=0,\) (b) the bright fringe for which \(m=1,\) (c) the dark fringe for which \(m=1,\) and \((\mathrm{d})\) the bright fringe for which \(m=2\)

Short Answer

Expert verified
(a) \(10.70^\circ\), (b) \(21.82^\circ\), (c) \(33.87^\circ\), and (d) \(48.17^\circ\).

Step by step solution

01

Understanding the Problem

In a Young's double-slit experiment, the angle for dark and bright fringes are calculated using different formulas. For dark fringes, the formula is \( d \sin \theta = (m + 0.5) \lambda \), and for bright fringes, it is \( d \sin \theta = m \lambda \). Here, \( d = 1.4 \times 10^{-6} \text{ m} \), and \( \lambda = 520 \text{ nm} = 520 \times 10^{-9} \text{ m} \).
02

Calculate Angle for Dark Fringe with m=0

Using the formula for dark fringes: \( d \sin \theta = (m + 0.5) \lambda \). Substitute \( m = 0 \), \( d = 1.4 \times 10^{-6} \text{ m} \), and \( \lambda = 520 \times 10^{-9} \text{ m} \).\[ 1.4 \times 10^{-6} \sin \theta = (0 + 0.5) \times 520 \times 10^{-9} \]Solve for \( \sin \theta \):\[ \sin \theta = \frac{260 \times 10^{-9}}{1.4 \times 10^{-6}} \approx 0.1857 \]Calculate \( \theta \):\[ \theta = \arcsin(0.1857) \approx 10.70^\circ \]
03

Calculate Angle for Bright Fringe with m=1

Using the formula for bright fringes: \( d \sin \theta = m \lambda \). Substitute \( m = 1 \), \( d = 1.4 \times 10^{-6} \text{ m} \), and \( \lambda = 520 \times 10^{-9} \text{ m} \).\[ 1.4 \times 10^{-6} \sin \theta = 1 \times 520 \times 10^{-9} \]Solve for \( \sin \theta \):\[ \sin \theta = \frac{520 \times 10^{-9}}{1.4 \times 10^{-6}} \approx 0.3714 \]Calculate \( \theta \):\[ \theta = \arcsin(0.3714) \approx 21.82^\circ \]
04

Calculate Angle for Dark Fringe with m=1

Using the dark fringe formula again: \( d \sin \theta = (m + 0.5) \lambda \). Substitute \( m = 1 \), \( d = 1.4 \times 10^{-6} \text{ m} \), and \( \lambda = 520 \times 10^{-9} \text{ m} \).\[ 1.4 \times 10^{-6} \sin \theta = (1 + 0.5) \times 520 \times 10^{-9} \]Solve for \( \sin \theta \):\[ \sin \theta = \frac{780 \times 10^{-9}}{1.4 \times 10^{-6}} \approx 0.5571 \]Calculate \( \theta \):\[ \theta = \arcsin(0.5571) \approx 33.87^\circ \]
05

Calculate Angle for Bright Fringe with m=2

Use the bright fringe formula: \( d \sin \theta = m \lambda \). Substitute \( m = 2 \), \( d = 1.4 \times 10^{-6} \text{ m} \), and \( \lambda = 520 \times 10^{-9} \text{ m} \).\[ 1.4 \times 10^{-6} \sin \theta = 2 \times 520 \times 10^{-9} \]Solve for \( \sin \theta \):\[ \sin \theta = \frac{1040 \times 10^{-9}}{1.4 \times 10^{-6}} \approx 0.7429 \]Calculate \( \theta \):\[ \theta = \arcsin(0.7429) \approx 48.17^\circ \]

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Key Concepts

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

Interference Patterns
In Young's double-slit experiment, light waves pass through two closely spaced narrow slits, creating an interference pattern on a screen. This pattern consists of alternating bright and dark spots, known as fringes. These occur because light waves from the two slits overlap and either reinforce or cancel each other out depending on the path difference between them. The bright fringes occur where waves are in phase (constructive interference), leading to increased intensity. Conversely, dark fringes happen where waves are out of phase (destructive interference), resulting in lower or zero intensity.

Interference is a fundamental phenomenon in wave optics and is a direct evidence of the wave nature of light. Understanding these patterns helps scientists and engineers to design precision optical instruments and technologies that utilize the principles of diffraction and interference.

In practical terms, elucidating interference patterns involves calculating specific angles or positions on the screen where these bright and dark fringes appear. Correctly predicting these positions requires a strong grasp of wave interactions and the mathematical formulas that describe them.
Wavelength
Wavelength is a crucial parameter in understanding light and optics. It is defined as the distance between consecutive peaks of a wave. In Young's double-slit experiment, wavelength (\(\lambda\)) influences the spacing and position of interference fringes. The experiment typically uses monochromatic light, meaning light of a single wavelength, to ensure clear and predictable patterns.

In a vacuum, light travels with a specific wavelength, denoted in nanometers (nm). For example, light with a wavelength of 520 nm appears green. This wavelength interacts with the slit separation and screen distance to determine the exact formation of the bright and dark fringes.

In practice, altering the wavelength changes the interference pattern. Increasing the wavelength generally results in more widely spaced fringes, as the path difference required for interference shifts. Consequently, understanding wavelength is fundamental for interpreting and predicting interference phenomena in double-slit experiments.
Fringe Angle Calculation
The angle at which the bright and dark fringes occur in a double-slit experiment can be calculated using established formulas. These calculations are critical to understanding the precise locations of fringes on a screen. For bright fringes, the formula used is:\[d \sin \theta = m \lambda\]where:
  • \(d\) is the distance between the slits.
  • \(\theta\) is the angle made with the normal of the slits.
  • \(m\) is the fringe order number.
For dark fringes, the calculation changes slightly to:\[d \sin \theta = (m + 0.5) \lambda\]
In this context, \(m\) is an integer representing the order of the fringe under consideration. Each specific value of \(m\) corresponds to either a crucial dark or bright fringe.

These formulas help determine how the wave crests align relative to the slits, allowing precise predictions of where each fringe will fall. Accurate angle calculations are vital for various optics applications, including those in engineering and scientific research.
Optics
Optics is the branch of physics that deals with the study of light. It examines its properties, behavior, and interactions with different materials. Young's double-slit experiment is a classic optics experiment highlighting the wave nature of light.

Understanding optics helps in grasping concepts such as refraction, reflection, diffraction, and interference, all of which are crucial for a wide range of scientific and practical applications. Optics knowledge enables the design and development of lenses, microscopes, telescopes, and fiber optics communication systems.

Young's experiment demonstrates how light behaves similarly to other waves, providing insights into the principles of wave-particle duality. It is fundamental in advancing fields like quantum physics, where understanding the properties and behavior of light is essential.

Moreover, knowing optics allows for the creation of devices that alter and control light for various uses, from enhancing images to increasing security through advanced optical systems.

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Most popular questions from this chapter

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and \(660 \mathrm{~nm}\). What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Concept Questions (a) What, if any, phase change occurs when light, traveling in air, reflects from the interface between the air and a soap film \((n=1.33) ?\) (b) What, if any, phase change occurs when light, traveling in a soap film, reflects from the interface between the soap film and a glass plate \((n=1.52) ?(\mathrm{c})\) Is the wavelength of the light in a soap film greater than, smaller than, or equal to the wavelength in a vacuum? Problem A soap film \((n=1.33)\) is \(465 \mathrm{nm}\) thick and lies on a glass plate \((n=1.52)\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm}\), travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00)\). A thin film of liquid floats on the gasoline and has a refractive index that is between \(1.00\) and \(1.40\). Light that has a wavelength of \(625 \mathrm{~nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{~nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{~nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{~m} ?\)

In a single-slit diffraction pattern on a flat screen, the central bright fringe is \(1.2 \mathrm{~cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{~m}\). When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of \(1.9 \mathrm{~cm}\). What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta\).

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