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Chapter 27: Problem 5

In a Young's double-slit experiment, the seventh dark fringe is located \(0.025 \mathrm{m}\) to the side of the central bright fringe on a flat screen, which is \(1.1 \mathrm{m}\) away from the slits. The separation between the slits is \(1.4 \times 10^{-4} \mathrm{m}\) What is the wavelength of the light being used?

Short Answer

Expert verified
The wavelength is approximately 500 nm.

Step by step solution

01

Understanding the Problem

We have a double-slit experiment where the distance to the seventh dark fringe and other parameters like screen distance and slit separation are given. We need to find the light's wavelength.
02

Applying the Dark Fringe Formula

For dark fringes in a double-slit experiment, the formula is \(d \sin \theta = (m + \frac{1}{2}) \lambda\), where \(d\) is the slit separation, \(m\) is the fringe number (7 in this case), and \(\lambda\) is the wavelength.
03

Finding the Angle \(\theta\)

Use \(\tan \theta \approx \sin \theta = \frac{x}{L}\), where \(x = 0.025\) m (distance to the dark fringe) and \(L = 1.1\) m (distance to screen). So, \(\sin \theta = \frac{0.025}{1.1}\).
04

Solving for Wavelength \(\lambda\)

After calculating \(\sin \theta\), use \(d \sin \theta = (7 + \frac{1}{2}) \lambda\). Substitute \(d = 1.4 \times 10^{-4}\) m and \(\sin \theta\) from the previous step to solve for \(\lambda\).
05

Final Calculation

Calculate \(\lambda = \frac{d \cdot \sin \theta}{7.5}\) using calculated values to find the 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.

Interference Patterns
In Young's double-slit experiment, creating interference patterns is like weaving light into a marvelous tapestry. These patterns occur because light waves from the two slits meet on the screen and either add up or cancel each other out. This happens due to the principle of superposition. When waves overlap, they form bright and dark spots known as fringes.

Interference patterns consist of:
  • Bright Fringes: These are the locations where the waves meet in phase and reinforce each other, creating maxima of light.
  • Dark Fringes: These appear where the waves meet out of phase and cancel each other out, creating minima of light. The seventh dark fringe, for example, is a point where destructive interference results in darkness.
Without interference patterns, the interaction of light waves in such a setup wouldn't be perceptible as distinct bright and dark bands on the screen.
Fringe Separation
Fringe separation refers to the distance between these interference bands on the screen. In Young's experiment, understanding fringe separation is crucial to insights into how different physics parameters interact.

Key considerations around fringe separation include:
  • Impact of Slit Separation (d): Smaller slit separation increases fringe spacing, while wider slits result in closer fringes.
  • Distance to Screen (L): Increasing the distance to the screen results in wider separation of fringes, making them easier to measure and analyze. In our example, with the screen 1.1 meters away, the fringes appear clearly distinct.
These factors allow us to manipulate the experimental setup to analyze fringe properties under different physical conditions.
Wavelength Calculation
Calculating the wavelength of light in a double-slit experiment involves measuring and applying several important parameters. This calculation links physical observations to the wave nature of light.

To calculate the wavelength, we often use the formula specific to dark fringes:\[ d \sin \theta = \left(m + \frac{1}{2}\right) \lambda \]Where:
  • d: The separation between the slits
  • m: Fringe order; for the seventh dark fringe, it is 7
  • \(\lambda\): The wavelength, which is what we aim to find
Step by step, we follow these actions:
  • Find the angle \(\theta\) using \(\sin \theta = \frac{x}{L}\), based on the measured distance to a specific fringe and the distance to the screen.
  • Substitute the values into the formula to solve for \(\lambda\).
  • In the corresponding scenario, with a known dark fringe and slits, we find that the entire setup yields the precise wavelength of the light being measured.
This reveals how mathematical relationships and empirical observations blend to understand light's intrinsic properties.
Optics
Optics is the branch of physics that deals with the study of light and its interactions. Young's double-slit experiment is a pivotal concept within optics. It helped verify the wave nature of light, challenging the previously dominant particle theory of light.

Important principles at play in optics are illustrated by this experiment:
  • Wave Particles Duality: Light behaves as both a wave and a particle, with interference demonstrating wave properties.
  • Coherent Light Sources: For the experiment to succeed, light from the two slits must be coherent. This means the waves must have a constant phase difference and the same frequency.
Through these concepts, optics shows us how intricate and profound the behavior of light is, enhancing technologies from simple lenses to complex systems in modern devices.

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Most popular questions from this chapter

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 (d) the bright fringe for which \(m=2\).

A hunter who is a bit of a braggart claims that from a distance of \(1.6 \mathrm{km}\) he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{mm},\) the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{nm}\).

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 \(3.76 \times 10^{-6} \mathrm{m} ?\)

Light waves with two different wavelengths, 632 nm and \(474 \mathrm{nm},\) pass simultaneously through a single slit whose width is \(7.15 \times\) \(10^{-5} \mathrm{m}\) and strike a screen \(1.20 \mathrm{m}\) from the slit. Two diffraction patterns are formed on the screen. What is the distance (in \(\mathrm{cm}\) ) between the common center of the diffraction patterns and the first occurrence of a dark fringe from one pattern falling on top of a dark fringe from the other pattern?

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?
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