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Chapter 35: Problem 17

Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fright fright fright fringe is at 4.84 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?

Short Answer

Expert verified
The wavelength is 400 nm.

Step by step solution

01

Understanding the Problem

We have been given the wavelength of coherent light, 600 nm, producing a first-order bright fringe at a specified distance on the screen. We need to find the wavelength that places a first-order dark fringe at the same position.
02

Identifying Key Formulas

For bright fringes: \(y = \frac{m \lambda D}{d} \), and for dark fringes: \(y = \frac{(m+\frac{1}{2}) \lambda' D}{d} \), where \(y\) is the fringe position, \(m\) is the order of the fringe, \(\lambda\) is the wavelength, \(\lambda'\) is the wavelength for the dark fringe, \(D\) is the distance from slits to screen, and \(d\) is the slit separation.
03

Setting Up the Equation for Bright Fringe

Using the formula for the first-order bright fringe, \(y = \frac{1 \cdot 600 \times 10^{-9} \cdot 3.00}{d} = 4.84 \times 10^{-3} \), we can solve for \(d\).
04

Finding the Slit Separation, d

From the bright fringe formula, \(d = \frac{(1 \cdot 600 \times 10^{-9} \cdot 3.00)}{4.84 \times 10^{-3}} = 3.72 \times 10^{-4} \) m. This value of \(d\) will be used in the equation for the dark fringe.
05

Equating Dark Fringe Formula

Set up the equation for the dark fringe at the same \(y\): \(4.84 \times 10^{-3} = \frac{(0 + \frac{1}{2}) \lambda' 3.00}{3.72 \times 10^{-4}} \).
06

Solving for New Wavelength \(\lambda'\)

Rearranging the equation gives \(\lambda' = \frac{4.84 \times 10^{-3} \times 3.72 \times 10^{-4}}{(0.5 \times 3.00)} \). Calculating gives \(\lambda' \approx 400\) nm.

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Key Concepts

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

Coherent Light
Coherent light is essential for creating clear interference patterns. This type of light consists of waves where the crests and troughs are aligned and maintain a constant phase relationship.
This means that the waves are synchronized, which is crucial for studying interference effects.
Coherent light, usually produced by lasers, ensures that the interference pattern is stable and observable over time.
Without coherence, the patterns would fluctuate or be weak, making them difficult to analyze.
  • Coherent light waves have consistent amplitude and frequency.
  • They produce sharp and distinct interference patterns.
  • Incoherent light, on the other hand, would create blurry or non-existent patterns.
Understanding this concept is key to the physics of light interference experiments.
Wavelength
Wavelength is the distance between consecutive crests (or troughs) of a wave.
It determines many of the wave's characteristics, including color for visible light, and plays a crucial role in interference patterns.
For the problem at hand, the initial wavelength is 600 nm, which corresponds to visible red light.
The wavelength affects where bright and dark fringes appear on a screen when light passes through slits.
  • Shorter wavelengths result in closer fringe spacing.
  • Longer wavelengths create wider gaps between fringes.
  • In our example, finding a new wavelength (around 400 nm) for the dark fringe involves re-calculating the interference conditions.
Wavelength directly influences interference patterns, linking distinct colors to specific fringe behaviors.
Interference Pattern
An interference pattern consists of alternating bright and dark fringes.
These patterns emerge when coherent light waves overlap and either reinforce or cancel each other out.
To form an interference pattern, the light passes through two narrow slits, allowing the waves to overlap on the other side.
  • Constructive interference occurs when waves align perfectly, causing bright fringes.
  • Destructive interference happens when crests meet troughs, resulting in dark fringes.
  • The pattern of fringes depends on the wavelength, slit distance, and distance to the screen.
Interference patterns provide invaluable insights into wave properties and the behavior of light.
Bright Fringe
A bright fringe is a region on the screen where light waves have undergone constructive interference.
This means their amplitudes add together, producing a reinforced or more intense light.
In our problem, the first-order bright fringe for 600 nm light appears at a specific screen position.
  • The position of bright fringes is determined by \( y = \frac{m \lambda D}{d} \), where \( m \) is the order of the fringe.
  • Brightness and spacing depend on the wave's wavelength and the experimental setup.
  • First-order fringes are typically the brightest and most easily distinguished.
Bright fringes exemplify the constructive interference aspect of wave physics.
Dark Fringe
A dark fringe occurs where destructive interference takes place.
Here, the waves are out of phase, leading crests to align with troughs, effectively cancelling each other out.
In the exercise scenario, the dark fringe needs to match the same location as the observed bright fringe but with a different wavelength.
  • The equation \( y = \frac{(m+\frac{1}{2}) \lambda' D}{d} \) helps find dark fringe positions.
  • Dark fringes mark areas of minimal light intensity.
  • These are crucial for validating interference models and calculations.
The formation of dark fringes highlights the principle of wave cancellation in interference patterns.

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Most popular questions from this chapter

Jan first uses a Michelson interferometer with the 606 -nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502 -nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

Two speakers that are 15.0 \(\mathrm{m}\) apart produce in-phase sound waves of frequency 250.0 \(\mathrm{Hz}\) in a room where the speed of sound is 340.0 \(\mathrm{m} / \mathrm{s} .\) A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

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?
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