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Chapter 35: Problem 10

Coherent light with wavelength 450 nm falls on a double slit. On a screen 1.80 \(\mathrm{m}\) away, the distance between dark fringes is 4.20 \(\mathrm{mm} .\) What is the separation of the slits?

Short Answer

Expert verified
The separation of the slits is approximately 193 碌m.

Step by step solution

01

Introduction to the Problem

We have coherent light with wavelength \( \lambda = 450 \text{ nm} \) (nanometers), which falls on a double slit, producing an interference pattern on a screen. The distance between consecutive dark fringes on the screen is given as 4.20 mm, and the distance from the slits to the screen is 1.80 m. We need to find the separation between the slits, denoted as \( d \).
02

Understanding Fringe Spacing

The distance between dark fringes in a double slit experiment is given by the formula: \[ \Delta y = \frac{\lambda L}{d} \] where \( \Delta y \) is the distance between two consecutive dark fringes, \( \lambda \) is the wavelength of the light, \( L \) is the distance from the slits to the screen, and \( d \) is the separation of the slits.
03

Solving for Slit Separation

We can rearrange the formula to solve for \( d \): \[ d = \frac{\lambda L}{\Delta y} \] Substitute the given values into this formula: \( \lambda = 450 \text{ nm} = 450 \times 10^{-9} \text{ m} \), \( L = 1.80 \text{ m} \), and \( \Delta y = 4.20 \text{ mm} = 4.20 \times 10^{-3} \text{ m} \).
04

Calculation of Slit Separation

Plug the values into the equation: \[ d = \frac{450 \times 10^{-9} \times 1.80}{4.20 \times 10^{-3}} \]First, calculate the product in the numerator:\[ 450 \times 10^{-9} \times 1.80 = 810 \times 10^{-9} \]Then divide by \( 4.20 \times 10^{-3} \):\[ d = \frac{810 \times 10^{-9}}{4.20 \times 10^{-3}} \approx 1.93 \times 10^{-4} \text{ m} \].
05

Converting to Micrometers

The resultant slit separation \( d \) is in meters. To make it more comprehensible, convert it to micrometers (1 m = 1,000,000 碌m):\[ d = 1.93 \times 10^{-4} \times 1,000,000 = 193 \text{ 碌m} \]

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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 crucial for the double slit experiment. Coherence refers to the waves being in phase with each other. This means they have a constant phase difference. To visualize this, imagine waves of light traveling side by side, never differing in their peaks and troughs as they move.

For a double slit experiment to work effectively, coherence ensures the interference of the light waves can occur predictably. This creates a clear and steady interference pattern of alternating bright and dark fringes on the screen.
  • Coherent waves originated from a single source.
  • Important for producing discernible patterns in experiments.
Understanding coherence helps grasp why only certain patterns occur during the experiment.
Wavelength
Wavelength is a measure of the distance between consecutive peaks of a wave. In this context, it is represented by the Greek letter \( \lambda \). For the double slit experiment, we used a wavelength of 450 nm, which is in the visible light spectrum, giving it a bluish color.
The wavelength determines the spacing and nature of the fringes seen in the interference pattern:
  • Shorter wavelengths lead to closer fringe spacing.
  • Longer wavelengths spread the fringes further apart.
In our problem, the wavelength was key to calculating how the light interacted with the slits, ultimately determining the fringe pattern on the screen.
Interference Pattern
An interference pattern arises when waves overlap and combine. In the double slit experiment, this combination is due to the coherent light passing through two slits. The light waves coming from each slit interfere with each other, creating regions of constructive and destructive interference.

Constructive interference leads to bright fringes, where waves peak together, amplifying their brightness. Conversely, destructive interference results in dark fringes, where the peak of one wave meets the trough of another, canceling each other out. This alternating sequence of bright and dark bands forms the characteristic interference pattern.
  • Bright fringes = Constructive interference.
  • Dark fringes = Destructive interference.
By studying the position of these fringes, as seen in experiments, we can learn about the properties of waves and how different factors influence wave behavior.
Fringe Spacing
Fringe spacing is the distance between two consecutive similar fringes, such as bright to bright or dark to dark. This is crucial for calculating any unknowns in the double slit experiment, like the separation of the slits.
In the exercise, the formula \( \Delta y = \frac{\lambda L}{d} \) was used, where:
  • \( \Delta y \) is fringe spacing, 4.20 mm in this case.
  • \( \lambda \) is the light's wavelength, 450 nm.
  • \( L \) is the distance from slits to the screen, 1.80 m.
  • \( d \) is the slit separation.
Substituting these values helps pinpoint the space between the slits, calculated here as 193 碌m. Understanding fringe spacing helps interpret the results and make calculations in wave-based experiments.

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Most popular questions from this chapter

A source \(S\) of monochromatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass and are separated by a horizontal distance \(x .\) Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q ?\)

One round face of a 3.25 -m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ},\) what is the intensity? (b) What is the path difference for 480 -nm light from the two slits at a point where the phase angle is \(60.0^{\circ} ?\)

Two speakers, 2.50 \(\mathrm{m}\) apart, are driven by the same audio oscillator so that each one produces a sound consisting of two distinct frequencies, 0.900 \(\mathrm{kHz}\) and 1.20 \(\mathrm{kHz}\) . The speed of sound in the room is 344 \(\mathrm{m} / \mathrm{s}\) . Find all the angles relative to the usual centerline in front of (and far from) the speakers at which both frequencies interfere constructively.
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