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\(\bullet\) Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wave- length. It then falls on two slits separated by 0.460 \(\mathrm{mm} .\) In the resulting interference pattern on a screen 2.20 \(\mathrm{m}\) away, adjacent bright fringes are separated by 2.82 \(\mathrm{mm}\) . What is the wave- length of the light that falls on the slits?

Short Answer

Expert verified
The wavelength of the light is 589 nm.

Step by step solution

01

Understanding the Problem

We have light from a sodium-vapor lamp passing through two slits separated by 0.460 mm. The light forms an interference pattern on a screen 2.20 m away, and the distance between adjacent bright fringes is 2.82 mm. We need to find the wavelength of the light.
02

Applying Interference Formula

For a double-slit interference pattern, the path difference between the two beams of light is related to the bright fringe separation. The formula is given by \( y = \frac{\lambda L}{d} \), where \( y \) is the fringe separation, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the slit separation.
03

Substitute Known Values

We are given \( y = 2.82 \times 10^{-3} \: \mathrm{m} \), \( L = 2.20 \: \mathrm{m} \), and \( d = 0.460 \times 10^{-3} \: \mathrm{m} \). Substitute these values into the equation: \[ 2.82 \times 10^{-3} = \frac{\lambda \times 2.20}{0.460 \times 10^{-3}} \]
04

Solve for Wavelength

Rearrange the equation to solve for \( \lambda \): \[ \lambda = \frac{2.82 \times 0.460 \times 10^{-3}}{2.20}\] Calculate this expression to find the wavelength.
05

Calculating the Wavelength

Performing the calculation in the previous step, we find: \( \lambda = \frac{2.82 \times 0.460}{2.20} \times 10^{-3} = 0.589 \times 10^{-6} = 589 \: \mathrm{nm} \).

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Key Concepts

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

Double-Slit Experiment
The double-slit experiment is a fundamental demonstration of the wave nature of light. Imagine a light source that emits waves of light that travel in straight lines. These waves encounter a barrier with two small slits cut into it.
As the light passes through these slits, something interesting happens. Instead of continuing straight, the waves spread out and overlap on the other side.
This overlapping is known as interference. When the light waves intersect, they can either amplify each other, creating brighter regions called fringes or cancel each other out, creating darker regions. These patterns are projected onto a screen to form an interference pattern.
  • Constructive interference results in bright fringes where waves crest and trough align.
  • Destructive interference occurs when a crest meets a trough resulting in dark fringes.
Understanding this process is crucial as it illustrates how wave behavior, such as interference, provides evidence of the wave properties of light.
Fringe Separation
Fringe separation refers to the distance between successive bright or dark lines in the interference pattern created during a double-slit experiment. This separation is not random but instead relates closely to the properties of the light and the setup of the experiment. It is a key measurement because it gives insights into the wave characteristics of light.
In an experimental setup, fringe separation can be influenced by a few factors:
  • Distance from slits to screen: A greater distance results in larger fringe separations since the waves have more space to diverge and overlap.
  • Slit separation: Smaller slit separation can lead to wider-spaced fringes because the paths over which interference occurs are different.
  • Wavelength of light: Longer wavelengths produce wider separations between fringes.
Knowing the fringe separation allows the calculation of the wavelength when other parameters are known, which is pivotal in experiments where direct wavelength measurement is challenging.
Wavelength Calculation
Wavelength calculation in the context of the double-slit experiment involves using the interference pattern to find the light's wavelength. The relationship between the wavelength and fringe separation helps understand the wave properties of the involved light.
The formula used is \[ y = \frac{\lambda L}{d} \] where:
  • \( y \) is fringe separation.
  • \( \lambda \) is the wavelength of the light.
  • \( L \) is the distance from the slits to the screen.
  • \( d \) is the separation between the slits.
For solving, one would rearrange the equation to derive the wavelength as: \[ \lambda = \frac{y \, d}{L} \] Substitute the known values into the equation and calculate to find the light's precise wavelength. This method is fundamental in optics for measuring wavelengths accurately, providing a window into the characteristics of light that cannot be seen directly but can be inferred from the behavior of its wave interactions.

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Most popular questions from this chapter

\(\bullet\) The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of 525 \(\mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

\(\bullet\) A laser beam of unknown wavelength passes through a dif- fraction grating having 5510 lines/cm after striking it perpen- dicularly. Taking measurements, you find that the first pair of bright spots away from the central maximum occurs at \(\pm 15.4^{\circ}\) with respect to the original direction of the beam. (a) What is the wavelength of the light? (b) At what angle will the next pair of bright spots occur?

\(\bullet\) If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

\(\bullet\) Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)

\(\bullet\) A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a sec- ond plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interfer- ence fringes are observed per centimeter in the reflected light?

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