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Chapter 35: Problem 11

35.11. Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except light of a single wavelength. 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 wavelength?

Short Answer

Expert verified
The wavelength of the light is 589 nm.

Step by step solution

01

Understanding the Setup

The coherent light passes through two slits separated by a distance (\(d\)) and creates an interference pattern on a screen that is a certain distance (\(L\)) away from the slits. The bright fringes in this pattern are separated by a distance (\(\Delta y\)). We need to find the wavelength (\(\lambda\)) of the light.
02

Identify the Formula

For this double-slit experiment, the formula for the separation between adjacent bright fringes (\(\Delta y\)) is given by:\[\Delta y = \frac{\lambda L}{d}\]where:- \(\lambda\) is the wavelength of the light,- \(L = 2.20 \, \mathrm{m}\) is the distance from the slits to the screen,- \(d = 0.460 \, \mathrm{mm} = 0.460 \times 10^{-3} \, \mathrm{m}\) is the separation between the slits,- \(\Delta y = 2.82 \, \mathrm{mm} = 2.82 \times 10^{-3} \, \mathrm{m}\) is the separation between adjacent bright fringes.
03

Solve for the Wavelength

Rearrange the formula to solve for the wavelength (\(\lambda\)):\[\lambda = \frac{\Delta y \cdot d}{L}\]Substitute the known values:\[\lambda = \frac{2.82 \times 10^{-3} \, \mathrm{m} \cdot 0.460 \times 10^{-3} \, \mathrm{m}}{2.20 \, \mathrm{m}}\]
04

Calculate the Wavelength

Now compute \(\lambda\):\[\lambda = \frac{2.82 \times 10^{-3} \times 0.460 \times 10^{-3}}{2.20}\]\[\lambda \approx 5.89 \times 10^{-7} \, \mathrm{m}\]
05

Convert to Nanometers

Since wavelengths of light are often expressed in nanometers (nm), convert the wavelength from meters to nanometers:\[\lambda = 5.89 \times 10^{-7} \, \mathrm{m} = 589 \, \mathrm{nm}\]

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Key Concepts

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

Wavelength Calculation
Calculating the wavelength of light in a double-slit experiment is crucial for understanding the characteristics of the light used. In this context, the sodium-vapor lamp emits coherent light of a single wavelength, which we need to determine. The formula used here is derived from the interference pattern created by light passing through two slits and is expressed as:\[\lambda = \frac{\Delta y \cdot d}{L}\]The variables in the equation are:
  • \(\Delta y\): the separation between adjacent bright fringes
  • \(d\): the distance between the two slits
  • \(L\): the distance from the slits to the screen where the pattern is observed
To find the wavelength \(\lambda\), you multiply the fringe separation \(\Delta y\) by the slit separation \(d\), then divide by the distance \(L\) from the slits to the screen. This calculation tells us how long each wave is from peak to peak, which we then convert from meters to nanometers鈥攃ommonly used for light wavelengths. In our scenario, the calculated wavelength is 589 nm, fitting into the visible spectrum.
Double-Slit Experiment
The double-slit experiment, originally conducted by Thomas Young, is a powerful demonstration of the wave nature of light. It shows how light can interfere with itself, creating a pattern of bright and dark bands known as fringes. When monochromatic light, like that from our sodium-vapor lamp, passes through two closely spaced slits, it spreads out and overlaps. This overlap causes interference, where:
  • Constructive interference occurs where waves peak together, forming bright fringes.
  • Destructive interference happens where peaks meet troughs, leading to dark fringes.
The distance between these fringes provides valuable information about the light's wavelength. By analyzing this interference pattern, scientists confirm that light behaves as a wave, explaining phenomena like diffraction and other wave-like behaviors.
Fringe Separation
Fringe separation is a critical concept in understanding interference patterns produced during the double-slit experiment. It refers to the distance between consecutive bright or dark bands on the screen.In mathematical terms, the fringe separation \(\Delta y\) in the interference pattern can be expressed as:\[\Delta y = \frac{\lambda L}{d}\]Where:
  • \(\lambda\) is the wavelength of the light.
  • \(L\) is the distance from the slits to the screen.
  • \(d\) is the distance between the slits.
Understanding fringe separation helps in determining not just the wavelength, but also validates the wave properties of light. The spacing between fringes reveals how coherent the light source is, as well as the geometry of the slit setup. By fine-tuning these values, we can produce precise interference patterns, valuable in various applications like the design of lenses and other optical technologies.

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Most popular questions from this chapter

35.51. 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.)

35.61. The index of refraction of a glass rod is 1.48 at \(T=20.0^{\circ} \mathrm{C}\) and varies linearly with temperature, with a coefficient of \(2.50 \times 10^{-5} / \mathrm{C}^{\circ} .\) The coefficient of linear expansion of the glass is \(5.00 \times 10^{-6} / \mathrm{C}^{\circ}\) . At \(20.0^{\circ} \mathrm{C}\) the length of the rod is \(3.00 \mathrm{cm} . \mathrm{A}\) Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 \(\mathrm{C}^{\circ} / \mathrm{min}\) . The light source has wavelength \(\lambda=589 \mathrm{nm},\) and the rod initially is at \(T=20.0^{\circ} \mathrm{C}\) . How many fringes cross the field of view each minute?

35.31. A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 \(\mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO_ 2 film.

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ? )\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

35.39. The radius of curvature of the convex surface of a planoconvex lens is 95.2 \(\mathrm{cm}\) . The lens is placed convex side down on a perfectly flat glass plate that is illuminated from above with red light having a wavelength of 580 \(\mathrm{nm}\) . Find the diameter of the second bright ring in the interference pattern.
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