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Chapter 9: Problem 2

A double slit of slit separation \(0.5 \mathrm{~mm}\) is illuminated by a parallel beam from a helium-neon laser that emits monochromatic light of wavelength \(6328 \AA\). Five meters beyond the slits is a screen. What is the separation of the interference fringes on the screen?

Short Answer

Expert verified
The separation of the interference fringes is 6.328 mm.

Step by step solution

01

Understand the Problem

We need to calculate the separation between the interference fringes formed on a screen when light passes through a double slit within certain geometric and light conditions. We know the slit separation, the wavelength of light, and the distance to the screen.
02

Recall the Formula

The formula for the fringe separation (also known as fringe width) in a double-slit experiment is given by \( \Delta y = \frac{\lambda L}{d} \), where \( \lambda \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the slit separation.
03

Convert Units

Convert the wavelength from Angstroms to meters. Since \(1 \text{ 脜} = 10^{-10} \text{ m}\), \( 6328 \text{ 脜} = 6328 \times 10^{-10} \text{ m} = 6.328 \times 10^{-7} \text{ m} \).
04

Plug Values into the Formula

Now substitute the given values into the formula: \( \Delta y = \frac{6.328 \times 10^{-7} \text{ m} \times 5 \text{ m}}{0.5 \times 10^{-3} \text{ m}} \).
05

Solve for Fringe Separation

Calculate \( \Delta y = \frac{6.328 \times 10^{-7} \times 5}{0.5 \times 10^{-3}} \). Simplifying gives \( \Delta y = \frac{3.164 \times 10^{-6}}{0.5 \times 10^{-3}} = 6.328 \times 10^{-3} \text{ m} \).
06

Interpretation of Result

The calculated fringe separation is \(6.328 \times 10^{-3} \text{ m} \), which translates to about \(6.328 \text{ mm} \). This is the distance between adjacent bright or dark fringes on the screen.

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Key Concepts

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

Interference fringes
In the fascinating world of light and optics, interference fringes are crucial patterns that arise when light waves overlap and interact with each other. This phenomenon occurs in the double-slit experiment, as beams of light pass through two closely spaced slits and hit a screen. On the screen, light waves from each slit overlap, creating patterns of bright and dark bands known as interference fringes.

These fringes appear due to the constructive and destructive interference of light waves coming from the slits.
  • Bright fringes occur where the light waves are in phase and add up, reinforcing each other.
  • Dark fringes appear where the waves are out of phase and cancel each other out.
This alternating pattern of light and dark is what we see as interference fringes, and understanding how they form helps us delve deeper into wave behaviors of light.
Wavelength conversion
When dealing with problems in physics, units of measurement need to be consistent, especially when solving equations. In the context of the double-slit experiment, initial light measurements are often given in Angstroms (\( \text{脜} \)), a common unit for expressing wavelengths of light.

Since most calculations require metric units like meters, it's essential to convert Angstroms to meters. Here's how you do it:
  • Recognize that 1 Angstrom is equal to \( 10^{-10} \) meters.
  • Multiply the wavelength by this conversion factor. For instance, light with a wavelength of 6328 脜 translates to 6328 脳 \( 10^{-10} \) meters, which results in 6.328 脳 \( 10^{-7} \) meters.
This conversion is vital for ensuring accuracy and consistency in calculations related to light properties.
Fringe separation calculation
Calculating the separation between interference fringes in a double-slit experiment might seem daunting at first, but it's quite simple once you know the right formula. The formula to find fringe separation is:\[\Delta y = \frac{\lambda L}{d}\]where:
  • \( \Delta y \) is the fringe separation.
  • \( \lambda \) stands for the wavelength of the light used.
  • \( L \) is the distance from the slits to the screen.
  • \( d \) is the slit separation distance.
Substitute the known values for these variables to compute \( \Delta y \). Using our example, substitute the converted wavelength, the given slit separation, and the distance to the screen into the formula. After performing the arithmetic, you find that the fringe separation is approximately 6.328 mm. Understanding and applying this formula allows you to interpret the pattern of interference fringes accurately.
Monochromatic light
Monochromatic light is light with a single wavelength or color. In our exercise, the light source is a helium-neon laser emitting light at a wavelength of 6328 脜. This kind of light source is excellent for interference experiments because it ensures consistency in the fringe patterns formed.

This uniformity in wavelength is key to producing clear and predictable interference patterns.
  • Monochromatic sources provide consistent phase relationships between waves.
  • This helps achieve distinct and regular interference fringes on the screen.
  • Such sources are particularly useful in precision optical experiments, like the double-slit experiment.
The precision and stability of monochromatic light allow researchers and students alike to explore the wave properties of light in a controlled setting.

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Most popular questions from this chapter

When you put your face under water and try looking without a face mask, everything looks blurred, because the change of index of refraction in going from water to eye is not very great. As a simplification, assume there is no change in index. Also assume your eye lens has very little effect, as if all the focusing were done at the first air-to-eye interface. (This is a crude approximation. Actually, you can see underwater to same extent.) Assume the focal length of that first surface is \(3 \mathrm{~cm}\), and that a parallel beam of light in air is brought to a focus at the retina. When you look underwater, you lose that focusing action. Design glasses that can be worn underwater so as to enable you to see clearly. Use glass with index of refraction 1.5. Show that if the focal length when used underwater is \(3 \mathrm{~cm}\), then the focal length when used in air is about \(1 \mathrm{~cm} .\) If one of these glass lenses is used as an ordinary magnifying glass what is its magnification? Suppose you use an ordinary glass marble for the lens. You want to form an image (of a parallel beam in water) \(3 \mathrm{~cm}\) behind the rear surface of the marble. What should be the diameter of the marble?

Show that a plane wave normally incident on one face of a wedge-shaped prism of angle \(A\) is deviated by an amout \(\theta_{\text {dev }}\), where $$ n \sin A=\sin \left(A+\theta_{\text {dev }}\right) \text { . } $$

Pour some table salt on a wet knife or spoon (one that you don't mind ruining). Set the knife in the flame of a gas stove. Look at the yellow flame through your diffraction grating (this is easiest at night in a darkened room). Notice that the first-order (and higher-order) images of the yellow sodium flame are as sharp and clear as the zeroth-order "direct" image. That is because the yellow light is a "spectral line" having narrow bandwidth. (Actually the yellow light from sodium is a "doublet" of two lines with wavelengths 5890 and \(5896 \AA\).) Now look at a candle. In zeroth order, it does not look terribly different from the sodium flame; they are both yellow. But in the first-order diffraction image, the candle is very much spread out in color, whereas the sodium remains sharp. The "yellow" of the candle, which is due to hot particles of carbon, has a wavelength spectrum extending over (and beyond) the entire visible range. Here are other convenient sources of sharp spectral lines; look at them through your grating: Mercury vapor: Fluorescent lamps, mercury-vapor street lights, sunlamps. (A sunlamp is convenient in that it screws directly into an ordinary 110 -volt AC socket. It is probably the cheapest source of mercury-vapor spectral lines; the cost is about \(\$ 10 .\).) Neon: Many advertising signs. Neon has a profusion of lines; you see "many signs." A cheap broad monochromatic source is a G.E. bulb NE-34 which screws directly into a 110 -volt AC socket (the cost is about \(\$ 1.60)\). Others are a "circuit continuity tester," which plugs into any wall receptacle and which costs about \(\$ 1\) (at a hardware store), and a neon "night light." Strontium: Strontium chloride salt (available at a chemical supply house for about 25 cents \(/ \mathrm{oz}\) ); dissolve a little in a few drops of water and put it in the gas flame on your ruined spoon. The wavelength of the red line is a famous length standard. Copper: Copper sulfate; availability and technique as for strontium chloride. It gives a beautiful green color. Hydrocarbon: Look at your gas flame in the first-order spectrum. There are a sharp, clear blue image and a sharp, clear green image. The "blue" color of the flame is therefore due to one or more almost monochromatic spectral lines.

Which side has the scratches? One side of the plastic of your diffraction grating is smooth; the other side has the scratches. You can find out which side has the scratches by looking through it at a white source after rubbing one side of the grating with an oily finger; then clean it and try the other side. What is the explanation?

The world's cheapest broad, almost monochro- matic light source is obtained by burning a wad of toilet paper. You can use this source to see Fabry-Perot fringes. Burn the paper. (The room should be dark-perhaps also you should have some water handy!) Look through the flame at the image of the flame at near- normal incidence in a piece of glass-a microscope slide or picture- frame glass. You will see finger-print-like fringes. If the glass is optically flat, the fringes will be circles centered on your eyeballs; in any case, you can see them easily. If you have a gas stove or bunsen burner, you can get a brighter monochromatic sodium source by sprinkling salt on a wet knife and immersing it in the flame. Then you can see the Fabry-Perot fringes even in the daytime. For a nice, steady, broad monochromatic source with which to look at the fringes, use the G.E. neon bulb NE-34.
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