2 Classroom Practice Problem The distance between the two slits is mm. Find the angles of the zeroth-, first-, second- and third-order bright fringes of interference produced by light with a wavelength of 550 nm.Answers: 0°, 6.3°, 13°, 19°For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.It is important that students use the same units for wavelength and distance between slits. It may be easiest to convert both to meters.
3 Classroom Practice Problem When monochromatic light (light of a single wavelength) falls on two slits with a separation of mm, the zeroth-order dark fringes are observed at a 2.0° angle. Find the wavelength.Answer: mm or 7.0 10-7 m or 7.0 102 nmFor problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.Students can calculate the wavelengths in mm but generally, wavelengths of light are expressed in nm. One possible approach is to convert the mm into meters to start the problem, then convert to nm afterward.
4 DiffractionA change in direction for waves when they encounter an obstacle or pass through an openingSound waves around a tree or through a doorHuygens’ principle states each point on a wave is a source for new waves.Diffraction occurs to a greater extent if the wavelength and opening size are appropriate.Openings much larger than the wavelength show very little diffraction.Light waves through a doorPoint out that diffraction requires an opening that is small enough for the wave. Since sound waves have wavelengths about times greater than those of light waves, the opening for light waves must be much smaller. Openings as large as 1 mm simply produce sharp shadows with little spreading out of the wave.Try the following web site to help students visualize the effect of changing the opening size and the wavelength. It will also allow them to see a single slit diffraction pattern.Go to “Easy Java Simulations,” then “Optics,” then choose either “Single slit diffraction” or “Single/multiple slit diffraction/interference.”If using the latter, set it for one slit and adjust parameters to see different patterns.The first simulator shows the slit size. You can adjust the width and the wavelength of the light to observe changes.
5 Single Slit Interference Occurs because one part of the wave interferes with the other partsThe diagram shows 5 sample points, each producing waves.The center line will be a point of constructive interference.Waves 1 and 5 travel the same distance to this line as do waves 2 and 4.The viewing screen is shown close but is actually distant. The waves angle toward the point in the center of the screen.
6 Single Slit Interference In order to reach a point above the center line, waves travel different distances.Wave 3 travels 1/2 farther than wave 1, so destructive interference occurs.Similarly, wave 5 travels 1/2 farther than wave 3.Thus, at the angle shown, destructive interference occurs and the screen is dark.Obviously wave 4 travels 1/2 farther than wave 2, and so on. The technique used is to divide the slit in half, and compare the distances traveled for corresponding points in each half.
7 Diffraction Around Obstacles Suppose a laser shines on a screen. The round head of a pin (like those used to pin clothing) is placed between the laser and screen. You would expect to see a round shadow on the screen.What will you see in the center of the shadow? Why?You will see a bright spot if you look very carefully. The waves diffract and all waves travel the same distance to the center of the shadow, creating constructive interference.This is a relatively easy demonstration to set up. You might need to try different sizes of pin heads (the “pearl” heads seem to work well). It is best if the room is dark and the students’ eyes have some time to adjust. They will need to be close to the screen to see the bright spot. They may even be able to see bright rings showing other path length differences that lead to constructive interference.
8 Diffraction Gratings Many slits very closely spaced Behaves like the double slit, but maxima and minima are much brighter.Monochromatic light produces bright and dark fringes.White light produces a full spectrum.Similar to reflection off a CD surface
9 Diffraction GratingsDiffraction gratings are used in spectrometers to separate light into its component colors.Used to study the makeup of distant starsHelium’s spectrum was first observed on the sun, and helium was later discovered on Earth.One of the spectral lines for helium (Helios was a Greek god of the sun) was observed during an eclipse of the sun. This did not match any spectral line for known elements. Helium was later discovered on Earth.
10 Function of a Spectrometer Click below to watch the Visual Concept.Visual Concept
11 Problem Solving - Diffraction Gratings Equations are identical to those for two slit interference.d is the distance between adjacent slits or lines.If there are 8000 lines per cm, then d = (1/8000) cm
12 Instrument Resolution If observing two distant objects, the diffraction patterns could overlap as shown.Without diffraction, there would be two bright spots on the screen.Try the following web site to help students visualize the resolving power of circular openings.Go to “Easy Java Simulations,” then “Optics,” then choose “Resolution of circular apertures.”With this simulation you can change the size of the opening, the distance to the sources, or the distance between the sources.
13 Resolving PowerThe wavelength () and the opening size (D) determine the resolving power. is the limiting angle between the two resolved objects measured in radians.Longer wavelengths require a larger aperture (D) to resolve distant objects.Because radio waves are long waves, radio telescopes are very large to accommodate the need for a large aperture.
14 Instrument Resolution Imagine the opening is the pupil of your eye and the two sources are adjacent red and green pixels on your television screen.What will you see?How would it change if your pupil opening was larger?How would it change if you were closer?How would it change if the pixels were farther apart?These questions are designed to help students apply resolving power to the eye. There are several web sites that provide information about the theoretical resolving power (see the equation on the previous slide) and the actual resolving power due to imperfections in the eye, astigmatism, and the ability of the retina to create an image.If the eye is as shown, students will see yellow because the two images overlap extensively (red + green = yellow). All the following questions would lead to greater resolution, making it easier to see the individual pixels.
15 Practice ProblemsA new astronomical facility is set for completion on Mount Wilson, in Georgia, by Named the Center for High Angular Resolution Astronomy (CHARA) Array, it comprises five telescopes whose individual images will be analyzed by computer. The resulting data will then be used to form a single high-resolution image of distant galaxies. The technique is similar to what is currently being done with radio telescopes. The CHARA Array will be used to observe radiation in the infrared portion of the spectrum with wavelengths as short as 2200 nm. Suppose 2200 nm light passes through a diffraction grating with 64 103 lines/m and produces a fringe at an angle of 34.0. What order maximum will the fringe be?
16 Practice ProblemsA new technology to improve the image quality of large-screen televisions has been developed recently. The entire screen would consist of tiny cells, each equipped with a movable mirror. The mirrors would change positions relative to the incoming signal, providing a brighter image. The entire screen would look like a diffraction grating with lines/m. Imagine shining red light with a wavelength of 750 nm onto this grating. What order maximum would be observed at an angle of 48.6?
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