Physics for Scientists and Engineers, 6e Chapter 38 - Diffraction Patterns and Polarization

Suppose the slit width in the figure below is made half as wide Suppose the slit width in the figure below is made half as wide. The central bright fringe becomes wider remains the same becomes narrower 1 2 3 4 5

Equation 38.1 shows that a decrease in a results in an increase in the angles at which the dark fringes appear.

If a classroom door is open slightly, you can hear sounds coming from the hallway. Yet you cannot see what is happening in the hallway. Why is there this difference? Light waves do not diffract through the single slit of the open doorway. Sound waves can pass through the walls, but light waves cannot. The open door is a small slit for sound waves, but a large slit for light waves. The open door is a large slit for sound waves, but a small slit for light waves. 1 2 3 4 5

The space between the slightly open door and the doorframe acts as a single slit. Sound waves have wavelengths that are larger than the opening and so are diffracted and spread throughout the room you are in. Because light wavelengths are much smaller than the slit width, they experience negligible diffraction. As a result, you must have a direct line of sight to detect the light waves.

Consider the central peak in the diffraction envelope in the figure of question 3. Suppose the wavelength of the light is changed to 450 nm. What happens to this central peak? The width of the peak decreases and the number of interference fringes it encloses decreases. The width of the peak decreases and the number of interference fringes it encloses increases. The width of the peak decreases and the number of interference fringes it encloses remains the same. The width of the peak increases and the number of interference fringes it encloses decreases. The width of the peak increases and the number of interference fringes it encloses increases. The width of the peak increases and the number of interference fringes it encloses remains the same. 1 2 3 4 5

In Equation 38.7, the ratio d/a is independent of wavelength, so the number of interference fringes in the central diffraction pattern peak remains the same. Equation 38.1 tells us that a decrease in wavelength causes a decrease in the width of the central peak.

headlights on a distant car Cat’s eyes have pupils that can be modeled as vertical slits. At night, cats would be more successful in resolving headlights on a distant car vertically-separated lights on the mast of a distant boat 1 2 3 4 5

The effective slit width in the vertical direction of the cat’s eye is larger than that in the horizontal direction. Thus, the eye has more resolving power for lights separated in the vertical direction and would be more effective at resolving the mast lights on the boat.

Suppose you are observing a binary star with a telescope and are having difficulty resolving the two stars. You decide to use a colored filter to maximize the resolution. (A filter of a given color transmits only that color of light.) What color filter should you choose? blue green yellow red 1 2 3 4 5

We would like to reduce the minimum angular separation for two objects below the angle subtended by the two stars in the binary system. We can do that by reducing the wavelength of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing the resolving power. Thus, we should choose a blue filter.

If laser light is reflected from a phonograph record or a compact disc, a diffraction pattern appears. This is due to the fact that both devices contain parallel tracks of information that act as a reflection diffraction grating. Which device results in diffraction maxima that are farther apart in angle? record compact disc 1 2 3 4 5

The tracks of information on a compact disc are much closer together than on a phonograph record. As a result, the diffraction maxima from the compact disc will be farther apart than those from the record.

parallel perpendicular A polarizer for microwaves can be made as a grid of parallel metal wires about a centimeter apart. The electric field vector for microwaves transmitted through this polarizer is _____ to the metal wires. parallel perpendicular 1 2 3 4 5

Electric field vectors parallel to the metal wires cause electrons in the metal to oscillate parallel to the wires. Thus, the energy from the waves with these electric field vectors is transferred to the metal by accelerating these electrons and is eventually transformed to internal energy through the resistance of the metal. Waves with electric-field vectors perpendicular to the metal wires pass through because they are not able to accelerate electrons in the wires.

at an intermediate distance from you You are walking down a long hallway that has many light fixtures in the ceiling and a very shiny, newly waxed floor. In the floor, you see reflections of every light fixture. Now you put on sunglasses that are polarized. Some of the reflections of the light fixtures can no longer be seen. (Try this!) The reflections that disappear are those nearest to you farthest from you at an intermediate distance from you 1 2 3 4 5

At some intermediate distance, the light rays from the fixtures will strike the floor at Brewster’s angle and reflect to your eyes. Because this light is polarized horizontally, it will not pass through your polarized sunglasses. Tilting your head to the side will cause the reflections to reappear.