ism_ch28 - Chapter 28 Physical Optics: Interference and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
211 Chapter 28 Physical Optics: Interference and Diffraction Answers to Even-numbered Conceptual Questions 2. If the slit spacing, d , were less than the wavelength, λ , the condition for a bright fringe (Equation 28-1) could be satisfied only for the central bright fringe ( m = 0). For nonzero values of m there are no solutions, because sin θ cannot be greater than one. In addition, Equation 28-2 shows that if d is greater than /2, though still less than , there will be only one dark fringe on either side of the central bright fringe. If d is less than /2, no dark fringes will be observed. 4. The locations of bright and dark fringes depends on the wavelength of light. Therefore, if white light is used in a two-slit experiment, each bright fringe will show some separation into colors, giving a “rainbow” effect. 6. Submerging the two-slit experiment in water would reduce the wavelength of the light from to / n , where n = 1.33 is the index of refraction of water. Therefore, the angles to all the bright fringes would be reduced, as we can see from Equation 28-1. It follows that the two-slit pattern of bright fringes would be more tightly spaced in this case. 8. The soap film in the photograph is thinnest near the top (as one might expect) because in that region the film appears black. Specifically, light reflected from the front surface of the film has its phase changed by 180 ˚ ; light that reflects from the back surface of the film has no change in phase. Therefore, light from the front and back surfaces of the film will undergo destructive interference as the path length between the surfaces goes to zero. This is why the top of the film, where the film is thinnest, appears black in the photo. 10. One possible reason is that one of the films may have an index of refraction greater than that of glass, whereas the other may have an index of refraction that is less than that of glass. If this is the case, the phase change in reflection from the film-glass interface will be different for the two films. This, in turn, would result in different colors appearing in the reflected light. 12. Light reflected from the top of the film has a phase change of 180 ˚ ; light reflected from the film-water surface also has a phase change of 180 ˚ , since the film’s index of refraction is less than that of water. It follows that the film appears bright (constructive interference) where the film’s thickness goes to zero. 14. The location of dark fringes in a single-slit diffraction pattern is given by Equation 28-12. Notice that if W is decreased, the angle must increase to compensate, and to maintain the equality. This is why the dark fringes move outward. As the angle approaches 90 ˚ for m = ±1, all of the dark fringes have moved outward to infinity. Clearly, this occurs in Equation 28-12 when the width W is equal to the wavelength, .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/08/2008 for the course PHYS 104 taught by Professor Pengyi during the Fall '08 term at UNC.

Page1 / 28

ism_ch28 - Chapter 28 Physical Optics: Interference and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online