1. Read the section in your textbook on double slit interference and diffraction (Chapter 36, Section 4). Suppose monochromatic light is incident on a barrier with two slits, as in Fig. 19. Each slit has width 11:, and the center-to-center separation of the slits is d. What happens to the number of bright fringes visible within the central diffraction enve10pe as d increases and w is xed? What happens to the number of bright fringes visible within the central diffraction envelope as w increases and d is xed? 2. Read the sectiOn in your textbook on diffraction gratings (Chapter 36, Section 5). Suppose monochro- matic light is incident on a diffractiOn grating. How does the width of a bright fringe observed on a distant screen depend on the number of illuminated slits N ? How will the pattern of bright fringes created by the grating be similar to that of a double-slit interference pattern? How will the patterns differ? A) Diffraction from a Reective Grating In this part of the experiment we shall use diffraction in a novel application to measure the wavelength of the light emitted by a helium-neon laser. The method is quite simple in principle and capable of accuracy exceeding 0.1% if care is taken in the measurements. If laser light is reected off of a metal ruler at a grazing angle (angle of incidence nearly 90 deg) as illustrated in Fig. 17, an interference pattern will be formed on the screen. Incident laser light Metal ruler Point where light hits if ruler is removed Figure 17: Schematic diagram of an experiment to measure the wavelength of laser light using a metal ruler (grating). The diagram is not to scale; in practice L >> 33m for all m. The markings on the ruler scatter the laser diffusely whereas there is specular reflection from the shiny metal between the markings. The net result is effectively a reflection type diffraction grating with distance between the adjacent grating lines equal to d, the distance between the fine markings on the ruler. A principal maximum will occur when the path difference is an integral number of wavelengths for rays reflected from the space between adjacent ruler markings. From Fig. 18 we see that the path difference is AD-CB. But AD=ABcos 0 and CB=ABcoso. Furthermore, AB is equal to the distance d between adjacent ruler markings. Thus from Fig. 17 and 18, we have for a principal maximum L L d cos 0 - d cos d = d = ma. (45) Co+ L2 2 + 12B) Interference from Single and Double Slits When a plane wavefront strikes two narrow slits, Spherical wavelets of the same wavelength as the wavefront Spread from each slit as shown in Fig. 19, and a characteristic pattern appears on the screen due to constructive and destructive interference. Some points, such as A, C, and E, are bright while others, Such as B and D, are dark. At a point such as A the wavelets from each slit are in phase and combine to reinforce each other and produce a bright Spot. Note that the light from slit 1 has traveled a distance 5A to get to point A, whereas the light from slit 2 has traveled 6A to reach the same point. The difference in the distance traveled along each path to point A is 1). The canditiOn for a point of mamimum brightness (i.e., constructive interference) is that the path difference be an integral number of wavelengths. Thus at point C the difference is 0A, at E 1)., at F 2A, and in general, nA for the nth bright band. The integer n is called the order of interference. Between the bright bands are the dark bands (points B and D), where the Plane wavefront of outgoing light incoming light Expanded view of adjacent marks on ruler Figure 18: Diffraction from a metal ruler illuminated by laser light at grazing incidence. The magnitudes of angles 5 and 3 are exaggerated for clarity. waves are completely out of phase; for these the path difference is (1/2)), (3/2)}, etc. It can be shown that for a double slit pattern the general condition for a point of maximum brightness is given by HR = dsin 6,.1 , (maximatwo slit interference} (46) where d is the distance between the slits and is 3\" is the corresponding angle shown in Fig. 19. The previous discussion assumed that all the rays passing through a given slit were in phase with each other, and this is true if the slit width is small compared with the wavelength. However, for a wider slit, one must take account of phase differences between the innitely many rays passing through various parts of the slit. If we consider the diffraction pattern formed by rays leaving the slit at an angle 6', and choose 6' so that /\\ = wsin 6', where w is the width of the slit, then we can show that the resultant amplitude of all the rays is zero. In this case each ray in the bottom half of the slit cancels out a corresponding ray in the upper half which is A/Z out of phase. Thus the general condition for a minimum brightness (a dark point) of a single slit diffraction pattern is ml = wsin 6m . [minimasingle slit diffraction] [47) Between these minima (but not exactly half-way}, the pattern has bright maxima. A special case occurs if m = 0 and 6m = 0: no path differences exist, all of the rays are in phase, and in this case they add up to give an intense central maximum. Notice that although Equations (46) and (47) have exactly the same mathematical form, their physical meanings are different. Second order First order Zeroth order C D E First order F Screen Second order Figure 19: Geometry for double-slit interference