Homework #9
Conceptual Solution

1. (b) The central fringe, which is brightest, is at , the same distance from both slits.
2. (a) One slit is 300 nm, or one half-wavelength, closer. Thus this is the first point of destructive interference, or the first dark fringe, which is A.
3. (e) Fringe B is the central maximum, at . Fringe E is the third order maximum (m=3), which occurs for a path difference

4. (c) If the wavelength of light is decreased, the angle to the m-th order bright fringe decreases:

   (Note that the zeroth order fringe at remains fixed.) Similarly, the angle to the m-th dark fringe decreases

   The angular separation between all features is thus being reduced, with the bright fringes coming closer together, and the dark fringes coming closer together. Answer (c) best captures this.

5. (c) The top of slide A is nonreflective because of its coating: light reflecting from the top of the coating destructively interferes with light reflecting from the bottom of the coating. The layers of optical material are air (n=1) then coating (n=1.4) then glass (n=1.5). Thus light reflecting at the air-coating interface, and light reflecting at the coating-glass interface, are both reflecting from a larger index of refraction, and so both have the same phase shift, an effective path shift of . These phase shifts cancel, so that the net path difference between the 2 rays is just the physical path length difference of 2t, where t is the thickness of the coating. The smallest t for which this path length difference will give destructive interference occurs when . However, since the path difference occurs in the thin film, the appropriate is the wavelength in the thin film, . Thus

   

6. (e) In this situation, light reflecting from the bottom of slide A and light reflecting from the top of slide B interfere (the other 2 surfaces are non-reflecting). Because the 2 slides are parallel, these 2 paths have the same difference in path length at every point on the slide. Thus, whether interference is constructive, destructive, or in between, it is the same, giving the same intensity (or brightness), at every point on the slide. This eliminates choices (a) and (b). As slide A is moved up, the path length difference between the 2 light waves increases. At it increases, it passes through a point of constructive interference (where the path difference is ), then a little later through a point of destructive interference (with path difference ). This cycle of constructive then destructive interference repeats many times as we raise slide A, so that we see a uniform field which brightens (constructive interference) and darkens (destructive interference) as we raise slide A.
7. (b) Again light reflecting from the bottom of slide A and from the top of slide B interfere. Here, though, the thickness of the air wedge -- determining the path length difference of the 2 reflected rays -- varies as we move left to right, so that we go through the cycles of constructive and destructive interference as we move left to right, producing bright and dark fringes. Thus the answer must be (a) or (b) (note that we have chosen the thickness at the right edge so that the right edge will have the same intensity as the left). To distinguish the two choices, we note that the layers of optical material are glass A (n=1.5), air (n=1), then glass B (n=1.7). Thus light reflecting at the bottom of slide A, at the glass A-air interface, is reflecting from a lower n and has no phase change; while light reflecting at the top of slide B, at the air-glass B interface, is reflecting from a higher n and has a phase change of . At the left edge, there is no physical path length difference between the two reflected light waves, so the whole phase change comes from the effective path difference due to reflection. Since this is a half-wavelength, destructive interference occurs, and the left edge is dark.
8. (a) This differs from question 7 only in having a different index of refraction in the intermediate wedge between the slides. Again the path length difference of the 2 reflected rays varies as we move left to right, so that we go through the cycles of constructive and destructive interference as we move left to right, producing bright and dark fringes. Here though we have layers of optical material which are glass A (n=1.5), sassafras oil (n=1.6), then glass B (n=1.7). Thus light reflecting at the bottom of slide A, at the glass A-sassafras oil interface, is reflecting from a higher n and has a phase change of ; as does light reflecting at the top of slide B, at the sassafras oil-glass B interface. These phase shifts cancel, giving no net phase change due to reflection. At the left edge, there is no physical path length difference either. Thus both reflected rays are in phase at the left edge, interfering constructively to produce a bright fringe.
9. (e) The edges are both bright for the case of sassafras oil between the slides. Here the left edge is the zeroth order bright fringe, corresponding to a path difference of 0. The right edge has a path difference of . Since the wavelength of this light in sassafras oil is , the right edge has order

   that is, the path difference is 160 wavelengths at the right edge. This means that bright fringes of order 1, 2, 159 lie in between the left and right edges. There are 159 of them.

• About this document ...