Chapter 34 The Wave Nature of Light; Interference Chapter 34 Opener. The beautiful colors from the surface of this soap bubble can be nicely explained by the wave theory of light. A soap bubble is a very thin spherical film filled with air. Light reflected from the outer and inner surfaces of this thin film of soapy water interferes constructively to produce the bright colors. Which colors we see at any point depends on the thickness of the soapy water film at that point and also on the viewing angle. Near the top of the bubble, we see a small black area surrounded by a silver or white area. The bubble’s thickness is smallest at that black spot, perhaps only about 30 nm thick, and is fully transparent (we see the black background). We cover fundamental aspects of the wave nature of light, including two-slit interference and interference in thin films.

34-4 Intensity in the Double-Slit Interference Pattern The electric fields at the point P from the two slits are given by . Figure 34-12. Determining the intensity in a double-slit interference pattern. Not to scale: in fact l >> d and the two rays become essentially parallel. where

34-4 Intensity in the Double-Slit Interference Pattern The two waves can be added using phasors, to take the phase difference into account: Figure 34-11. Phasor diagram for double-slit interference pattern.

34-4 Intensity in the Double-Slit Interference Pattern The time-averaged intensity is proportional to the square of the field:

34-4 Intensity in the Double-Slit Interference Pattern This plot shows the intensity as a function of angle. Figure 34-14. Intensity I as a function of phase difference δ and position on screen y (assuming y << l).

34-4 Intensity in the Double-Slit Interference Pattern Example 34-5: Antenna intensity. Two radio antennas are located close to each other, separated by a distance d. The antennas radiate in phase with each other, emitting waves of intensity I0 at wavelength λ. (a) Calculate the net intensity as a function of θ for points very far from the antennas. (b) For d = λ, determine I and find in which directions I is a maximum and a minimum. (c) Repeat part (b) when d = λ/2. Figure 34-15: Example 34–5.The two dots represent the antennas. Solution: a. The points of constructive and destructive interference are the same as those for the double slit; eq. 34-6 or 34-7 can be used to find the intensity as a function of angle. b. When d = λ, I = I0 cos2(π sin θ). I is a maximum when sin θ = +1, 0, or -1, and a minimum when sin θ = +1/2 or -1/2. c. When d = λ/2, I is a maximum when θ = 0 and 180°, and a minimum when θ = 90° and 270°.

ConcepTest 34.3a Double Slits I In a double-slit experiment, when the wavelength of the light is increased, the interference pattern 1) spreads out 2) stays the same 3) shrinks together 4) disappears

ConcepTest 34.3a Double Slits I In a double-slit experiment, when the wavelength of the light is increased, the interference pattern 1) spreads out 2) stays the same 3) shrinks together 4) disappears d sin  = m If  is increased and d does not change, then  must increase, so the pattern spreads out.

34-5 Interference in Thin Films Another way path lengths can differ, and waves interfere, is if they travel through different media. If there is a very thin film of material – a few wavelengths thick – light will reflect from both the bottom and the top of the layer, causing interference. This can be seen in soap bubbles and oil slicks. Figure 34-16. Thin film interference patterns seen in (a) a soap bubble, (b) a thin film of soapy water, and (c) a thin layer of oil on wet pavement.

34-5 Interference in Thin Films The wavelength of the light will be different in the oil and the air, and the reflections at points A and B may or may not involve phase changes. Figure 34-17. Light reflected from the upper and lower surfaces of a thin film of oil lying on water. This analysis assumes the light strikes the surface nearly perpendicularly, but is shown here at an angle so we can display each ray.

34-5 Interference in Thin Films A similar effect takes place when a shallowly curved piece of glass is placed on a flat one. When viewed from above, concentric circles appear that are called Newton’s rings. Figure 34-18. Newton’s rings. (a) Light rays reflected from upper and lower surfaces of the thin air gap can interfere. (b) Photograph of interference patterns using white light.

34-5 Interference in Thin Films A beam of light reflected by a material with index of refraction greater than that of the material in which it is traveling, changes phase by 180° or ½ cycle. Figure 34-19. (a) Reflected ray changes phase by 180° or ½ cycle if n2 > n1, but (b) does not if n2 < n1.

34-5 Interference in Thin Films Example 34-6: Thin film of air, wedge-shaped. A very fine wire 7.35 x 10-3 mm in diameter is placed between two flat glass plates. Light whose wavelength in air is 600 nm falls (and is viewed) perpendicular to the plates and a series of bright and dark bands is seen. How many light and dark bands will there be in this case? Will the area next to the wire be bright or dark? Figure 34-20. (a) Light rays reflected from the upper and lower surfaces of a thin wedge of air interfere to produce bright and dark bands. (b) Pattern observed when glass plates are optically flat; (c) pattern when plates are not so flat. See Example 34–6. Solution: The path lengths are different for the rays reflected from the upper and lower surfaces; in addition, the ray reflected from the lower surface undergoes a 180° phase change. Dark bands will occur when 2t = (m + ½)λ. At the position of the wire, t is 24.5 wavelengths. This is a half-integer; the area next to the wire will be bright, and there will be 25 dark bands between it and the other edge. Including the band next to the wire, there will also be 25 light bands.

34-5 Interference in Thin Films Example 34-7: Thickness of soap bubble skin. A soap bubble appears green (λ = 540 nm) at the point on its front surface nearest the viewer. What is the smallest thickness the soap bubble film could have? Assume n = 1.35. Figure 34-21: Example 34–7.The incident and reflected rays are assumed to be perpendicular to the bubble’s surface. They are shown at a slight angle so we can distinguish them. Solution: The path length difference is twice the thickness of the film; the ray reflecting from the first surface undergoes a 180° phase change, while the other ray does not. The smallest thickness where the green light will be bright is when 2t = λ/2n, or t = 100 nm. Constructive interference then occurs for every additional 200 nm in thickness.

34-5 Interference in Thin Films Problem Solving: Interference Interference occurs when two or more waves arrive simultaneously at the same point in space. Constructive interference occurs when the waves are in phase. Destructive interference occurs when the waves are out of phase. An extra half-wavelength shift occurs when light reflects from a medium with higher refractive index.

34-5 Interference in Thin Films Example 34-8: Nonreflective coating. What is the thickness of an optical coating of MgF2 whose index of refraction is n = 1.38 and which is designed to eliminate reflected light at wavelengths (in air) around 550 nm when incident normally on glass for which n = 1.50? Figure 34-23. Example 34–8. Incident ray of light is partially reflected at the front surface of a lens coating (ray 1) and again partially reflected at the rear surface of the coating (ray 2), with most of the energy passing as the transmitted ray into the glass. Solution: To eliminate reflection, the two rays should interfere destructively. Both rays change phase on reflection, so the net result is no phase difference due to reflection. The condition for destructive interference is then 2t = λ/2n, or t = 99.6 nm.

ConcepTest 34.6d Parallel Slides IV Two identical microscope slides in air illuminated with light from a laser are creating an interference pattern. The space between the slides is now filled with water (n = 1.33). What happens to the interference fringes? 1) spaced farther apart 2) spaced closer together 3) no change ray 1 ray 2 ray 3 t

ConcepTest 34.6d Parallel Slides IV Two identical microscope slides in air illuminated with light from a laser are creating an interference pattern. The space between the slides is now filled with water (n=1.33). What happens to the interference fringes? 1) spaced farther apart 2) spaced closer together 3) no change The path difference between ray 2 and ray 3 is 2t (in addition, ray 3 experiences a phase change of 180°). Thus, the dark fringes will occur for: 2t = mlwater where lwater = lair/n Thus, the water has decreased the wavelength of the light. ray 1 ray 2 ray 3 t

34-6 Michelson Interferometer The Michelson interferometer is centered around a beam splitter, which transmits about half the light hitting it and reflects the rest. It can be a very sensitive measure of length. Figure 34-24. Michelson interferometer.

34-7 Luminous Intensity The intensity of light as perceived depends not only on the actual intensity but also on the sensitivity of the eye at different wavelengths. Luminous flux: 1 lumen = 1/683 W of 555-nm light Luminous intensity: 1 candela = 1 lumen/steradian Illuminance: luminous flux per unit area

34-7 Luminous Intensity Example 34-9: Lightbulb illuminance. The brightness of a particular type of 100-W lightbulb is rated at 1700 lm. Determine (a) the luminous intensity and (b) the illuminance at a distance of 2.0 m. Solution: a. Assume the light output is uniform in all directions. Then the luminous intensity is 1700 lm/4π sr = 135 cd. b. At d = 2.0 m, the illuminance is 1700 lm/(4πd2) = 34 lm/m2.

Summary of Chapter 34 The wave theory of light is strengthened by the interference and diffraction of light. Huygens’ principle: every point on a wave front is a source of spherical wavelets. Wavelength of light in a medium with index of refraction n: Young’s double-slit experiment demonstrated interference.

Summary of Chapter 34 In the double-slit experiment, constructive interference occurs when and destructive interference when Two sources of light are coherent if they have the same frequency and maintain the same phase relationship.

Summary of Chapter 34 Interference can occur between reflections from the front and back surfaces of a thin film. Light undergoes a 180° phase change if it reflects from a medium of higher index of refraction.

Chapter 35 Diffraction and Polarization Chapter 35 opener. Parallel coherent light from a laser, which acts as nearly a point source, illuminates these shears. Instead of a clean shadow, there is a dramatic diffraction pattern, which is a strong confirmation of the wave theory of light. Diffraction patterns are washed out when typical extended sources of light are used, and hence are not seen, although a careful examination of shadows will reveal fuzziness. We will examine diffraction by a single slit, and how it affects the double-slit pattern. We also discuss diffraction gratings and diffraction of X-rays by crystals. We will see how diffraction affects the resolution of optical instruments, and that the ultimate resolution can never be greater than the wavelength of the radiation used. Finally we study the polarization of light.

Units of Chapter 35 Diffraction by a Single Slit or Disk Intensity in Single-Slit Diffraction Pattern Diffraction in the Double-Slit Experiment Limits of Resolution; Circular Apertures Resolution of Telescopes and Microscopes; the λ Limit Resolution of the Human Eye and Useful Magnification Diffraction Grating

Units of Chapter 35 The Spectrometer and Spectroscopy Peak Widths and Resolving Power for a Diffraction Grating X-Rays and X-Ray Diffraction Polarization Liquid Crystal Displays (LCD) Scattering of Light by the Atmosphere

35-1 Diffraction by a Single Slit or Disk If light is a wave, it will diffract around a single slit or obstacle. Figure 35-1. If light is a wave, a bright spot will appear at the center of the shadow of a solid disk illuminated by a point source of monochromatic light.

35-1 Diffraction by a Single Slit or Disk The resulting pattern of light and dark stripes is called a diffraction pattern. Figure 35-2. Diffraction pattern of (a) a circular disk (a coin), (b) razor, (c) a single slit, each illuminated by a coherent point source of monochromatic light, such as a laser.

35-1 Diffraction by a Single Slit or Disk This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would. Figure 35-3. Analysis of diffraction pattern formed by light passing through a narrow slit of width D.

35-1 Diffraction by a Single Slit or Disk The minima of the single-slit diffraction pattern occur when Figure 35-4. Intensity in the diffraction pattern of a single slit as a function of sin θ. Note that the central maximum is not only much higher than the maxima to each side, but it is also twice as wide (2λ/D wide) as any of the others (only λ/D wide each).

35-1 Diffraction by a Single Slit or Disk Example 35-1: Single-slit diffraction maximum. Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm wide. How wide is the central maximum (a) in degrees, and (b) in centimeters, on a screen 20 cm away? Solution: a. The first minimum occurs at sin θ = λ/D = 0.75, or θ = 49°. The full width is twice this, or 98°. b. The width is 46 cm.

35-1 Diffraction by a Single Slit or Disk Conceptual Example 35-2: Diffraction spreads. Light shines through a rectangular hole that is narrower in the vertical direction than the horizontal. (a) Would you expect the diffraction pattern to be more spread out in the vertical direction or in the horizontal direction? (b) Should a rectangular loudspeaker horn at a stadium be high and narrow, or wide and flat? Solution: a. The pattern will be more spread out in the vertical direction, as the slit is narrower there. b. A pattern that is wider than it is high is desired, so the speaker should be high and narrow.

35-2 Intensity in Single-Slit Diffraction Pattern Light passing through a single slit can be divided into a series of narrower strips; each contributes the same amplitude to the total intensity on the screen, but the phases differ due to the differing path lengths: Figure 35-7. Slit of width D divided into N strips of width Δy. .

ConcepTest 35.1a Diffraction I The diffraction pattern below arises from a single slit. If we would like to sharpen the pattern, i.e., make the central bright spot narrower, what should we do to the slit width? 1) narrow the slit 2) widen the slit 3) enlarge the screen 4) close off the slit

ConcepTest 35.1a Diffraction I The diffraction pattern below arises from a single slit. If we would like to sharpen the pattern, i.e., make the central bright spot narrower, what should we do to the slit width? 1) narrow the slit 2) widen the slit 3) enlarge the screen 4) close off the slit d q The angle at which one finds the first minimum is: The central bright spot can be narrowed by having a smaller angle. This in turn is accomplished by widening the slit. sin  =  /d

35-2 Intensity in Single-Slit Diffraction Pattern Phasor diagrams give us the intensity as a function of angle. Figure 35-8. Phasor diagram for single-slit diffraction, giving the total amplitude Eθ at four different angles θ.

35-2 Intensity in Single-Slit Diffraction Pattern Taking the limit as the width becomes infinitesimally small gives the field as a function of angle: Figure 35-9. Determining amplitude Eθ as a function of θ for single-slit diffraction.

35-2 Intensity in Single-Slit Diffraction Pattern Finally, we have the phase difference and the intensity as a function of angle: and .

35-2 Intensity in Single-Slit Diffraction Pattern Example 35-3: Intensity at secondary maxima. Estimate the intensities of the first two secondary maxima to either side of the central maximum. Solution: The secondary maxima occur approximately halfway between the minima. Equation 35-7 then gives, for m = 1, I = 0.045 I0, and for m = 2, I = 0.016 I0.

35-3 Diffraction in the Double-Slit Experiment The double-slit experiment also exhibits diffraction effects, as the slits have a finite width. This means the amplitude at an angle θ will be modified by the same factor as in the single-slit experiment: The intensity is, as usual, proportional to the square of the field.

35-3 Diffraction in the Double-Slit Experiment The diffraction factor (depends on β) appears as an “envelope” modifying the more rapidly varying interference factor (depends on δ). Figure 35-10. (a) Diffraction factor, (b) interference factor, and (c) the resultant intensity plotted as a function of θ for d = 6D = 60λ.

35-3 Diffraction in the Double-Slit Experiment Example 35-4: Diffraction plus interference. Show why the central diffraction peak shown, plotted for the case where d = 6D = 60λ, contains 11 interference fringes. Solution: The first minimum in the diffraction pattern occurs where sin θ = λ/D. Since d = 6D, d sin θ = 6λ. This means that the m=6 interference fringes will not appear; the central diffraction peak contains the fringes for m = 0 (one) and m = 1 through 5 (two each for a total of 10), making 11 fringes.