Chapter 24 Wave Optics Conceptual questions: 3, 4, 13, 14, 17, 18 Quick Quizzes: 1, 2, 3, 4 Problems: 10, 17, 34

Interference Light waves interfere with each other much like mechanical waves do Two conditions which must be met The sources must be coherent. They must maintain a constant phase with respect to each other The waves must have identical wavelengths

Producing Coherent Sources Old method: Light from a monochromatic source is allowed to pass through a narrow slit The light from the single slit is allowed to fall on a screen containing two narrow slits The first slit is needed to insure the light comes from a tiny region of the source which is coherent New method: use a laser

Young’s Double Slit Experiment The narrow slits, S1 and S2 act as sources of waves The waves emerging from the slits originate from the same wave front and therefore are always in phase

Interference Patterns Constructive interference occurs at the center point The two waves travel the same distance and they arrive in phase

Interference Patterns, 2 The upper wave travels one wavelength farther Therefore, the waves arrive in phase A bright fringe occurs

Interference Patterns, 3 The upper wave travels one-half of a wavelength farther than the lower wave The trough of the bottom wave overlaps the crest of the upper wave This is destructive interference A dark fringe occurs

Interference Equations The path difference, δ, is found from the tan triangle δ = r2 – r1 = d sin θ This assumes the paths are parallel

Interference Equations, 2 For a bright fringe, produced by constructive interference, the path difference must be δ = m λ m = 0, ±1, ±2, … δ = d sin θbright = m λ m is called the order number When m = 0, it is the zeroth order maximum When m = ±1, it is called the first order maximum

Interference Equations, 3 When destructive interference occurs, a dark fringe is observed This needs a path difference of an odd half wavelength; δ = (m + ½) λ δ = d sin θdark = (m + ½) λ m = 0, ±1, ±2, …

Interference Equations, 4 The positions of the fringes can be measured vertically from the zeroth order maximum y = L tan θ ~ L sin θ Approximation θ is small and therefore tanθ ~ sin θ For bright fringes For dark fringes

Quick quiz 24-1 In a two slit interference pattern projected on a screen the fringes are equally spaced on the screen A. everywhere B. only for large angles C. only for small angles

Problem 24.10 A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of λ = 643 nm. An interference pattern is observed on a screen 140 cm from the slits. Consider a point on the screen located at y = 1.80 cm from the central maximum of this pattern. What is the path difference δ for the two slits at the location y? (b) Express this path difference in terms of the wavelength. (c) Will the interference correspond to a maximum, a minimum, or an intermediate condition?

Phase Changes Due To Reflection An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling Analogous to a reflected pulse on a string

Phase Changes Due To Reflection, cont There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction Analogous to a pulse in a string reflecting from a free support

Interference in Thin Films Rules to remember An electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1 There is no phase change in the reflected wave if n2 < n1 The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum

Interference in Thin Films, 2 Ray 1 undergoes a phase change of 180° with respect to the incident ray Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave Ray 2 also travels an additional distance of 2t before the waves recombine

Interference in Thin Films, 3 For constructive interference 2 n t = (m + ½ ) λ m = 0, 1, 2 … This takes into account both the difference in optical path length (2t) for the two rays and the 180° phase change (1/2 λ) For destruction interference 2 n t = m λ m = 0, 1, 2 …

Interference in Thin Films, 4 An example of different indices of refraction A coating on a solar cell

Quick quiz 24-2 Supposed Young’s experiment is carried out in air, and then, in a second experiment, the apparatus is immersed in water. In what way does the distance between bright fringes change? A. they move further apart B. they move closer together C. no change

Problem 24.17 A coating is applied to a lens to minimize reflections. The index of refraction of the coating is 1.55, and that of the lens is 1.48. If the coating is 177.4 nm thick, what wavelength is minimally reflected for normal incidence in the lowest order?

Newton’s rings

Conceptual questions 3. Consider a dark fringe in an interference pattern, at which almost no light energy is arriving. Light from both slits is arriving at this point, but the waves are canceling. Where does the energy go? 4. If Young’s double slit experiment were performed under water, how would the observed interference pattern be affected? 13.Would it be possible to place a nonreflective coating on an airplane to cancel radar waves of wavelength 3 cm?

Reading a CD As the disk rotates, the laser reflects off the sequence of lands and pits into a photodector The photodector converts the fluctuating reflected light intensity into an electrical string of zeros and ones The pit depth is made equal to one-quarter of the wavelength of the light land

Diffraction Huygen’s principle requires that the waves spread out after they pass through slits This spreading out of light from its initial line of travel is called diffraction

Fraunhofer Diffraction Fraunhofer Diffraction occurs when the rays leave the diffracting object in parallel directions A bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side

Single Slit Diffraction According to Huygen’s principle, each portion of the slit acts as a source of waves The light from one portion of the slit can interfere with light from another portion The resultant intensity on the screen depends on the direction θ Wave 1 travels farther than wave 3 by an amount equal to the path difference (a/2) sin θ destructive interference occurs when sin θdark = mλ / a

Single Slit Diffraction, 2 A broad central bright fringe is flanked by much weaker bright fringes alternating with dark fringes The points of constructive interference lie approximately halfway between the dark fringes

Problem 34 A screen is placed 50.0 cm from a single slit, which is illuminated with light of wavelength 680 nm. If the distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit?

Quick quiz 24.3 In a single-slit diffraction experiment, as the width of the slit is made smaller, the width of the central maximum of the diffraction pattern becomes (a) smaller, (b) larger, (c) remains the same.

Diffraction Grating The condition for maxima is d sin θbright = m λ m = 0, 1, 2, … The integer m is the order number of the diffraction pattern If the incident radiation contains several wavelengths, each wavelength deviates through a specific angle

QUICK QUIZ 24.4 If laser light is reflected from a phonograph record or a compact disc, a diffraction pattern appears. This occurs because both devices contain parallel tracks of information that act as a reflection diffraction grating. Which device, record or compact disc, results in diffraction maxima that are farther apart?

Diffraction Grating in CD Tracking A diffraction grating can be used in a three-beam method to keep the beam on a CD on track The central maximum of the diffraction pattern is used to read the information on the CD The two first-order maxima are used for steering

Polarization of Light Waves Each atom produces a wave with its own orientation of E This is an unpolarized wave

Polarization of Light, cont A wave is said to be linearly polarized if the resultant electric field vibrates in the same direction at all times at a particular point Polarization can be obtained from an unpolarized beam by selective absorption reflection scattering

Polarization by Selective Absorption The most common technique for polarizing light Uses a material that transmits waves whose electric field vectors in the plane parallel to a certain direction and absorbs waves whose electric field vectors are perpendicular to that direction Malus’ law: I = Io cos2 θ

Polarization by Reflection The angle of incidence for which the reflected beam is completely polarized is called the polarizing angle, θp θp is also called Brewster’s Angle Brewster’s Law relates the polarizing angle to the index of refraction for the material

Polarization by Scattering The horizontal part of the electric field vector in the incident wave causes the charges to vibrate horizontally The vertical part of the vector simultaneously causes them to vibrate vertically Horizontally and vertically polarized waves are emitted

Conceptual question 14. Certain sunglasses use a polarizing material to reduce intensity of light reflected from shiny surfaces, such as water or a hood of a car. What orientation of the transmission axis should the material have to be most effective? 18. Can a sound wave be polarized? 17. When you receive a chest x-ray at a hospital, the ex-ray passes through a series of parallel ribs in your chest. Do the ribs act as a diffraction grating for x-rays?

Optical Activity Certain materials display the property of optical activity A substance is optically active if it rotates the plane of polarization of transmitted light

Liquid Crystals Rotation of a polarized light beam by a liquid crystal when the applied voltage is zero Light passes through the polarizer on the right and is reflected back to the observer, who sees the segment as being bright

Liquid Crystals When a voltage is applied, the liquid crystal does not rotate the plane of polarization The light is absorbed by the polarizer on the right and none is reflected back to the observer The segment is dark

MCAD Two light sources produce light with wavelength l. The sources are placed 22.5 l and 45 l away from point P. When both sources are turned on and their intensities, I, at point P are equal, the resultant intensity at point P will be A. 0 B. 0.5 I C. I D. 2I

The process discussed in the previous question is called Diffraction b. Refraction Interference d. Dispersion Which of the following would result in greatest diffraction? Small wavelengths moving through a small opening Large wavelengths moving through a small opening Small wavelengths moving through a large opening Large wavelengths moving through a large opening