Reminder: Our final exam is on Tuesday, December 21, 6:00pm to 8:30pm.

Summary of sections covered this semester.

Chap.	Title	Sections
1	Introduction	all
2	Motion in One Dimension	all
3	Vectors and Two-Dimensional Motion	1-5
4	The Laws of Motion	all
5	Work and Energy	all
6	Momentum and Collisions	all
7	Circular Motion and the Law of Gravity	1-9
8	Rotational Equilibrium and Rotational Dynamics	all
9	Solids and Fluids	3-6
10	Thermal Physics	all
11	Heat	1-8
12	The Laws of Thermodynamics	1-4, 6-7
13	Vibrations and Waves	1-6, 9-13
14	Sound	1-3, 5-8, 10
22	Reflection and Refraction of Light	3-6, 8, 9
23	Mirrors and Lenses	1-3, 6
24	Wave Optics	1, 2, 5, 6

Chapter 24: Wave Optics

Geometric optics makes use of the reflecting and refracting properties of light. But, light is a wave, and can therefore interfere. Optics that involves interference, and superposition of waves is called wave optics.

Conditions for Interference

Interference of light is not a common occurrence because the wavelength is small (»400 nm to 700 nm) and the frequency is high. For interference to be observable, 3 conditions must be met:

1. The sources must be coherent, that is, they must maintain a constant phase with respect to each other.
2. The sources must have identical wavelengths.
3. The superposition principle must apply.

Young's Double-Slit Interference

An experimental arrangement with one source illuminating two thin, closely spaced slits was used by Young at the beginning of the 19^th century. Huygen's principle tells us that each slit can be treated as a source of outgoing spherical waves. Emerging from each slit will be a spherical wave which we will represent by the arc of a circle centered on the slit -- the two dimensional projection of a spherical wave.

The phase of the incident light at the two slits may not be identical, but any difference will remain constant. The wavelength is not affected by the slits, so it is identical at each slit. Therefore the first 2 conditions are satisfied.

The third condition is satisfied if light is a wave, since all waves obey the superposition principal. The original intent of this experiment was to determine if light is a wave. We know that the answer is yes, so we can go ahead and work out what the results will be, that is, what will be the pattern of light on a screen placed a distance L from the slits.

The pattern we expect to see is called an interference pattern, and consists of light and dark bands called fringes. The light bands are formed by constructive interference and the dark bands by destructive interference.

Constructive interference occurs when the waves from each slit arrive in phase. For simplicity, let's assume that the waves are in phase when they emerge from each slit. Then at a point on the screen halfway between the two slits, the light from each slit has traveled equal distances. Over equal distances, the light will have the same number of oscillations, and is therefore in phase at the screen. At this point the waves add constructively and we get a bright fringe.

If we move along the screen a short distance, we will come to a point where the light from slit 2 travels one half wavelength (l/2) further than the light from slit 1. At this point, the waves are 180° out of phase and destructive interference occurs. The waves will cancel out and we get a dark fringe.

What is important is the difference in path lengths. From the geometry of the experiment, the path difference from the two slits to the point on the screen at an angle q, we find d = d sinq, where d is the distance between the two slits.

When the path difference is a whole multiple of a wavelength, constructive interference occurs, and a bright fringe appears on the screen:
d = d sinq = ml    m = 0, ±1, ±2, ...
The integer m is called the order number of the bright fringe.

When the path difference is a multiple plus 1/2 of a wavelength, then destructive interference occurs and a dark fringe appears on the screen:
d = d sinq = (m + ½)l    m = 0, ±1, ±2, ...
Again, the integer m is call the order number, but now for the dark fringe.

Often we want to know the location of the fringe on the screen. We'll call the point halfway between the two slits y=0. Then, the location of the m^th bright fringe is
y_bright = (lL/d)m
where L is the distance from the slits to the screen. The location of the dark fringes is given by
y_dark = (lL/d)(m + ½).

Example: P24.1

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.100 mm is illuminated by light have a wavelength of 589 nm and the interference pattern observed on a screen 4.00 m from the slits. (a) What is the difference in path lengths from each of the slits to the screen location of a third-order bright fringe? (b) What is the difference in path lengths from the two slits to the screen location of the third dark fringe away from the the center of the pattern?

(a) The zeroeth order bright fringe is where the path lengths are equal, at the first order fringe the path lengths differ by one wavelength (constructive interference), at the second order fringe the path lengths differ by two wavelengths, and at the third order fringe the path lengths differ by three wavelengths. In general d = ml for an m^th order fringe. For the third order bright fringe, d = 3l = 3(589nm) = 1767nm = 1.767µm.

(b) The first dark fringe occurs where the path lengths differ by half a wavelength (destructive interference). The m^th dark fringe occurs where d = (m + ½)l. At the third dark fringe, d = 3.5l = 3.5(589nm) = 2062nm = 2.06µm.

Example: P24.2

A pair of narrow, parallel slits separated by 0.250 mm are illuminated by the green component from a mercury vapor lamp (l = 546.1 nm). The interference pattern is observed on a screen 1.20 m from the plane of the parallel slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands in the interference pattern.

(a) We want the distance between the m=0 and m=1 bright fringes on the screen. The location of the m^th bright fringe on the screen is given by y = (lL/d)m. The zeroeth fringe is at y=0. The distance to the first bright fringe is y = lL/d = (546.1nm)(1.20m)/(0.250mm) = 2.62×10^-3m = 2.62mm.

(b) The location of the m^th dark fringe on the screen is y = (lL/d)(m + ½). The distance between the first and second dark fringes is Dy = (lL/d)(2.5 - 1.5) = lL/d = 2.62mm, same as in (a).

Single-Slit Diffraction

In double slit interference, we treated the slits like point sources, that is, we ignored the width of each slit. But, a slit will have a finite width, and so it can "act like" a series of very closely spaced, infinitely thin slits. Therefore a single slit can produce an interference pattern, which we call a diffraction pattern.

Diffraction is important because it limits the ultimate resolution of optical instruments like microscopes and telescopes.

The resulting diffraction pattern is shown in Figure 24.16. The characteristics are a large central maximum, with alternating dark and bright fringes. The first dark fringes occur at angles given by:
sinq = l/a
where a is the width of the slit. The width of the central maximum is given by the distance between the first dark fringes.

Example: P24.27

Light of wavelength 600nm falls on a 0.40mm wide slit and forms a diffraction pattern on a screen 1.5m away. (a) Find the position of the first dark band on each side of the central maximum. (b) Find the width of the central maximum.

(a) The first dark fringe occurs at sinq = l/a. Since the screen is L=1.5m away, the fringe will be at a location y = L tanq » L sinq = lL/a = (600nm)(1.5m)/(0.40mm) = 2.25mm.

(b) The width of the central maximum is the distance between the dark fringes on either side = 2(2.25mm) = 4.50mm.