CHAPTER 1 PHYSICAL OPTICS: INTERFERENCE • Introduction

CHAPTER 1

PHYSICAL OPTICS: INTERFERENCE What is “physical optics”? • Introduction The methods of physical optics are used when the • Waves wavelength of light and dimensions of the system are of • Principle of superposition a comparable order of magnitude, when the simple ray • Wave packets approximation of geometric optics is not valid. So, it is • Phasors • Interference intermediate between geometric optics, which ignores • Reflection of waves wave effects, and full wave electromagnetism, which is a precise theory. • Young’s double-slit experiment

• Interference in thin films and air gaps In General Physics II you studied some aspects of geometrical optics. Geometrical optics rests on the assumption that light propagates along straight lines and is reflected and refracted according to definite laws, such Or the use of a convex lens as a magnifying lens: as Fermat’s principle and Snell’s Law. As a result the positions of images in mirrors and through lenses, etc. can be determined by scaled drawings. For example, the s ! s production of an image in a concave mirror.

s Object object

y Image • F • C f f y ! image 2 s ! 1 But many optical phenomena cannot be adequately The colors you see in a soap bubble are also due to an explained by geometrical optics. For example, the interference effect between light rays reflected from the iridescence that makes the colors of a hummingbird so front and back surfaces of the thin film of soap making brilliant are not due to pigment but to an interference the bubble. The color depends on the thickness of film, effect caused by structures in the feathers. ranging from black, where the film is thinnest, to magenta, where the film is thickest. Likewise, the colors you see in a thin film of oil floating on water are due to an interference effect between light rays reflected from the front and back surfaces of the oil Another example is the “spectrum of colors” you see film. when you reflect light from the active side of a CD. The effect that produces these colors is closely related to interference; it is called diffraction. Here is an example of diffraction caused by the edges of the razor blade when viewed from behind with Waves monochromatic blue light.

In physics,we come across three main types of waves:

Mechanical waves: water waves, sound waves, seismic waves, waves on a string ...

Electromagnetic waves: visible and ultraviolet light, radio and television waves, microwaves, x-rays and radar waves. All e-m waves travel in vacuum with a speed of c = 2.99792458 !108 m s. The pattern of “fringes” is easily observable with monochromatic light. With white light, fringes due to Matter waves: the waves associated with electrons, the different wavelengths overlap making them more protons and other fundamental particles, atoms and difficult to observe. molecules.

Diffraction and interference cannot be expained by However, all waves have features in common. geometrical optics; instead, light has to be treated as waves. Take the ripples (waves) in a pond caused by a small y $ stone being dropped into the water ... A

x x " A !

Shown is a snapshot of the wave at some time t. It is described by the general equation: y (t,x) = A cos(!t " kx) At a position, x say, the disturbance y( t) varies x ! sinusoidally with time (i.e., simple harmonic motion). Similarly, at a time t, as shown here, the disturbance varies sinusoidally with position, x. The parameter y ! A k 2# is called the wavevector, where is the = $ $ x wavelength. The product kx is called the phase angle

" A (= %). A negative (positive) sign indicates the wave is traveling to the right (left). Vertical and horizontal illustrations of the wave

Also, ! is the angular frequency of the wave given by To describe the “disturbance” of such a wave we need ! = 2#f , and T = 1 is the periodic time of the wave. two variables, t and x. f y ! x y ! A x x wave at t wave at t + !t $ A

Note that the disturbance at some fixed time t !, To find the speed of a wave, we take two snapshots at a time interval t apart. If the wave (i.e., the red dot) y (t!,x) = y(t!,x ± n!), ! i.e., the wave is reproduced at displacements of n !, travels a distance ! x in that time, then the speed of the where n is an integer. wave is v !x . y T = !t A Since the disturbances (y) are equal t y (t,x) = y(t + !t, x + !x), T = 1 = 2" so "t # kx = "(t + !t) # k(x + !x), $ A f # i.e., "! t = k!x, Similarly, the disturbance at some fixed position x !, v " . $ = k y(t,x ) = y(t ± T, x ) = y(t ± m2" ,x ), ! ! # ! Note also v " 2%f f . = k = 2% & = & i.e., the wave is reproduced after time intervals of The velocity of a fixed point on a wave (such as the red m2" , where m is an integer. # dot) is called the phase velocity. Principle of superposition

“If two or more waves are traveling through a medium, y 1 = 6sin x the resultant disturbance at any point is the algebraic sum of the individual disturbances.” y 2 = 5sin 2x

Waves that obey this principle are called linear waves. y 3 = 4sin 3x One consequence is that two waves can “pass” through each other! y 4 = 3sin 4x y 2 y 5 = 2sin 5x y1 y2 y 1 y 6 = sin 6x

y 1 + y2 y 1 + y2

y 1 + y2 y 2 y y 2 1 y 1 (Waves that do not obey this principle are called non- y = y1 + y2 + y3 + y4 + y5 + y6 linear waves.) Consider the superposition of two waves of equal So, the superposition of several waves of differing amplitude but slightly different frequencies and wavelengths and amplitudes produces complex wavelengths. Then, the resultant is waveforms. For example, to produce a square wave ... y (x,t) = Asin(!1t " k1x) + Asin(!2t " k2x) $ #! #k ' = 2Acos& t " x) sin(! t " k x), % 2 2 (

(!1 + !2) where #! = !1 " !2, # k = k1 " k2, ! = 2

(k1 + k2) and k = 2 . When plotted, at some time t, we WBX06VD1.MOV get A square wave can be expressed as a so-called Fourier y (x) series: A x ! 1 $ n# ' f(x) = " sin& x) , n=1,3,5… n % L ( y (x) * 2A where L = 2, i.e., one-half of the wavelength.

x A Fourier series decomposes a periodic function into a sum of simple oscillating functions, i.e., sines and/or cosines. i.e., the waves are separated into “groups”. y (x) y (x) 2A 2A

x x x 1 x 2 x 1 x 2

! x ! x

$ !" !k ' So, successive minima (at time t) occur at x 2 and x 1 where y = (x,t) = 2Acos& t # x) sin(" t # k x) % 2 2 ( $ !" !k ' $ !" !k ' & t # x ) # & t # x ) = *, % 2( % 1( The first term is the envelope, i.e., the green curve. The 2 2 2 2 2* second term is the wave within the envelope. Both the i.e, x # x = !x = . 2 1 k envelope and wave within the envelope are traveling ! #1 waves. Thus, the spatial extent of the group is ! x + !k . If we plot the resultant as a function of t at a fixed point, we get

The envelope moves with velocity v !" , called y (t) g = !k 2A the group velocity. The wave inside the envelope moves with velocity v = " , called the phase velocity. p k t The amplitude of the envelope is zero when $ !" !k ' * & t # x ) = (2n +1) , In the case of sound waves, this waveform produces the % 2 2 n( 2 phenomenon of “beats”. where n = 0,±1,±2 !. y (t) The group velocity (the velocity of the envelope) is 2A " d kv % d! ( p) " dvp % vg = = $ ' = $ vp + k ' . [ dk]k $ dk' # dk& t # & k k t 1 t 2 If the phase velocity is the same at all frequencies and ! t wavelengths, i.e., there is no dispersion then Successive minima (at point x) occur at t and t where dvp 2 1 dk = 0 i.e., v g = vp. $ !" !k ' $ !" !k ' & t2 # x) # & t1 # x) = *, % 2 2 ( % 2 2 ( 2* i.e, t2 # t1 = !t = . !" Hence, the temporal extent ! t + !"#1. So, we find that ! x.!k , constant ( 2*) and ! t.!" , constant ( 2*).

The phase velocity of the individual harmonic waves is " v p = k , " = vpk. Case A.mpg Wave Packets dvp A medium in which dk = 0 is said to be non- dispersive. (An example is an electromagnetic waves in A “wave packet” can be created by superposing many vacuum.) Glass, for instance, is a dispersive medium. waves spanning a wavevector range k ± !k . ! 2 Shown here is a plot of the phase velocity in flint glass. a (kn)

v !108 m/s y p( ) 1 .98 k " !k k + !k x Red ! 2 ! 2 dv p 2 k n 1 .97 # "39.3 m s k ! dk ! k

1 .96 To generate the wave packet shown above, we put 25 Violet y = #a(kn )cos(knx) 1 .95 n="25 # 7 "1 k(!10 m ) where k k n !k and the amplitudes a(k ) are n = ( ! + 50) n 0 1 .0 1 .2 1 .4 1 .6 a Gaussian distribution, i.e., dv 2 2 p a(k ) = e"n $ . In this case, dk < 0, so v p > vg. n dv We set k ! = 5, ! k = 3 and $ = 10. Thus, there are 51 If p 0, then v v . dk > g > p equally spaced values of k n in the sum. In constructing a wave packet in this way we find the For example, if we use the full-width at half maximum interesting result that the spatial extent or “length” of the (FWHM) for ! x and ! k we find that ! x " !k # 5.54. wave packet, ! x, is inversely proportional to ! k, viz: ! k = 0.333

! k = 1 x x a (k) ! x ! x ! k = 2 ! k ! x ! x" !k ! k x ! k = 0.666 1 $ 45 ±1 $ 45 k k ! x x ! 2 $ 22 ±1 $ 44 ! k = 3 x ! k ! x ! x" !k 3 $ 15 ±1 $ 45 ! x 0.333 16.65 5.54 ! x 4 $ 11 ±1 $ 44 ! k = 1.00 0.666 8.33 5.55 ! k = 4 x arbitrary units x 1.00 5.52 5.52 ! x 1.33 4.16 5.53 ! x arbitrary units Although it is difficult to locate exactly the start and end ! k = 1.33 points of the wave packets, it is clear that x The important point is that ! x " !k # constant, there is a relationship a result we found earlier. The actual value of the ! x between ! x and ! k, i.e., constant depends on the amplitude function a( kn ) and the range of k values and the width of the packet. We the precise definitions of ! x and ! k. will see the significance of this result in chapter 6. Again, if we had we summed originally over a frequency range, ! f , instead of ! k, we would have found similar condition also occurs between the frequency range and the temporal extent, ! t, i.e., ! f " !t # !$ " !t # constant.

v g

x ! x

Furthermore, a wave packet is composed of many Question 1: individual component waves of differing wavevector (hence wavelength and frequency). So, a wave packet For the wave packet shown, (a) estimate the range in traveling through a dispersive medium, i.e., one in which wavevectors ! k. (b) If ! t is the time it takes the wave the phase velocity depends on frequency (and packet to pass a point in space, what is the range of wavevector) - such as glass, water, etc. - will broaden frequencies ! f in the wave packet? because the component waves travel with different velocities. (b) The frequency is given by the number of oscillations passing the point in time ! t, i.e., N f = !t. !N 1 " !f = !t = !t. ! x But # = 2$f , i.e., !# = 2$!f , !# 1 (a) First, we determine how many oscillations there are " 2$ = !t, in the wave packet; but it is difficult to say definitively i.e., !# % !t = 2$ (a constant). where the waves start and stop. We find there are between 12 and 13 oscillations, so, let us take N = 12 but Earlier we found that ! x % !k = 2$ ( = !#.!t) with an uncertainty ! N = 1. !# !x " !k = !t, !x 16 1.33 units. " # = N = 12 = which the speed of the wave packet; it is called the 2$ 2$N But k = = . group velocity ( v g). We note that # !x ! x % !k = 2$ and !# % !t = 2$ 2$ 2$ "!k = !N = %1 = 0.393 inverse units. represent the minimum values, with ! N = 1, i.e., the !x 16 uncertainty in the number of oscillations. Note that ! x & !k = 2$, i.e., a constant whose value depends on the uncertainty ! N. Phasors A monchromatic source is defined as one with a constant wavelength (or frequency). However, if a source Returning to single frequency (i.e., monochromatic) produces a single wave train of finite length, i.e., for a waves. Suppose we add two sine waves. What is the finite length of time, the disturbance cannot be resultant? monchromatic. P Consider two separated y r 1 sources, S and S , with t r 2 1 2 identical frequency, S 1 S 2 t ! wavelength and amplitude. Such sources are described as As the wave train shown above is not a simple harmonic coherent. Then at P we have: wave - it has a start and an end - and it is non-repeating, y = A cos(!t " kr ) and y = A cos(!t " kr ) it must be represented as an infinite sum of harmonic 1 1 2 2 Thus, the resultant disturbance at P is: waves of a range of different frequencies, centered on y + y = Acos(!t " kr ) + Acos(!t " kr ) the principal frequency. Thus, strictly speaking, unless 1 2 1 2 the wave train produced by a source is infinitely long, it How can we work this out ?? cannot be truly monchromatic. ... remember phasors ?? You might think that we could use trigonometric However, in what follows, we will ignore such effects. identities but, in general, that is not the best approach. y(t) = A cos!t = Acos"(t) Let’s solve this problem using phasors ...

( ) A " + # $ % ! A Put y 1 = A cos(/t $ kr1) = Acos# " R and y A t kr A % 2 = cos(/ $ 2) = cos% y(t) i.e., # = /t $ kr1 and % = /t $ kr2. ! A # y 1 y 2

The length of the resultant phasor is R = 2Acos! . y(t) (# $ %) But, 2 ! + (" + # $ %) = " & ! = $ 2 . ( ( ) + 'R = 2Acos* $ # $ % - . ) 2 , But cos. is an even function, i.e., cos ! = cos($! ). ( ( ) + 'R = 2Acos* # $ % - . ) 2 , t Thus, the projection of this phasor onto the x-axis is A “phasor” is a rotating vector. Phasors can be added just y 1 + y2 = Rcos(# + ! ), ( ( ) + like ordinary vectors. = 2Acos* # $ % - cos(# + ! ). ) 2 , http://www.animations.physics.unsw.edu.au/jw/phasor-addition.html amplitude Substituting for !, " and # we get: & $2 % $1) & $2 + $1) y1 + y2 = 2Acos( + cos( ,t % + , ' 2 * ' 2 * where ( $2 % $1) = (kr2 % kr1) = -, i.e., the phase difference between the two waves.

We can also get the result trignometrically, viz: Question 2: y 1 + y2 = Acos(,t % kr1) + Acos(,t % kr2) = A cos(,t % $1) + Acos(,t % $2). Using the trignometric relationship: Find the resultant of the two waves whose disturbances & X % Y) & X + Y) at a given location vary with time as follows: cosX + cosY = 2cos( + cos( + , ' 2 * ' 2 * y 4cos t and y 3cos t " . 1 = ! 2 = (! + 3) we find, as before, (see top of page), & $2 % $1) & $2 + $1) y1 + y2 = 2Acos( + cos( ,t % + . ' 2 * ' 2 * Although this seems like much less work, it is only applicable if the amplitudes of y 1 and y 2 are equal. The method using phasors is more generally applicable ... see the next example ... Interference Draw the two waves as phasors. P Earlier, we found the resultant r 1 r 2 at P for two identical sources S 1 S 1 S 2 and S 2 is: 3sin ! 2.60 ( 3) = 3 R ! 3 # ! " ! & # ! + ! & y + y = 2Acos% 2 1( cos% )t " 2 1( , 3cos ! 1.50 1 2 ( 3) = $ 2 ' $ 2 ' $ 4 2 # t + where (!2 " !1) = k(r2 " r1) = k*r = *r. Here, ,

The length of the resultant phasor is given by • ( r2 " r1) = *r is called the path difference, and 2 ! 2 ! 2 • ( !2 " !1) = - is the phase difference R = 4 + 3cos( 3) + 3sin( 3) = 37, ( ) ( ) between the two waves. Note, the resultant disturbance " R = 6.08. ( y1 + y2) is a maximum when The resultant wave ( y1 + y2) is given by the projection (! " ! ) - k + 2 1 = n+ = = *r = *r, of R onto the horizontal axis, i.e., 2 2 2 , EVEN R cos(#t + $), i.e., when * r = n, or - = 2n+. %1& 2.60 ) ! Thus, when the path difference is an integral number (n) where $ = tan ( + = 25.3 , 0.44 rad. ' (4 +1.5)* of wavelengths or the phase difference is an even number " y1 + y2 = 6.08cos(#t + 0.44 rad). of +, ( y1 + y2) is a maximum. This condition is known as constructive interference. P r 1 r 2 S 1 S 2

The resultant at P is: Question 3: # !2 " !1& # !2 + !1& y1 + y2 = 2Acos% ( cos% )t " ( . $ 2 ' $ 2 ' If a transparent sheet of plastic of thickness 1.00mm and (! " ! ) * 3* 5* (2n +1) refractive index 1.25 is placed in the path of a ray, But, if 2 1 = , , , etc = *, µ = 2 2 2 2 2 what is the optical path difference ( ! r) it introduces? the resultant disturbance ( y1 + y2) = 0. That occurs when (! " ! ) + k * (2n +1) 2 1 = = ,r = ,r = * 2 2 2 - 2 ODD # 1& 1 .00mm i.e., ,r = % n + ( - or + = (2n +1)* $ 2' So, when the path difference is a half-integer number of wavelengths or the phase difference is an odd number of *, ( y1 + y2) = 0. This is destructive interference. The general wave equation L ( a) Earlier we wrote: y (t,x) = A cos(!t " kx) ( b) to describe the wave. However, this is not the only t expression we could have chosen. For example, Let the overall distance be L, the thickness of the plastic y(t,x) = Bsin(!t " kx) and y(t,x) = Ce"i(!t" kx) be t and the wavelength of the light be !. Then, the number of wavelengths in case (a) is are also waves. They may look different but they are all L solutions of the “general wave equation” Na = , ! #2y 1 #2y 2 = 2 2 . and the number of wavelengths in case (b) is # x v #t L " t t L " t µt L + (µ "1)t Nb = + = + = . ! ! # ! ! ! Example: if y (t, x) = A cos(!t " kx), 2 So, N > N , and the optical path difference is #y # y 2 b a = kA sin(!t " kx) and 2 = "k Acos(!t " kx), #x #x % L + (µ "1)t L( r N N 1 t. 2 $ = ( b " a )! = ' " * ! = (µ " ) #y # y 2 & ! !) = "!Asin(!t " kx) and 2 = "! Acos(!t " kx). #t #t "3 Using t = 1.00 +10 m and µ = 1.25, we get #2y $ k' 2 #2y #2y 1 #2y 2 = & ) 2 , i.e., 2 = 2 2 , $ r = 0.25+1.00 +10"3m = 0.25mm. # x % !( #t # x v #t where v is the phase velocity of the wave. Preliminary observations prior to discussing interference Standing waves and diffraction effects.

Consider two waves with the same amplitude traveling in Look at what happens opposite directions along the x-axis, when plane water waves, i.e., y R = a cos(!t " kx) and y L = a cos(!t + kx). traveling from the left, Then applying the superposition principle, the resultant is encounter a barrier with a y R + yL = a cos(!t " kx) + a cos(!t + kx) small opening. The waves Using the trignometric relationship: on the right are circular and centered on the opening just # A + B& # A " B& as if there was a point source of waves at the opening. cosA + cosB = 2cos% ( cos% ( , $ 2 ' $ 2 ' we get

y (x,t) = yR + yL = 2a cos(!t)cos("kx) d = 2a cos(!t)cos(kx). This is a standing wave; because the crests and troughs “stand in-place” but the amplitude varies sinusiodally ( a) ! << d ( b) ! " d ( c) ! >> d with a maximum of 2a! If we vary the wavelength and the size of the opening we find that when ! << d, as in (a), the ray approximation is See animation ... valid but when ! " d the waves spread out. The http://www.walter-fendt.de/ph14e/stwaverefl.htm spreading effect in (b) and (c) is called diffraction. P Two-slit interference pattern (Young, 1801) r 1 y n P S 1 r r 1 2 d ! y n ! S 1 S 2 r 2 d * * S d sin ! 2 L L >> d

d sin * Thus, we have interference maxima at angle !, when L L >> d " r = dsin !n = n#, and interference minima when If the path difference to P between the two rays is $ 1' ! r = (r2 " r1) = n#, "r = dsin !n = & n + ) #. % 2( where n is an integer, we have constructive interference, The phase difference at point P is i.e., maximum disturbance at the point P on the screen. 2+ However, if * = k"r = dsin !. # $ 1' !r = (r2 " r1) = & n + ) #, If L >> d, y = L tan! . For small angles, tan! , sin !, % 2( n n so the condition for maximum disturbance at P is we have destructive interference, i.e., zero disturbance $ #L' at the point P. yn = n& ) . ( n = 0, ±1, ... ). % d ( Let E 1 be the electric field at P due to the waves from S 1 and E 2 be the electric field at P due to the waves from S 2. For zero disturbance at P, Since both electric fields result from the same single source ! 1$ 'L yn = # n + & . ( n = 0, ±1, ... ). that illuminates the two slits, they have the same frequency " 2% d ! and amplitude, and the E -vectors will be parallel. When So, providing ( ) small, the positions of the maxima and they reach P, they will have a phase difference ! = k"r. minima are separated by a distance So, we can represent the individual wave functions at P as 1 'L *y = , 2 d E 1 = A" sin #t and E 2 = A" sin(#t + !). i.e., they are equally spaced. The resultant is E = E1 + E2 = A" sin #t + A" sin(#t + !). If light is used as the source, what is the intensity Using the identity ($ & %) ($ + %) distribution across the screen? sin $ + sin % = 2cos sin , 2 2 we find With all e-m radiation (like light) the “disturbance” or ! ' !* “wave function” is the electric field vector, i.e., E = 2A" cos .sin) #t + , . ! ! 2 ( 2+ E = A " sin +t. ! Thus the amplitude of the wave is 2A" cos . 2 The intensity of e-m radiation is proportional to the square of the amplitude, i.e., 2 ! I = 4I! cos , 2

where I ! is the intensity of the light on the screen from 2# either slit separately and ! = k"r = dsin %. $ Intensity Question 4: 4I !

Using a conventional two-slit apparatus and light that has I av = 2I! a 589nm wavelength, 28 bright fringes per centimeter are

y observed near the center of a screen 3.00m away. What is the slit separation?

n = , 4 , 3 , 2 , 1 0 1 2 3 4 An alternating intensity pattern is observed on the screen, called an interference pattern. Maximum intensity occurs & $L) where y = n( + , with n = 0, ±1, ± 2, etc. ' d * From earlier, when ! is small, the spacing between bright 1$ #L' and dark fringes is "y = & ) . Since bright and dark 2% d ( fringes are equally spaced, the spacing between Class discussion problem: At the dark fringes there is neighboring bright fringes is zero light intensity and so no energy is arriving. When +2 #L 1*10 m/cm +4 light waves interfere and produce an interference pattern "yn,n+1 = = = 3.57 *10 m . d 28 fringes/cm what happens to the energy in the light waves? +9 #L (589 *10 m)(3 m) +3 ,d = = +4 = 4.95 *10 m "yn,n+1 3.57 *10 m = 4.95mm. At the dark fringes there is zero light intensity and so no Conditions for interference energy is arriving. When light waves interfere and produce an interference pattern what happens to the To observe interference, energy in the light waves? the sources must be coherent, i.e., they must maintain a The energy is re-distributed non-uniformly. The energy constant phase difference with respect to each other. in the dark regions is less than the average; the energy in the bright regions is above the average. Class discussion problem: Which of the following are Intensity coherent sources: 4I ! (a) two candles, I av = 2I! (b) a point source and its image in a plane mirror, (c) two pinholes illuminated by the same source, y (d) two headlights of a car, So, the energy “missing” in the dark fringes is (e) two images of a point source reflecting from the “transferred” to the bright fringes. top and bottom surfaces of a glass block. Class discussion problem: If you were to blow smoke into the region between the slits and the viewing screen, would the smoke show evidence of the interference?

Class discussion problem: What would the interference P pattern look like if the light was composed of two r 1 different wavelengths; say, red light with ! R = 750nm S and blue light with 430nm? 1 ! B = r 2

S 2 Note: we cannot use the small angle approximation

400m since d ! L ! y. 300m 1000m

Question 5: (a) From earlier the condition for a maximum is " r = dsin # = n$. At the n = 2 maximum, Two radio antennas, separated by 300m, simultaneously %1 ! # = tan (400 m 1000 m) = 21.8 . broadcast identical coherent signals. A radio in a car dsin # (300 m) sin 21.8! traveling north receives the &$ = = = 55.7 m. n 2 signals. (a) If the car is at (b) After the n = 2 maximum, the next minimum occurs 400m the position of the second when the path length is one-half wavelength greater, 300m 1000m maximum, what is the i.e., when wavelength of the signals? ' 1* 5 (b) How much farther north does the car have to travel dsin # = ) n + , $ = $. ( 2+ 2 to reach the next minimum? 1' 5$* 1' 5 - 55.7 m* &# = sin % ) , = sin % ) , = 27.7!. ( 2d+ ( 2 - 300 m + ! So at the minimum: y = (1000 m) tan27.7 = 525 m. Therefore, the car must travel an additional " y = (525% 400) m = 125 m. Interference in thin films and air gaps

Consider a thin film (of water or oil) that is viewed at a small angle to the normal. Some of the light from the Reflection of waves source is (directly) reflected from the top surface (#1). Most light enters the film and is refracted. Some of that When a traveling wave or light is reflected from the Incident pulse pulse encounters a change in lower surface and refracted at the medium, part of the wave #1 #2 the upper surface (#2). At the (or all of the wave, if it is a upper surface, a change of hard medium) is reflected. phase of occurs on t ! However, the reflected pulse reflection. At the lower or wave is inverted, i.e., it surface, there is no change of phase on reflection. If the Reflected pulse undergoes a phase change of rays are nearly parallel, the path difference is " r = 2t so ! (180 !). the phase difference is k #" r = 2! 2t, ( $ # ) http://www.kettering.edu/~drussell/Demos/reflect/reflect.html where $ is the wavelength of light in the medium $ # = n (of refractive index n). n 1 = 1 Because of the phase change on reflection at the upper n 2 " 1.33 surface, the total phase difference between #1 and #2 is n 3 " 1.5 ! = 2" 2t + ". ( $ # ) So, if a film of water is on a glass slide, say, rays • if 2t = m$ # , i.e., a whole number of wavelengths, reflected at the air-water interface and at the water- then glass interface both undergo a phase change of !, so the ! = " + 2"m = (2m +1)" ( = ", 3", 5" …), phase difference between these rays when they emerge and we get destructive interference. in air is determined solely by the thickness of the water film, i.e., 2t. $ # • if 2t = (2m +1) 2, i.e., an odd number of half- Air wedge wavelengths, then ! = " + (2m +1)" = 2(m +1)" ( = 2", 4", 6" …) 1 Consider a wedge between and we get constructive interference. 2 t two glass slides. Ray #1 is t # = x reflected from the upper Note: there is only a change in phase of " if the x glass- air interface, where reflection occurs at an interface where the incident and there is no change of phase. Ray #2 is reflected from reflected rays are in the less dense medium; a so-called the lower air-glass interface, where there is a change of hard reflection. phase of !. Therefore, the total phase difference between rays #1 Newton’s rings and #2 is ! = k"r + # = 2# 2t + #. ( $) When 2t = m$, then ! = 2#m + # = (2m +1)# ( = #, 3#, 5# …), i.e., destructive interference occurs. t But, since % = x, that condition occurs when m 2t 2x% , i.e., x m$ . = $ = $ = 2% So, dark fringes appear that are equally spaced in x. The linear density of dark fringes, i.e., the number per unit They are caused by interference length is between light reflected from the m 2% . x = $ bottom of the curved surface and air gap the light reflected from an optically optical flat Shown alongside are the flat surface. If the curved surface fringes produced by two very is spherical the fringes are circular. Since the thickness flat glass plates inclined at a of the air gap increases away from the point of contact, very small angle. the fringe spacing decreases. 2r L

From the notes, dark fringes appear when 2t = m!. Therefore, if dark fringes appear when 2t = m!, bright fringes occur when " 1% Question 6: 2t = $ m + ' !, ( m = 0, 1, 2, 3 …). # 2& 2t 1 Light, with wavelength 600nm, is used to illuminate two (m = ) . ! 2 glass plates at normal incidence. The plates are 22.0cm The maximum value of m occurs when t = 2r. Then in length. They touch at one end but are separated at the 2(2r) 1 4 * 0.025*10)3 1 other end by a wire of radius 0.025mm. How many m = ) = ) = 166.2. ! 2 600 *10)9 2 bright fringes appear along the length of the plates? Hence, m max = 166, since m must be a whole number.

Note, the spacing of bright fringes is L 22.0 *10)2 = = 1.33*10)3m (1.33mm) m max 166 Class discussion question:

Class discussion question: Consider the thin film we looked at previously. What difference, if any, would it make to the appearance of Consider the air-wedge shown below, where the air gap the fringes if white light was used instead of increases linearly. Is the first band (fringe) at the right monochromatic light? hand edge - where the gap is zero - bright or dark?

#1 #2