ENGINEERING PHYSICS

PHYSICAL OPTICS

QUESTION BANK

Prepared By:Dr. Sripathi Punchithaya K,Professor, Dept. of Physics, MUJ.

Interference patterns are not standing waves. The interference patterns in are not standing waves, though they have some similarities to the standing-wave patterns. In a standing wave, the interference is between two waves propagating in opposite directions; a stationary pattern of antinodes and nodes appears, and there is no net energy flow in either direction (the energy in the wave is left "standing"). In Interference pattern, there is likewise a stationary pattern of anti nodal and nodal curves, but there is a net flow of energy outward from the two sources. From the energy standpoint, all that interference does is to "channel" the energy flow so that it is greatest along the anti nodal curves and least along the nodal curves. Interference patterns are not standing waves !!

Let monochromatic light fall on to a thin film. If the film has thickness t, the light is at normal incidence and has wavelength in the film; if neither or both of the reflected waves from the two surfaces have a half-cycle reflection phase shift, the conditions for constructive and destructive interference are 2t = m . (m = 0, 1,2,... ) (constructive reflection from thin film, no relative phase shift) 2t=(m+) (m = 0,1,2, ... ) (destructive reflection from thin film, no relative phase shift)

If one of the two waves has a half-cycle reflection phase shift, the conditions for constructive and destructive interference are reversed: 2t=(m+) (m = 0,1,2, ... ) (constructive reflection from thin film, half - cycle relative phase shift). 2t = m . (m = 0, 1,2,... ) (destructive reflection from thin film, half - cycle relative phase shift). Interference due to the transmitted light.

Chapter One InterferenceShort Answer Questions If we are to observe interference in a thin film, why must the film not be very thick (with thickness only on the order of a few wavelengths)?What is the necessary condition on the path length difference between two waves that interfere (a) constructively and (b) destructively?

Why only thin films? In order for two waves to cause a steady interference pattern, the waves must be coherent, with a definite and constant phase relationship. However, the sun and light bulbs emit light in a stream of short bursts, each of which is only a few micrometers long (1 micrometer = l,m = 10- 6 m). If light reflects from the two surfaces of a thin film, the two reflected waves are part of the same burst (Fig. 1Left)

Why only thin films?... Hence these waves are coherent and interference occurs as we have described. If the film is too thick, however, the two reflected waves will belong to different bursts (Fig. 2Right). There is no definite phase relationship between different light bursts, so the two waves are incoherent and there is no fixed interference pattern. That's why one can see interference colors in light reflected from an oil slick a few micrometers thick, but one can not see such colors in the light reflected from a pane of window glass with a thickness of a few millimeters (a thousand times greater).

*The wavelength of light n in a medium whose index of refraction is n - wavelength of the light in free space.

Concept of path length difference, phase and index of refractionPath length difference = Phase difference = Rays are in phase if where m=1, 2, 3..Rays are out of phase if where m=1,2,31 l is the same as 2p radian (rad), l/2 is the same as p rad, etc.Optical Path length = Refractive index x Actual path length

Interference: ExampleA red light beam with wavelength l=0.625mm travels through glass (n=1.46) a distance of 1mm. A second beam, parallel to the first one and originally in phase with it, travels the same distance through sapphire (n=1.77). How many wavelengths are there of each beam inside the material? In glass, lg=0.625mm/1.46= 0.428 mm and Ng=D/ lg=2336.45In sapphire, ls=0.625mm/1.77= 0.353 mm (UV!) and Ns=D/ ls=2832.86What is the phase difference in the beams when they come out?The difference in wavelengths is Ns-Ng=496.41. Each wavelength is 360o, so DN=496.41 means Df=DNx360o=0.41x360o=148oHow thick should the glass be so that the beams are exactly out of phase at the exit (destructive interference!)DN=D/ ls- D/ lg= (D/ l)(n2-n1)=0.31 (D/ l)=m+1/2A thickness D=(m+0.5) 2.02 mm would make the waves OUT of phase.For example, 1.008 mm makes them in phase, and 1.010 mm makes them OUT of phase.

Chapter One InterferenceShort Answer Questions #01What changes, if any ,occur in the interference fringes if the double slit arrangement is placed under water of refractive index n?Answer: Upon immersing the entire set up in water of Refractive index n fringe width decreases. [ Fringe width = D /n d ]; D= distance between the source and the screen, d= distance between the slits and = wavelength of the light used.

Chapter One InterferenceShort Answer Questions #01(a)If Youngs double-slit experiment were performed under water, how would the observed interference pattern be affected?

Chapter One InterferenceShort Answer Questions #01(a)Underwater, the wavelength of the light would decrease, water=air/nwater. Since the positions of light and dark bands are proportional to , the underwater fringe separations will decrease.

InterferenceShort Answer Question #02If one of the slits in the Youngs double slit experiment is covered with an opaque object, what change would occur in the intensity of the light at the center of the screen?Answer: Intensity will decrease to (1/4)th of the earlier value.

InterferenceShort Answer Question #02(a)If one of the slits in the Youngs double slit experiment is covered with an opaque object, what would happen to the intensity of the light at the location of the first minimum?Answer: Intensity will increase.

InterferenceShort Answer Question #02(b)In Youngs double slit experiment what will happen to the fringe with if the separation of the double slits decreases?

Answer: If the separation of the double slits decreases, the separation of the fringes increases.

InterferenceShort Answer Question #02(c)In Youngs double slit experiment what will happen to the fringe pattern with if the monochromatic light is replaced by white light?Answer: If white light is used the central fringe is white and the fringes on either side are coloured.

InterferenceShort Answer Question #02(d)In Youngs double-slit experiment, why do we use monochromatic light? If white light is used, how would the pattern change?

Every color produces its own pattern, with a spacing between the maxima that is characteristic of the wavelength. With several colors, the patterns are superimposed and it can be difficult to pick out a single maximum. Use of monochromatic light can eliminate this problem.

Coated lens reduces reflectionsthrough thin-film interference.

Blooming of Lenses (1)The process of coating a film on the lens is called blooming.A very thin coating on the lens surface can reduce reflections of light considerably.http://users.erols.com/renau/thinfilm.html

Blooming of Lenses (2)The amount of reflection of light at a boundary depends on the difference in refractive index between the two materials.Ideally, the coating material should have a refractive index so that the amount of reflection at each surface is about equal. Then destructive interference can occur nearly completely for one particular wavelength. http://www3.ltu.edu/~s_schneider/physlets/main/thinfilm.shtml

Blooming of Lenses (3)The thickness of the film is chosen so that light reflecting from the front and rear surfaces of the film destructively interferes.For cancellation of reflected light,

InterferenceShort Answer Question #03To minimize reflections, camera lenses will be coated with non-reflecting coatings and hence little light is reflected by the lens in far end of visible region. Hence they look purple.Why do coated lenses look purple by reflected light?

InterferenceShort Answer Question #04Why Newtons Rings are circular ?Answer: In a Newtons ring set up a plano - ( or double) convex lens is placed on a glass plate such that its convex surface touches the glass plate. The air film thus possesses a radial symmetry about the point of contact. When this is illuminated normally by a monochromatic light, interference fringes which are loci of points of equal optical thickness are observed. Thus because of this radial symmetry, Newtons rings are circular in nature.

InterferenceShort Answer Question #05In a Newtons Ring experiment, is the central spot, as seen by reflection, dark or light? Explain.

At the central point, the lens and the glass plate are in contact and hence the thickness of the air film is very small. Excessively thin films appear dark in the reflected light because of destructive interference. Hence the central spot is dark.

InterferenceShort Answer Question #06In Youngs double slit experiment, What will happen if the source slit is moved nearer to the double slits ?Answer:If the source slit is moved nearer to the double slits the separation of the fringes is unaffected but their brightness increases.

InterferenceShort Answer Question #07In Youngs double slit experiment, What will happen to the fringe width if the separation of the double slits decreases?Answer:If the separation of the double slits decreases, the separation of the fringes increases.

InterferenceShort Answer Question #08What will happen to the spacing between the interference fringes if the wavelength of laser light is changed from red to green in a double slit experiment? Why?Answer:

InterferenceShort Answer Question #09What should be the minimum thickness (in terms of wavelength used) and refractive index of a non- reflective coating on lens made up of glass?

Non-reflective CoatingsSince both paths have the same phase change at the interfaces, take only the path differences into account.tFor destructive interferenceExample: l = 550 nm, no reflection

Non-reflective coatings for lens surfaces make use of thin-film interference. A thin layer or film of hard transparent material with an index of refraction smaller than that of the glass is deposited on the lens surface. Light is reflected from both surfaces of the layer. In both reflections the light is reflected from a medium of greater index than that in which it is traveling, so the same phase change occurs in both reflections. If the film thickness is a quarter (one-fourth) of the wavelength in the film (assuming normal incidence), the total path difference is a half-wavelength. Light reflected from the first surface is then a half-cycle out of phase with light reflected from the second, and there is destructive interference. The thickness of the non-reflective coating can be a quarter-wavelength. (t = /4) Answer to Question # 09

InterferenceShort Answer Question #10In a Michelsons Interferometer will the fringe pattern change if the index of refraction of the compensator plate changes? Why?

InterferenceAnswer to Question #10Yes, Changing the index of refraction changes the wavelength of the light inside the compensator plate, and so changes the number of wavelengths within the thickness of the plate. Hence this has the same effect as changing the distance L1 from the beam splitter to mirror M1 which would change the interference pattern.

InterferenceShort Answer Question #11Monochromatic coherent light passing through two thin slits is viewed on a distant screen. Are the bright fringes equally spaced on the screen? If so, why? If not, which ones are closest to being equally spaced?

InterferenceShort Answer Question #12 Soapy water is colorless, but when blown into bubbles it shows vibrant colors. How does the thickness of the bubble walls determine the particular colors that appear?

Interference Answer to Question #12The colors appear due to constructive interference between light waves reflected from the outer and inner surfaces of the soap bubble. The thickness of the bubble walls at each point determines the wavelength of light for which the most constructive interference occurs and hence the color that appears the brightest at that point

InterferenceShort Answer Question #13The top portion of a soapy water film on a vertical loop appears black when viewed by reflected light. Why?Answer: This is because the film by drainage, has become very thin and destructive interference occurs between light reflected from the front and back surfaces of the film.

If the film is very thin, then the interference is totally dominated by the 180 phase shift in the reflection. At the top the film is thinnest (due to gravity it lumps at the bottom), so one sees thefilm dark at the top.This film is illuminated with white light, therefore we see fringes of different colors corresponding to the various constructive interferences of the individual components of the white light, which change as we go down. The thickness increases steeply as we go down, which makes the width of the fringes become narrower and narrower.

InterferenceShort Answer Question #14Why is the lens on a good-quality camera coated with a thin film?A camera lens will have more than one element, to correct (at least) for chromatic aberration. It will have several surfaces, each of which would reflect some fraction of the incident light. To maximize light throughput the surfaces need antireflective coatings. The coating thickness is chosen to produce destructive interference for reflected light of some wavelength.

InterferenceShort Answer Question #15As a soap bubble evaporates, it appears black just before it breaks. Explain this phenomenon in terms of the phase changes that occur on reflection from the two surfaces of the soap film.

As water evaporates from the soap bubble, the thickness of the bubble wall approaches zero. Since light reflecting from the front of the water surface is phase-shifted 180 and light reflecting from the back of the soap film is phase-shifted 0, the reflected light meets the conditions for a minimum.Thus the soap film appears blackInterference Answer to Question #15

InterferenceShort Answer Question #16Explain why two flashlights held close together do not produce an interference pattern on a distant screen.

Interference Answer to Question #16The light from the flashlights consists of many different wavelengths (thats why its white) with random time differences between the light waves. There is no coherence between the two sources. The light from the two flashlights does not maintain a constant phase relationship over time. These three equivalent statements mean no possibility of an interference pattern.

InterferenceShort Answer Question #17Light of wavelength l1 illuminates a double slit, and interference fringes are observed on a screen behind the slits. When the wavelength is changed to l2, the fringes get closer together. How large is l2 relative to l1?ANSWER: l2 is smaller than l1

InterferenceShort Answer Question #18In Youngs double slit experiment, What will be the intensity on the screen if both the sources are replaced by incoherent sources?

InterferenceAnswer to Question #18If the two sources were incoherent, the intensity would be uniform over the screen and would be 2 I0 where I0 is the intensity due to single source.

InterferenceShort Answer Question #19

InterferenceShort Answer Question #20

ANSWERS to MCQs1.A 2.E 3.C 4.B 5.B 6.D 7.C 8.C 9.E 10.D 11.C 12.A 13.D 14.D 15.B 16.E 17.D 18.A 19.E 20.D 21.C 22.D 23.E 24.D 25.D 26.E 27.B 28.B 29.B 30.C31.D 32.D33D 34. A 35.E 36.C 37.B 38.A39.D 40.D

INTERFERENCE MCQ #01In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance they travel by a multiple of: A)a fourth of a wavelength B)a half a wavelength C)a wavelength D)three-fourths of a wavelength E)none of the above C

INTERFERENCE MCQ #02In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in phase by a multiple of: A)/4 B)/2 C) D)3/4 E)none of the above E

INTERFERENCE MCQ #03Waves from two slits are in phase at the slits and travel to a distant screen to produce the second side maximum of the interference pattern. The difference in the distance traveled by the waves is:A)half a wavelengthB)a wavelengthC)three halves of a wavelengthD)two wavelengthsE)five halves of a wavelength

D

INTERFERENCE MCQ #04A monochromatic light source illuminates a double slit and the resulting interference pattern is observed on a distant screen. Let d = center-to-center slit spacing, a = individual slit width, D = screen-to-slit distance, E = adjacent dark line spacing in the interference pattern. The wavelength of the light is then: A)dE /D B)Ld/a C)da/D D)E D/a E)Dd/E Answer: A

INTERFERENCE MCQ #05In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to: A)D/2 B)D/2 C)D2 D)2D E)4D Answer: D

INTERFERENCE MCQ #06In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: A)0.10 cm B)0.25 cm C)0.50 cm D)1.0 cm E)none of the above Answer: B

INTERFERENCE MCQ #07In a Young's double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. The number of bright fringes per unit length on the screen is: A)Dd/B)D/d C)D/dD)/Dd E)d/DAnswer: E

INTERFERENCE MCQ #08In a Young's double-slit experiment, the slit separation is doubled. This results in: A)an increase in fringe intensity B)a decrease in fringe intensity C)a halving of the wavelength D)a halving of the fringe spacing E)a doubling of the fringe spacing Answer: D

INTERFERENCE MCQ #09A light wave with an electric field amplitude of 2E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?A)wave A has an amplitude of E 0 and a phase constant of zero B)wave B has an amplitude of E0 and a phase constant of C)wave C has an amplitude of 2E0 and a phase constant of zero D)wave D has an amplitude of 2E0 and a phase constant of E)wave E has an amplitude of 3E0 and a phase constant of Answer: C

INTERFERENCE MCQ #10A light wave with an electric field amplitude of 2E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the least intensity?A)wave A has an amplitude of E 0 and a phase constant of zero B)wave B has an amplitude of E0 and a phase constant of C)wave C has an amplitude of 2E0 and a phase constant of zero D)wave D has an amplitude of 2E0 and a phase constant of E)wave E has an amplitude of 3E0 and a phase constant of Answer: D

INTERFERENCE MCQ #11One of the two slits in a Young's experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result: A)the fringe system disappears B)the bright fringes get brighter and the dark ones get darker C)the fringes just get dimmer D)the dark fringes just get brighter E)the dark fringes get brighter and the bright ones get darker

Answer: E

INTERFERENCE MCQ #12In a Young's double-slit experiment, a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern (on the screen) shifts by an amount corresponding to 30 dark bands. The wavelength of the light in this experiment is 480 mm and the index of the mica is 1.60. The mica thickness is: A)0.090 mm B)0.012 mm C)0.014 mm D)0.024 mm E) 0.062 mmAnswer: D

INTERFERENCE MCQ #13Monochromatic light, at normal incidence, strikes a thin film in air. If denotes the wavelength in the film, what is the thinnest film in which the reflected light will be a maximum? A)much less than B) /4 C) /2 D)3 /4 E) Answer: B

INTERFERENCE MCQ #14A soap film, 4 105 cm thick, is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Which wavelengths will be intensified in the reflected beam? A)400 nm and 600 nm B)480 nm and 800 nm C)360 nm and 533 nm D)400 nm and 800 nm E)510 nm and 720 nm Answer: B

INTERFERENCE MCQ #15A soap film, 4 105 cm thick, is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Which wavelengths will be intensified in the reflected beam? A)400 nm and 600 nm B)480 nm and 800 nm C)360 nm and 533 nm D)400 nm and 800 nm E)510 nm and 720 nm Answer: B

INTERFERENCE MCQ #16A liquid of refractive index n = 4/3 replaces the air between a fixed wedge formed from two glass plates as shown. As a result, the spacing between adjacent dark bands in the interference pattern: A)increases by a factor of 4/3 B)increases by a factor of 3 C)remains the same D)decreases to 3/4 of its original value E)decreases to 1/3 of its original value

Answer: D

INTERFERENCE MCQ #17In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th? A)2400 nm B)4800 nm C)240 nm D)480 nm E)none of these Answer: A

INTERFERENCE MCQ #18A glass (n = 1.6) lens is coated with a thin film (n = 1.3) to reduce reflection of certain incident light. If is the wavelength of the light in the film, the least film thickness is: A)less than /4 B) /4 C) /2 D)E)more than ANSWER: B

INTERFERENCE MCQ #19If two light waves are coherent: A)their amplitudes are the same B)their frequencies are the same C)their wavelengths are the same D)their phase difference is constant E)the difference in their frequencies is constant ANSWER: D

1. They get brighter but otherwise do not change.2. They get brighter and closer together.3. They get brighter and farther apart.4. They get out of focus.5. They fade out and disappear.20. Suppose the viewing screen in the figure is moved closer to the double slit. What happens to the interference fringes?

1. They get brighter but otherwise do not change.2. They get brighter and closer together.3. They get brighter and farther apart.4. They get out of focus.5. They fade out and disappear.20. Suppose the viewing screen in the figure is moved closer to the double slit. What happens to the interference fringes?

What was the first experiment to show that light is a wave? 1. Youngs double slit experiment2. Galileos observation of Jupiters moons3. The Michelson-Morley interferometer4. The Pound-Rebka experiment5. Millikans oil drop experimentINTERFERENCE MCQ # 21

What was the first experiment to show that light is a wave? 1. Youngs double slit experiment2. Galileos observation of Jupiters moons3. The Michelson-Morley interferometer4. The Pound-Rebka experiment5. Millikans oil drop experimentINTERFERENCE MCQ # 21

A Michelson interferometer using light of wavelength l has been adjusted to produce a bright spot at the center of the interference pattern. Mirror M1 is then moved distance l toward the beam splitter while M2 is moved distance l away from the beam splitter. How many bright-dark-bright fringe shifts are seen?1. 02. 13. 24. 3 5. 4INTERFERENCE MCQ # 22

A Michelson interferometer using light of wavelength l has been adjusted to produce a bright spot at the center of the interference pattern. Mirror M1 is then moved distance l toward the beam splitter while M2 is moved distance l away from the beam splitter. How many bright-dark-bright fringe shifts are seen?1. 02. 13. 24. 3 5. 4INTERFERENCE MCQ # 22

Radar waves have a wavelength of 3cm.Suppose the plane is made of metal(speed of propagation=0, n is infinite andreflection on the polymer-metal surfacetherefore has a 180 degree phase change).The polymer has n=1.5. Same calculation as in previous example gives,On the other hand, if one coated a plane with the same polymer(for instance to prevent rust) and for safety reasons wanted to maximizeradar visibility (reflective coating!), one would haveAnti-Reflective CoatingsStealth Fighter

Example 1: Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if light of wavelength 600 nm is used?x = 2 m; d = 0.08 mml = 600 nm; y = ?The third dark fringe occurs when n = 5d sin q = 5(l/2)

Example 1 (Cont.): Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if l = 600 nm?y = 3.75 cm

Example 2: Light (600 nm) strikes a grating ruled with 300 lines/mm. What is the angular deviation of the 2nd order bright fringe?To find slit separation, we take reciprocal of 300 lines/mm:Lines/mm mm/line

Example (Cont.) 2: A grating is ruled with 300 lines/mm. What is the angular deviation of the 2nd order bright fringe?q2 = 21.10Angular deviation of second order fringe is:

Example 3: Monochromatic light shines on a single slit of width 0.45 mm. On a screen 1.5 m away, the first dark fringe is displaced 2 mm from the central maximum. What is the wavelength of the light?l = 600 nm

Example 4: The tail lights (l = 632 nm) of an auto are 1.2 m apart and the pupil of the eye is around 2 mm in diameter. How far away can the tail lights be resolved as separate images?p = 3.11 km

Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.382 mm, causing the interferometer pattern to reproduce itself 1 700 times. Determine the wavelength of the light. What color is it?

Youngs double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slitsTwo narrow, parallel slits separated by 0.250 mm are illuminated by green light ( = 546.1 nm). The interference pattern is observed on a screen 1.20 m away from the plane of the 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 a double slit arrangement let L = 120 cm and d = 0.250 cm. The slits are illuminated with coherent 600-nm light. Calculate the distance y above the central maximum for which the average intensity on the screen is 75.0% of the maximum.

A thin film of oil (n = 1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no blue light at 512 nm. How thick is the oil film?

A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating? A beam of 580-nm light passes through two closely spaced glass plates, as shown in Figure P37.37. For what minimum nonzero value of the plate separation d is the transmitted light bright?

An oil slick (n = 1.40) on water (n = 1.33) appears yellow ( = 469 nm) when illuminated by the sun directly overhead. What is the minimum thickness of the oil slick?Constructive I/F when 2t = (m+1/2)n = (m+1/2)/nt = 83.8 nm

A soap bubble (n = 1.33) floats in the air. If the thickness of the bubble wall is 115 nm, what wavelength of light will be most visible?Constructive I/F when 2t = (m+1/2)n = (m+1/2)/n = 2nt/(m+1/2) = 612 nm (orange)

A thin wire is placed between the ends of two glass plates, forming an air wedge. When light with a wavelength of 600 nm falls nearly perpendicularly on the plates, 30 dark fringes are seen. What is the thickness of the wire?

Relative phase inversionDestructive interference: 2 t = mt = m / 2 = 8.7 m (m = 29: first fringe is m=0)

Wedge Film Example

Wedge Film Example 2A sheet of paper of thickness 41.0 m is used to separate two glass plates 5.00 cm long.

If light of wavelength 526 nm is used to illuminate the plates, how many dark fringes can be seen?What is the spacing of the dark fringes?Relative phase inversionDestructive interference: m = 2 t / = 155+1 = 156 (fringe at 0)Spacing = 5.00 cm / 156 fringes = .0321 cm/fringe

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Determine the wavelength of the light.Bright fringes:d

Example: A viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the distance between adjacent bright fringes.Bright fringes:d

Example: a viewing screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. Find the width of the bright fringes.Define the bright fringe width to be the distance between two adjacent destructive minima.d

Thin Film Interference: Path Length DifferenceExample: light of wavelength 600 nm in air is perpendicularly incident on a piece of glass 4.1 m thick. The index of refraction of glass is 1.5. Some of the light is reflected off the back surface of the glass. How many light waves are contained along the path of this light through the glass? AirGlasstAirLight enters the glasspasses through the glass and reflects off the back surfacepasses back through the glassand exits.Some probably reflects back off the front surface, but we are not interested in that light.Some probably passes through the second glass surface, but we are not interested in that light.Some probably reflects back into the glass, but we are not interested in that light.

Thin Film Interference: Path Length DifferenceHow many waves can fit in the path of length 2t?How many light waves are contained along the path of this light through the glass? AirGlasstAir

Thin Film Interference: Path Length DifferenceThe outgoing waves would differ in phase by wavelength from the incoming wavesAre the outgoing waves in phase or out of phase with the incoming waves AirGlasstAirexcept that you must also consider phase shift due to reflection.Note: if you look down at the glass, your eye sees only the reflected waves; you will not see interference of the incident and reflected waves, so you are not being asked if interference between incident and reflected waves will take place.

Example: a glass lens is coated on one side with a thin film of MgF2 to reduce reflection from the lens surface. The index of refraction for MgF2 is 1.38 and for glass is 1.50. What is the minimum thickness of MgF2 that eliminates reflection of light of wavelength = 550 nm? Assume approximately perpendicular angle of incidence for the light. Both rays and experience a 180 phase shift on reflection so the total phase difference is due to the path difference of the two rays.AirMgF2tn= 1.38nAir = 1.00180 phase change180 phasechangeglass, ng =1.50

The minimum thickness is for m=0.The reflected light is minimum when the two light rays meet the condition for destructive interference: the path length difference is a half-integral multiple of the light wavelength in MgF2.

Example: two glass plates 10 cm long are in contact on one side and separated by a piece of paper 0.02 mm thick on the other side. What is the spacing between the interference fringes? Assume monochromatic light with a wavelength in air of = 500 nm incident perpendicular to the slides. Ray is not phase shifted on reflection. Ray is shifted 180 on reflection.HtxL = 10 cmH = 2x10-5 mFor destructive interferenceThe light that is partly reflected at the bottom of the first glass surface and partly transmitted is responsible for the interference fringes.**This reference explains why there is no visible interference due to the relatively thick glass plates themselves.

Successive dark fringes are separated by 1.25 mm.x is the distance from the contact point to where destructive interference takes place.

Successive bright fringes occur for m+ and (m+1)+.For constructive interference

Successive bright fringes are separated by 1.25 mm.

Example: suppose the glass plates have ng = 1.50 and the space between them contains water (nw = 1.33). What happens now? Ray is not phase shifted on reflection. Ray is shifted 180 on reflection. Both are the same as before.HtxL = 10 cmH = 2x10-5 mFor destructive interferenceBut the path difference now occurs in water, where the light will have a wavelengthRepeat the calculation, using water.

Successive dark fringes are separated by 0.94 mm.For destructive interference, we now have

39. A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.30).(a) If you are looking straight down from an airplane while the Sun is overhead at a region of the slick where its thickness is L=460 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? Path difference between ray 1and ray 2 = 2L. Phase changes cancel outFor constructive interference path differencemust = integral number of wavelengthsWe note that only the 552 nm wavelength falls within the visible light range.

(b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest? (Hint: use figure (a) with appropriate indices of refraction.) Scuba diverFor transmission, ray 2 undergoes 180 deg phase shift upon reflection at theKerosene-water interface. Therefore, for constructive interference 2L= integral number of wavelengths in n2 plus half a wavelength.

We note that only the 441.6 nm wavelength (blue) is in the visible range, Visible spectrum is 430 nm - 690 nm

27. S1 and S2 in Fig. 36-29 are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power. (a) If a detector is moved to the right along the x-axis from source S1, at what distances from S1 are the first three interference maxima detected?detector

The solution for x of this equation is

For constructive interference we have

Solve for x

What about m = 4 ? This corresponds to x=0. Path difference =4 meters.

Although the amplitudes are the same at the sources, the waves travel different distances to get to the points of minimum intensity and each amplitude decreases in inverse proportion to the square of the distance traveled. The intensity is not zero at the minima positions.Where do the minima occur?m=0 x=15.75 mm=1 x =4.55 mm=2 x=1.95 mm=3 x= 0.55 m

Demo with speakers using sound wavesSet oscillator frequency to 1372 Hz,Then wavelength of sound is 343/1372 =0.25 mSet speakers apart by 1m. Then maxima occur at

Example Problem 1In a double slit interference experiment, the slits are 10 micron (10-6 meters) apart and the screen is 2 meters away. If 500nm wavelength light is used, find a) the location of the first dark fringe, b) the location of the 3rd bright fringe, c) the spacing between fringes, d) the theoretical number of bright fringes possible.

a) d sin = (m+1/2), 1E-5 sin =(1/2) 5E-7 = 1.43y = L tan = 2.0 tan(1.43) = 0.05mb) d sin = (m), 1E-5 sin = (3) 5E-7 = 8.63y = L tan = 2.0 tan(8.63) = 0.30mc) d sin = (1), = 2.86y = 0.10md) let maximum = 90, d sin 90 = (m), m=20. This is for 1 side and there is a middle fringe total = 41

Example Problem 2What is the minimum thickness of a soap bubble film with index of refraction 1.33 that would reflect 650nm most brightly? b) What is the minimum thickness for an anti-reflecting coating of index of refraction 1.4 or a glass of index 1.5 which would reflect no green light of wavelength 550nm? c) what would be the color of the light that is reflected off the lens

n= 1.33, o = 650nm, = o /n = 488nm, m = 0 = m2-m1

m2 = 2d/ , since m1 =

(due to 180 deg. inversion not present at m2) our path difference can be wavelength. Get wavelength inside material and determine d = m2 * /2 = * 488 / 2=122nm

b) coating n=1.4, glass = 1.5 at 550nm = o /n = 393nm, m1 = , m2 = 2d/ + since both m1 and m2 reflect from greater index of refraction mediums m = = m2-m1 = m2 , m2 = 1 = 550/1.4 = 393,hence m2 =1= 2d/ + , 2d/ = 1/2, d = /4 = 393/4 = 98nm

c) green transmitted, blue & red reflected, = magenta

ENDInterference

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