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  • 1. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Interference Engineering Physics1. Introduction In the 17th century, the properties of light were explained by Sir Isaac Newtonand Christian Huygens. Sir Isaac Newton was explained the properties of light by introducingcorpuscular theory in 1675. It explains reflection, refraction and dispersion properties of light.It fails to explain interference, diffraction, polarization, photo electric effect and doublerefraction. In 1679, Christian Huygens proposed the wave theory of light. According to Huygen’swave theory, each point on the wave front is to be considered as a source of secondarywavelets. It explains reflection, refraction, dispersion, double refraction, diffraction,interference and polarization properties of light. It fails to explain, photo electric effect, black mbody radiation etc,.2. Coherence .co Coherence is a property of waves. Two produce interference pattern coherent sourcesare required. Coherent light sources cannot be obtained from the separate light sources, butthey are obtained from the single light source. blyTwo waves are said to be coherent, the waves must have • Same wavelength • Same amplitude and ee • Constant phase difference s.wDefinition:- A predictable correlation of the amplitude and phase at any one point with other point iscalled coherence. sicThere are two types of coherence i) Temporal coherence (or longitudinal coherence) hy ii) Spatial coherence (or transverse coherence)Temporal coherence (or longitudinal coherence):- gp It is possible to predict the amplitude the amplitude and phase at a one point on the inwave w.r.t another point on the same wave, then the wave is in temporal coherence. To understand this, let us consider two points P1 and P2 on the same wave as in shown erin figure. Suppose the phase and amplitude at any one point is known, then we can easilycalculate the amplitude and phase for any other point on the same wave by using the wave eequation 2ߨ ‫ ݕ‬ൌ ܽ sin ቆ ሺܿ ‫ ݐ‬െ ‫ݔ‬ሻቇ gin ߣWhere ‘a’ is the amplitude of the wave and ‘x’ is the displacement of the wave at any instant oftime‘t’. en P1 P2Spatial coherence (or transverse coherence) It is possible to predict the amplitude the amplitude and phase at a one point on thewave w.r.t another point on the second wave, then the wave is in spatial coherence. P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 1
  • 2. Independent sources are not coherent Two independent sources of light with same power can never be treated as coherentsources; even it emits same wavelength and same amplitude waves. Generally the light isemitted from the source through spontaneous emission process. But the waves emittedtrough spontaneous emission process is in out of phase. So two waves generated by two lightsources are always in out of coherence. Even the waves are emitted from the different parts ofthe lamp, no longer they are coherent.3. Principle of superposition of wavesDefinition:- Principle of superposition of waves states that, the resultant displacement of two ormore waves traveling simultaneously in a medium at any point is equal to the algebraic sum ofthe displacements due to individual waves. m To understand this concept, Let us consider two waves traveling simultaneously in the .comedium. Let ܻଵ and ܻଶ are the displacements of first and second waves at any point in theabsence of other. According to principle of superposition, the resultant displacement ሺܻሻ of two waves blycan be written as ܻ ൌ ܻଵ േ ܻଶ ee Wave 1 + = s.w Wave 2Let two waves are in phase, their resultant amplitude is ܻ ൌ ܻଵ ൅ ܻଶ sicTwo waves are in opposite phase, their resultant amplitude is ܻ ൌ ܻଵ ~ ܻଶ hy Wave 1 gp + = Wave 2 in er4. Young’s double slit experiment The first experimental evidence of interference pattern was given by Thomas Young In e1801. He explained the interference pattern based on the wave nature of light. ginExperimental setup: His experimental arrangement is shown in figure. He passed the sun light through anarrow pin hole S on a opaque surface and then through closely spaced pin holes S1 and S2in another opaque surface. A screen XY was arranged in front of the pin holes S1 and S2. He enobserved few coloured bright and dark bands on the screen. Now a day, the pin holes arereplaced by slits and sun light is replaced by monochromatic light. X S1 S2 S Screen Y
  • 3. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Interference Engineering PhysicsExplanation of wave theory When sun light is passes through pin hole S, spherical waves are spread out.According to Huygen’s wave theory, each point on the wave front is to be considered as asource of secondary wavelets. Hence spherical waves also spread out from pin holes S1 andS2. The radii of the wave fronts increase as they move away from S1 and S2 and hencesuperimpose on each other. In the above figure, continuous circular arcs represent the wave crests while dottedcircular arcs represent the troughs of each wave. A crest fall on the other crest constructiveinterference will occurs and their resultant amplitude is equal to the sum of the amplitudesindividual waves. ܻ ൌ ܻଵ ൅ ܻଶ m A crest fall on the other trough destructive interference will occurs and their resultant .coamplitude is equal to the difference of the amplitudes of individual waves. ܻ ൌ ܻଵ െ ܻଶThus a number of dark and bright bands having equal width is observed on the screen. bly5. Interference of light The best evidence for the wave nature of light is interference phenomenon. This was eeexperimentally demonstrated by Thomas Young in 180, through double slit experiment. Dueto interference, we will observe many observations in our day today life, such as multiple s.wcolours on soap bubbles as well as on oil film when viewed under sun light. Interferenceconcept is explained on the basis of superposition of wave’s concept. When two light waves superimpose, then the resultant amplitude or intensity in sicthe region of superposition is different than the amplitude of individual waves.Definition:- hy The modification in the distribution of intensity in the region of superposition is knownas interference. gpIn case of interference pattern we observe two cases • Constructive interference in • Destructive interference erConstructive interference e The waves are reaching at a point are in phase constructive interference occurs In constructive interference, the resultant amplitude is always equal to the sum of the gin amplitudes of two individual waves. Condition en The path difference between the two waves is equal to the integral multiple of wavelength ሺߣሻ the constructive interference occurs.‫ ݁ܿ݊݁ݎ݂݁݅݀ ݄ݐܽ݌‬ൌ ݊ߣ Where n = 0, 1, 2, 3, 4 … Wave 1 + = Wave 2 P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 3
  • 4. Destructive interference The waves are reaching at a point are in out of phase desstructive interference occurs In Destructive interference, the resultant amplitude is always equal to the difference of the amplitudes of two individual waves. Condition The path difference between the two waves is equal to the odd integral multiple of λ/2destructive interference occurs ሺ2݊ െ 1ሻߣ‫ ݁ܿ݊݁ݎ݂݁݅݀ ݄ݐܽ݌‬ൌ Where n ൌ 1, 2, 3, 4 … ‫ݎ݋‬ 2 ሺଶ௡ାଵሻఒ‫ ݁ܿ݊݁ݎ݂݁݅݀ ݄ݐܽ݌‬ൌ Where n ൌ 0, 1, 2, 3, 4 … ଶ m Wave 1 + .co = Wave 2 bly6. Types of interference:- For the formation of interference pattern, two coherent light sources are required. To eeget two coherent sources form a single light source, two techniques are used. They are 1. Division of wave front 2. Division of amplitude s.wDivision of wave front The wave front from a single light source is divided into two parts using thephenomenon of reflection, refraction or diffraction. Young’s double slit experiment is belongs sicto this class of interference.Division of amplitude The amplitude of a single light beam is divided into two parts by parallel reflection or hyrefraction. Newton’s ring experiment, Michelson’s interferometer is belongs to this class ofinterference. gp7. Conditions for interference in 1) Two light sources of emitting light waves should be coherent. 2) Two sources must emit continuous light waves of same wavelengths or frequency. er 3) The separation between the two sources should be small. 4) The distance between the two sources and the screen should be large. 5) To view interference fringes, the background should be dark. e 6) The amplitude of light waves should be equal or nearly equal. gin 7) The sources should be narrow. 8) The sources should be monochromatic.8. Interference in thin films by reflection enPrinciple:- The formation of colours in thin films can explained as due to the phenomenon ofinterference. In this example, the formation of interference pattern is by the division ofamplitude. . Consider a thin film of uniform thickness ’‫ ’ݐ‬and refractive index ‘µ’. Let a monochromatic light ray AB is incident is on the upper surface of the film with an angle ‘i’. This ray is partially reflected along BC and the remaining part is refracted along BD with an angle ‘r’. The refracted ray BD is reflected into DE from the lower surface of the film and it emerges through the film in the form of light ray EF. The two light rays BC and EF are parallel and superimpose and produce interference. The intensity of interference fringe depends up on the path difference between the ray 1 and ray 2.
  • 5. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST.Unit –I OPTICS - Interference Engineering Physics Ray 1 Ray 2 m .co bly To find out the path difference between the light rays (1) and (2), draw one ee normal DG to BE and from point E to BC. From point E and H onwards the two light rays 1 and 2 travels equal distances. By the time the light ray 1 travels from B to H, the s.w transmitted ray has to travel from B to C and D to E. The path difference between the light rays (1) and (2) is ‫ ݁ܿ݊݁ݎ݂݂݁݅݀ ݄ݐܽ݌‬ൌ ߤሺ‫ ܦܤ‬൅ ‫ܧܦ‬ሻ ݅݊ ݂݈݅݉ െ ‫ݎ݅ܽ ݊݅ ܪܤ‬ (1) sic ஽ீ ௧ From ∆‫ܩܦܤ‬ cos ‫ ݎ‬ൌ ൌ ஻஽ ஻஽ ‫ݐ‬ hy ‫ ܦܤ‬ൌ cos ‫ݎ‬ gp ஽ீ ௧Similarly from ∆‫ܩܧܦ‬ cos ‫ ݎ‬ൌ ൌ ஽ா ஽ா ‫ݐ‬ in ‫ ܧܦ‬ൌ cos ‫ݎ‬ er ௧ ‫ ܦܤ ׵‬ൌ ‫ ܧܦ‬ൌ (2) ௖௢௦ ௥ e ஻ு ஻ு ∆‫ܪܧܤ‬ sin ݅ ൌ ൌ ஻ீାீா ginFrom ஻ா‫ ܪܤ ׵‬ൌ ሺ‫ ܩܤ‬൅ ‫ܧܩ‬ሻ. sin ݅From ∆‫ܩܧܦ∆ ݀݊ܽ ܩܦܤ‬ ‫ ܩܤ‬ൌ ‫ ܧܩ‬ൌ ‫ ݐ‬tan ‫ݎ‬ en‫ ܪܤ‬ൌ ሺ2‫ ݐ‬tan ‫ݎ‬ሻ. sin ݅From Snell’s law at point B sin ݅ ൌ ߤ sin ‫ݎ‬ ‫ ܪܤ ׵‬ൌ 2ߤ‫ ݐ‬tan ‫ . ݎ‬sin ‫ݎ‬ (3)Substituting the equations (2) and (3) in equation (1), we get ଶఓ௧ܲܽ‫ ݁ܿ݊݁ݎ݂݂݁݅݀ ݄ݐ‬ൌ ୡ୭ୱ ௥ െ 2ߤ‫ ݐ‬tan ‫ . ݎ‬sin ‫ݎ‬ P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 5
  • 6. 2ߤ‫ݐ‬ sin ଶ ‫ݎ‬ ൌ െ 2ߤ‫.ݐ‬ cos ‫ݎ‬ cos ‫ݎ‬ 2ߤ‫ݐ‬ ൌ ሺ1 െ sin ଶ ‫ݎ‬ሻ cos ‫ݎ‬ 2ߤ‫ݐ‬ ൌ ܿ‫ ݏ݋‬ଶ ‫ݎ‬ cos ‫ݎ‬ =2ߤ‫ ݐ‬cos ‫ݎ‬ At point B the light ray (1) is reflected at the surface of thin film (denser medium). So the light ray (1) undergoes a phase change π or an additional path difference λ/2. ఒ Total path difference ൌ 2ߤ‫ ݐ‬cos ‫ ݎ‬െ ଶ Constructive interference (or Bright fridge) m General condition; ‫ ݁ܿ݊݁ݎ݂݂݄݁݅݀ݐܽ݌‬ൌ ݊ߣ ߣ 2ߤ‫ ݐ‬cos ‫ ݎ‬െ ൌ ݊ߣ .co 2 ߣ 2ߤ‫ ݐ‬cos ‫ ݎ‬ൌ ݊ߣ ൅ 2 bly ሺ2݊ ൅ 1ሻߣ 2ߤ‫ ݐ‬cos ‫ ݎ‬ൌ 2 ee Destructive interference (or Dark fridge) ఒ General condition: ‫ ݁ܿ݊݁ݎ݂݂݄݁݅݀ݐܽ݌‬ൌ ሺ2݊ ൅ 1ሻ ଶ s.w ߣ ሺ2݊ െ 1ሻߣ 2ߤ‫ ݐ‬cos ‫ ݎ‬െ ൌ 2 2 2ߤ‫ ݐ‬cos ‫ ݎ‬ൌ ݊ߣ sic9. Newton’s rings hyPrinciple:- The formation of Newton’s rings can explained as due to the phenomenon of gpinterference. In this example, the formation of interference pattern is obtained by the divisionof amplitude. inExperimental arrangement er The experimental arrangement ofNewton’s rings is shown in figure. eThe Plano -convex lens (L) of large ginradius of curvature is placed with its Mconvex surface on the glass plate (P).The lens touches the glass plate at enO. A monochromatic light is allowedto fall normally on the lens with thehelp of glass plate M kept at 450 tothe incident monochromatic light. A part of light is reflected bythe curved surface of the lens ‘L’ anda part of light is transmitted is partly Oreflected back by the upper surface ofthe plane glass plate P as shown in figure. These reflected rays interfere and give rise to aninterference pattern in the form of circular fringes. These rings are seen through microscope.
  • 7. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Interference Engineering PhysicsExplanation of formation of Newton’s rings Newton’s rings are formed due to the interference between the light rays reflected fromthe lower surface of the lens and the upper surface of the glass plate (or top and bottomsurfaces of the air film). Let R be the radius of curvature of the lens and ‫ ݐ‬be the thickness of the air film at pointP. At point P a part of the incident monochromatic light is reflected from the bottom surface ofthe glass plate without phase change. The other part of the light is refracted from the uppersurface of glass plate with additional phase change of π or path difference λ/2.The path difference between the two rays is ߣ m 2ߤ‫ ݐ‬cos ‫ ݎ‬൅ 2For air film ߤ ൌ 1 and for normal incidence ‫ ݎ‬ൌ 0, so .co ఒThe path difference ൌ 2‫ ݐ‬൅ ଶ blyConstructive interference (or Bright fridge)General condition: ‫ ݁ܿ݊݁ݎ݂݂݄݁݅݀ݐܽ݌‬ൌ ݊ߣ ߣ 2‫ ݐ‬൅ ൌ ݊ߣ 2 ee ߣ 2‫ ݐ‬ൌ ሺ2݊ െ 1ሻ 2Where ݊ ൌ 0, 1, 2, … … … .. s.wDestructive interference (or Dark fridge) ఒGeneral condition: ‫ ݁ܿ݊݁ݎ݂݂݄݁݅݀ݐܽ݌‬ൌ ሺ2݊ ൅ 1ሻ ଶ sic ߣ ߣ 2‫ ݐ‬൅ ൌ ሺ2݊ ൅ 1ሻ 2 2 hy 2‫ ݐ‬ൌ ݊ ߣWhere ݊ ൌ 0, 1, 2, … … … .. gp ఒAt the point of contact ‫ ݐ‬ൌ 0, path difference is ଶ i.e., the reflected and incidence light are out inof phase and destructive interference occur. So the center fringe is always dark.Theory of Newton’s rings er To find the diameters of a dark and bright fringes construct a circle with the radius ofcurvature R of a lens L and let us choose a point P at a distance ‘r’ from the center of lens and e‫ ݐ‬be the thickness of air film at point ܲ. ginFrom the property of a circle ܰܲ. ܰ‫ ܤ‬ൌ ܱܰ. ܰ‫ܦ‬ ‫ ݎ .ݎ‬ൌ ܶ. ሺ2‫ ݎ‬െ ‫ݐ‬ሻ ‫ ݎ‬ଶ ൌ 2ܴ‫ ݐ‬െ ‫ ݐ‬ଶ enIf t is small ‫ ݐ‬ଶ is negligible. ‫ ݎ‬ଶ ൌ 2ܴ‫ݐ‬ ‫ݎ‬ଶ r r ‫ݐ‬ൌ 2ܴBright rings ఒFor bright ring, the condition is 2‫ ݐ‬ൌ ሺ2݊ െ 1ሻ ଶ 2 ‫ݎ‬ଶ ߣ ൌ ሺ2݊ െ 1ሻ 2ܴ 2 ሺ2݊ െ 1ሻߣܴ ‫ݎ‬ଶ ൌ 2 P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 7
  • 8. By replacing r by D/2 the diameter of the bright ring is ‫ܦ‬ଶ ሺ2݊ െ 1ሻߣܴ ൌ 4 2 ‫ܦ‬ଶ ൌ 2ሺ2݊ െ 1ሻߣܴ ‫ ܦ‬ൌ ඥ2ሺ2݊ െ 1ሻߣܴ ‫ ܦ‬ൌ ඥሺ2݊ െ 1ሻ√2ߣܴ ‫ ן ܦ‬ඥሺ2݊ െ 1ሻ‫ݎܾ݁݉ݑ݊ ݈ܽݎݑݐܽ݊ ݀݀݋√ ן ܦ‬Dark rings mFor dark rings, the condition is 2‫ ݐ‬ൌ ݊ ߣ .co 2 ‫ݎ‬ଶ ൌ݊ߣ 2ܴ bly ‫ ݎ‬ଶ ൌ ݊ ߣܴBy replacing r by D/2 the diameter of the nth dark ring is ee ‫ܦ‬ଶ ൌ ݊ ߣܴ 4 s.w ‫ ܦ‬ൌ √4݊ ߣܴ ‫ ܦ‬ൌ 2√݊ ߣܴ ‫݊√ ן ܦ‬ sic ‫ݎܾ݁݉ݑ݊ ݈ܽݎݑݐܽ݊√ ן ܦ‬ hyNote: suppose a liquid is taken in between the lens and glass plate having refractive index ߤ,then the diameter of the dark nth dark ring can be written as gp √4݊ ߣܴ ‫ܦ‬ൌ ߤ in10. Determination of wave length of sodium light using Newton’s rings er By forming Newton’s rings and measuring the radii of the rings formed, we can ecalculate the wavelength of the light used if the radius of curvature of the lens is known. Let Rbe the radius of curvature of the lens and ߣ is the wavelength of the light used. ginSo the diameter of the nth dark ring can be written as‫ܦ‬௡ ଶ ൌ 4 ݊ ߣ ܴ (1) enSimilarly the diameter of the (n+m)th dark ring is‫ܦ‬௡ା௠ ଶ ൌ 4 ሺ݊ ൅ ݉ሻ ߣ ܴ (2)Subtracting equation (1) from (2) we get ‫ܦ‬௡ା௠ ଶ െ ‫ܦ‬௡ ଶ ൌ 4 ሺ݊ ൅ ݉ሻߣ ܴ െ 4 ݊ ߣ ܴ ‫ܦ‬௡ା௠ ଶ െ ‫ܦ‬௡ ଶ ൌ 4 ݉ ߣ ܴ ‫ܦ‬௡ା௠ ଶ െ ‫ܦ‬௡ ଶ ߣൌ 4ܴ݉Using the above relation wavelength can be calculated
  • 9. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Interference Engineering Physics11. Determination of refractive index of a liquid using Newton’s rings By forming Newton’s rings and measuring the diameter of the rings formed, we cancalculate the refractive index of the liquid. In air film, the diameter of the nth and mth dark rings are ‫ܦ‬௡ and ‫ܦ‬௠ are measuredwith the help of travelling microscope.The diameter of the nth dark ring is‫ܦ‬௡ ଶ ൌ 4 ݊ ߣ ܴ m (1)The diameter of the mth dark ring is .co‫ܦ‬௠ ଶ ൌ 4 ݉ߣ ܴ (2)So ‫ܦ‬௠ ଶ െ ‫ܦ‬௡ ଶ ൌ 4 ݉ߣ ܴ െ 4 ݊ ߣ ܴ bly‫ܦ‬௠ ଶ െ ‫ܦ‬௡ ଶ ൌ 4 ሺ݉ െ ݊ሻ ߣ ܴ (3) The Newton’s rings setup is taken in a liquid. Now the air film is replaced by liquid film.In liquid film, the diameters of the same nth and mth dark rings are ‫ ′ܦ‬௡ and ‫ ′ܦ‬௠ are measured eewith the help of travelling microscope. ଶ ସ௡ఒோ‫ ′ܦ‬௡ ൌ s.w and ఓ ଶ 4݉ߣܴ ‫′ܦ‬௠ ൌ ߤ sic ଶ ଶ ସ ሺ௠ି௡ሻ ఒ ோSo ‫ ′ܦ‬௠ െ ‫ ′ܦ‬௡ ൌ (4) ఓ hyDividing equation (3) by (4) ‫ܦ‬௠ ଶ െ ‫ܦ‬௡ ଶ 4 ሺ݉ െ ݊ሻ ߣ ܴ ൌ ଶ ଶ 4 ሺ݉ െ ݊ሻ ߣ ܴ gp ‫′ܦ‬௠ െ ‫ ′ܦ‬௡ ߤ ‫ܦ‬௠ ଶ െ ‫ܦ‬௡ ଶ ൌߤ in ଶ ଶ ‫ ′ܦ‬௠ െ ‫ ′ܦ‬௡ erUsing the above relation ߤ can be calculated. e gin en P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 9
  • 10.        en gin ee rin gph ys ics .w ee bly .co m
  • 11. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Diffraction Engineering PhysicsIntroduction The wave nature of light is first confirmed by the phenomenon of interference.Further it is confirmed by the phenomenon of diffraction. The word ‘diffraction’ is derivedfrom the Latin word diffractus which means break to piece. When the light waves encounteran obstacle, they bend round the edges of the obstacle. The bending is predominant whenthe size of the obstacle is comparable with the wavelength of light. The bending of light mwaves around the edges of an obstacle is diffraction. It was first observed by Gremaldy. .co1. DiffractionDefinition:- When the light falls on the obstacle whose size is comparable with the wavelength of blylight then the light bends around the obstacle and enters in the geometrical shadow. Thisbending of light is called diffraction. ee When the light is incident on an obstacle AB, their corresponding shadow iscompletely dark on the screen. Suppose the width of the slit is comparable to the .wwavelength of light, then the shadow consists of bright and dark fringes. These fringes areformed due to the superposition of bended waves around the corners of an obstacle. Theamount of bending always depends on the size of the obstacle and wavelength of light used. ics ys g ph rin2. Types of diffractionThe diffraction phenomena are classified into two ways ee I. Fresnel diffraction II. Fraunhofer diffraction. ginFresnel diffraction:- In this diffraction the source and screen are separated at finite distance. To study thisendiffraction lenses are not used because the source and screen separated at finite distance.This diffraction can be studied in the direction of propagation of light. In this diffraction theincidence wave front must be spherical of cylindrical.Fraunhofer diffraction:- In this diffraction the source and screen are separated at infinite distance. To studythis diffraction lenses are used because the source and screen separated at infinitedistance. This diffraction can be studied in any direction. In this diffraction the incidencewave front must be plane.P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 1
  • 12. 3. Fraunhofer single slit diffraction:- Let us consider a slit AB of width ‘e’. Let a plane wavefront of monochromaticlight of wavelength λ is incident on the slit AB. According to Huygens principle, every pointon the wavefront is a source of secondary wavelets. Those secondary wavelets travellingnormal to the slit are brought to focus at point P0 on the screen by using the lens. Thesecondary wavelets traveling at an angle θ with the normal are focused at point P1. m .co bly Intensity at point P1 depends up on the path difference between the wavelets A and B eereaching to point P1. .w 2 2 sin icsLet the width of the slit is divided into ‘n’ equal parts and the amplitude of the wave fronteach part is ‘a’. Then the phase difference between any two successive waves from these ysparts would be 1 1 2 sin phUsing the vector addition method, the resultant amplitude R is sin g 2 rin sin 2 1 2 sin . sin ee sin sin 2 1 2 sin sin sin sin gin 2 sin sin sin en sin sin sin sinTherefore resultant intensity
  • 13. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Diffraction Engineering PhysicsPrincipal maximum:-The resultant amplitude R can be written as 3! 5! 7! 1 3! 5! 7! m 1 3! 5! 7!In the above expression for 0 values the resultant amplitude is maximum .coThen sin 0 bly sin 0 0For 0 value the resultant intensity is maximum at P0 and is known as principal eemaximum.Minimum intensity positions .w Will be minimum when sin 0 1,2,3,4,5 … … .. sin ics sin ys So we obtain the minimum intensity positions on either side of the principal maxima for all values. phSecondary maximum In between these minima secondary maxima positions are located. This can beobtained by differentiating the expression of w.r.t and equation to zero g sin 0 rin 2sin sin 0 ee 2sin cos sin · 0 ginIn the above expression can never equal to zero, soEither sin 0 or cos sin 0sin 0 gives the positions of minima.enThe condition for getting the secondary maxima is cos sin 0 cos sin tanThe values of satisfying the above equation are obtained graphically by plotting the curves and tan on the same graph. The plots of and tan is shownin figure.P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 3
  • 14. mFrom the graph intersecting points are 0, , , ………… .coFrom the above concepts the intensity distribution curve verses is shown in figure. bly ee .w ics ys ph4. Fraunhofer double slit diffraction Let S1and S2 be the two slits equal width e and separated by a distance d. The gdistance between the two slits is (e + d). A monochromatic light of wavelength λ is incidenton the two slits. The diffracted light from these slits is focused on the screen by using a rinlens. The diffraction of a two slits is a combination of diffraction and interference. Thewavelets travelling normal to the slit is focused at P0. The wavelets travelling at an angle θ eewith the incident light are focused at point P1.From Fraunhofer single slit experiment, the resultant amplitude is ginSo the amplitude of each secondary wavelet travelling with an angle θ can be taken sinas . These two wavelets interference and meet at point P1 on the screen. en
  • 15. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Diffraction Engineering Physics To find out the path difference between the two wavelets, let us draw a normal s1k tothe wavelet S2. S S SFrom ∆S S k sin S S S sin 2 2 sin m By using vector addition method, we can calculate the resultant amplitude at point P1 2 cos .co sin sin sin sin 2 cos sin sin 2 2 cos bly sin 2 1 cos ee sin 2 1 2 cos 1 sin 2 2 cos 4 sin .w cos ics sin 4 cos sin ys sin ph 4 cosThe resultant intensity 4 cos gFrom the above equation, it is clear that the resultant intensity is a product of two factors i.e., rin 1. represents the diffraction pattern due to a single slit. 2. cos represents the interference pattern due to wavelets from double slit. eeDiffraction effectIn diffraction pattern the variation of w.r.t as shown in figure. ginenP.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 5
  • 16. The diffraction pattern consists of central principal maxima for 0 value.The secondary maxima of decreasing intensity are present on either side of the centralmaxima for , , values.Between the secondary maxima the minima values are present for , 2 ,values.Interference effectIn interference pattern the variation of cos w.r.t as shown in figure. m .co bly eecos represents the interference pattern.Interference maximum will occur for cos 1 .w 0, 1, 2, 3, 4 … … … , 2 , 3 , 4 ……… ics sin sin ysInterference minima will occur for cos 0 ph 2 1 2 sin 2 1 2 g sin 2 1 rin 2The resultant intensity variation due both interference and diffraction patterns is shown infigure. Due to interference the resultant minimum in not exactly equal to zero. ee gin en
  • 17. θ GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Diffraction Engineering PhysicsConstruction of plane diffraction grating OrConstruction of grating A set of large number of parallel slits of same width and separated by opaque spacesis known as diffraction grating. Fraunhofer used the first grating consisting of a largenumber of parallel wires placed side by side very closely at regular separation. Now the mgratings are constructed by ruling the equidistance parallel lines on a transparent material .cosuch as glass with fine diamond point. The ruled lines are opaque to light while the spacebetween the two lines is transparent to light and act as a slit. bly Let ‘e’ be the width of line and‘d’ be the width of the slit. Then (e + d) is known asgrating element. If N is the number of lines per inch on the grating then ee 1 2.54 . .w ics ys g ph rin eePlane Diffraction grating ( or diffraction at N - slits) Let a plane wavefront of monochromatic light of wavelength λ is incident on the ginslits. Let ‘e’ be the width of line and ‘ ’ be the width of the slit. Then (e + d) is known asgrating element. According to Huygens principle, every point on the wavefront is a sourceof secondary wavelets. The wavelets spread out to the right in all directions. Thoseensecondary wavelets travelling normal to the slit are brought to focus at point P0 on thescreen by using the lens. . The wavelets travelling at an angle θ with the incident light arefocused at point P1.P.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 7
  • 18. sin The resultant amplitude due to diffraction of a single slit is , where sin . Let us assume that there are N slits in the grating. The path difference between thecorresponding points is sin 2 . The resultant of amplitude R of the N amplitudes of is sin sin 1 sin m And intensity is .co sin sin . 2 sin bly In the above equation shows the intensity distribution due to a single slitdiffraction and shows the combined intensity distribution due to interference at all eethe N slits.Principal Maxima: from equation 2, we know that the intensity would be maximum .wwhen sin 0, where 0, 2 , … . . and n= 0,1,2,3… Indeterminate icsTo find out its value, L- Hospitals rule is applied sin sin cos ys lim lim lim sin sin cos phHence the resultant intensity is can be written as sin g rinSo the maxima are obtained for sin or ee sinMinima:From equation 2 the resultant intensity is minimum for sin 0 but sin 0 gin sin en sin can be take integral values except 0, N, 2N,………..nN. For these values we bet maximaSecondary maxima:Between two adjacent principal maxima there are 1 minima and 2 other maximacalled secondary maxima.b to get the position so secondary maxima, differentiate equation2 w.r.t and then equate to zero.
  • 19. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - Diffraction Engineering Physics sin sin cos sin sin cos 2 . 0 sin 2 cos sin sin cos 0 or tan tanThe roots of the above equation (other than , this corresponds to principal maxima)give the secondary maxima. mThe variation of intensity due to the interference factor is shown in figure .co bly ee .wThe variation of intensity due to the diffraction factor of is shown in figure ics ys g ph rinThe variation of resultant intensity is shown in figure. ee ginenP.Sreenivasula Reddy M.Sc, (PhD) Website: www.engineeringphysics.weebly.com Page 9
  • 20. 5. Grating spectrum When white light is fall on the grating, the light gets diffracted through each slit. As aresult of both interference and diffraction, diffraction pattern will occur. This pattern is knownas grating spectrum or diffraction spectrum. m .co bly ee .w ics ys ph The condition for the principal maximum in grating spectrum is sin gWhere is the grating element and is the order of spectrum.From the grating equation, we observe that rin 1. For a particular wavelength λ the angle of diffraction θ is different for principal maxima and for different orders. ee 2. For white light and for a particular order , the light of different wavelengths will be diffracted in different directions. 3. Angle of diffraction increases with wavelength. gin 4. In the spectrum violet is in the innermost position and red at the outermost position. 5. Most of the energy is distributed for zero order and the rest is distributed among the other orders. en 6. Spectrum different orders are situated symmetrically on both sides of zero order. 7. Sharpness of maxima increases with increasing of number of lines in the grating. 8. The colour of central maxima is same as the incident light. 9. The condition for finding the maximum number of orders formed by the grating is
  • 21. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - polarisation Engineering PhysicsIntroduction From the previous chapter’s interference and diffraction, we can understand that the light isexhibiting wave nature. But these phenomenon don’t explain the character of these wave motion i.e.,whether it is longitudinal or transverse. In longitudinal waves, the particles of the medium move toand fro in the direction of propagation of the wave as in case of sound waves. In transverse waves,the particles of the medium vibrate up and down perpendicular to the direction of propagation of wave mas ripples on a water surface. .co1. Polarization Light is electromagnetic in nature, it consists of oscillating electric and magnetic fields ofvectors. They are perpendicular to each other and perpendicular to the propagation of light. In light, the electric field direction is always changes in random ways. So the unpolarized or blyordinary light is represented as shown in figure. ee If the direction of electric field vector is confined to one fixed direction, that light beam is .wcalled linearly polarized or plane polarized light. So the plane polarized light is represented as shownin figure ics ysDefinition: ph The process of restricting the vibration of electric field vector is called polarization.2. Plane of polarization and plane of vibration gPlane of vibration: - rin The plane which contains the vibrations of polarized is known as plan of vibration.Plane of vibration:- ee The plane perpendicular to the plane of vibration is known as plane of polarization. When an unpolarized light is incident on the polarizer, the emergent light beam always hasvibrations parallel to the optic axis. ginen In the figure the plane parallel to the optic axis of the polarizer and containing thevibrations of polarized light and hence it is known as plane of vibration the plane isperpendicular to the plane of vibration and hence it is known as plane of polarization.P.Sreenivasula Reddy M.Sc, (PhD) Page 1 of 9
  • 22. 3. Definitions:Blunt corners:- If any corner contain all obtuse angles (is an angle between 90 and 180 degrees),which is called blunt corner.Optic axis: - A line joining the two blunt corners is known as optic axis of the crystal. An optic axis is notan axis, but it is a direction of a crystal. Any line parallel to the optic is also known as optic axis.Uniaxial crystal: -If any crystal having only one optic axis is known as uniaxial crystal.Examples: - calcite, quartz. mBiaxial crystal: - .coIf any crystal having two optic axes is known as biaxial crystal.Examples: -borax, mica. blyPrincipal section: - A plane containing the optic axis and perpendicular to two opposite faces of the crystal isknown as the principal section of the crystal. eePrincipal planes: - The plane having the optic axis and ordinary ray of the crystal is known as principal plane of .wthe ordinary ray. The plane having the optic axis and extraordinary ray of the crystal is known asprincipal plane of the extraordinary ray. ics4. Production of polarized light For the production of plane polarized light different methods are followed, they are ys a) Polarization by reflection b) Polarization by transmission c) Polarization by absorption ph d) Polarization by double refraction ga) Polarization by reflection rin When the unpolarized light is fall on the smooth surface such as glass or water surface, the reflected light is partially plane polarized. The degree of polarization changes with the angle ee of incidence. For a particular angle of incidence the reflected light was completely polarized transmitted light ray is partially plane polarized. The vibration of the reflected light is gin perpendicular to the plane of incidence. The polarizing angle for glass is 57.50 At which angle of incidence the reflected ray is completely plane polarized, that angle is called polarizing angle or angle of polarization. en
  • 23. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - polarisation Engineering Physicsb) Polarization by transmission When the unpolarized light incident on smooth surface at polarizing angle the reflected light isplane polarized and the transmitted light ray is partially plane polarized. When the unpolarized light incident on the pile of plates at polarizing angle, the transmittedlight ray is purely plane polarized. The polarizing angle for glass is 57.50. The vibration of thetransmitted light is parallel to the plane of incidence. m .co bly eec) Polarization by absorption When un polarized light is passing through the tourmaline ( , , , it is split intohorizontal and vertical components. The horizontal components are completely absorbed by the .wcrystal and the vertical component comes out. The light coming from the crystal is linearly polarized. ics ys phd) Double refraction: - When an unpolarized light is passes through a calcite crystal, it is split up into two plane gpolarized lights. This phenomenon is known as double refraction. rin ee gin In which, one ray obeys the laws of refraction and has vibrations perpendicular to the principalsection is known as ordinary ray. The other ray does not obey the laws of refraction and hasenvibrations parallel to the principal section is known as extraordinary ray. The refractive index ofordinary and extraordinary rays for calcite crystal is sin sin          sin sin In calcite crystal the velocity light for O- ray inside the crystal is less than the E- ray. Therefractive index of O-ray is same for all angle of incidence and hence it is represented by sphericalwave front. The refractive index of E-ray is always varies with the angle of incidence and hence it isrepresented by ellipsoidal wave front. But both rays are travel with same velocity along the optic axisof the calcite crystal.P.Sreenivasula Reddy M.Sc, (PhD) Page 3 of 9
  • 24. 5. Polarization of light waves Let us consider an unpolarized light is passed through the tourmaline crystal as shown infigure. Even when we are rotating the tourmaline crystal by taking the incident ray as an axis, there isno intensity variation in the emergent beam. The tourmaline crystal only allows the vibrations of lightthat are parallel to the axis of the crystal, and it does not allow the other vibrations. So the emergentbeam from the tourmaline crystal is a polarized light. So the first tourmaline crystal is calledpolarizer. m .co bly If the second tourmaline crystal is rotated by taking the incident ray as an axis, the intensity ofemergent beam from the second tourmaline crystal has two times minimum intensity and two timesmaximum intensity is observed within one complete rotation. When the second tourmaline crystal is eeparallel to the first one the emergent beam from the second crystal is maximum. When the secondtourmaline crystal is perpendicular to the first one the emergent beam from the second crystal isminimum. Therefore the second crystal is called analyzer. .wSo this property of light of rays confirms the transverse nature of light rays.6. Nicol prism ics Nicol prism is an optical device made from calcite and is used to produce and analyze planepolarized light. It was invented by Nicol in 1828 and is known as Nicol prism. ysPrinciple:- Nicol prism is worked on the principle of double refraction exhibited by calcite crystal. Construction:- ph Nicol prism is constructed by using calcite crystal. Its length and breadth dimensions are inthe ratio 3:1. Consider a calcite crystal, in which is the principal section and B and D are blunt gcorners. The angles of the principal section are 710 and 1090. The angle of the principal section is rinreduced from 710 to 680 by grinding the faces of AB and CD. After grinding the new principal sectionis The crystal is cut into two pieces perpendicular to the principal section along  . The two eecut surfaces are highly polished into optical flat and joined together by means of Canada balsam. Therefractive index of Nicol prism for sodium wave length 5893 and the refractive indices of ginordinary ray, extraordinary ray and Canada balsam are 1.658, 1.46 and 1.55 respectively. The twofaces of the Nicol prism is kept open, while the two sides are blackened to absorb the totally internallyreflected rays. en   B C Canada balsam cement 68 71 A D
  • 25. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - polarisation Engineering PhysicsWorking of Nicol prismWhen an unpolarized light is passes through a calcite crystal, it is split up into two beams, namely,ordinary and extraordinary rays. m E ray .coWhen the ordinary ray passing through the calcite, the Canada balsam layer acts as a rarer medium. blyTherefore the ordinary ray is travelling from denser (calcite) to rarer (Canada balsam) medium. Forordinary ray, the angle of incidence at the interface of calcite and Canada balsam is greater than thecritical angle, it undergoes total internal reflection. eeThe critical angle for the ordinary ray is 1.55 sin 69.2 .w 1.658The angle of the principal section is greater than the critical angle, so the upper and lower surfaces ofthe Nicol prism are grounded from 710 to 680. For the extraordinary ray the Canada balsam is acts as icsa dense medium. Hence it travels in a straight line and emerges out from the Nicol prism. Theemergent beam is plane polarized light. ys7. Plane polarized light by Nicol prism. A Nicol prism is used to produce and analyze plane polarized light. When an unpolarized phlight is passed through a Nicol prism, it is split up into ordinary and extraordinary rays. The ordinaryray is cut of when it is passing from calcite to Canada balsam layer. The extraordinary ray is passed gthrough the Nicol prism. rin ee ginen If the second Nicol prism is rotated by taking the incident ray as an axis, the intensity ofemergent beam from the second Nicol prism has two times minimum intensity and two timesmaximum intensity is observed within one complete rotation. The first Nicol prism is known aspolarizer and the second Nicol prism is known as analyzer. Thus a Nicol prism can be used as apolarizer as well as analyzer.P.Sreenivasula Reddy M.Sc, (PhD) Page 5 of 9
  • 26. 8. Limitations of Nicol prism1. The Nicol prism acts as a polarizer only for the light rays having slightly convergent and slightly divergent beam.2. The Nicol prism does not acts as a polarizer for the light rays having highly convergent and highly divergent beam3. If the incident divergent beam is making larger angle with the front face, the extraordinary ray will become parallel to the optics axis of the crystal. Extraordinary ray also totally internally reflected by the Canada balsam layer. Therefore no light emerges our from the Nicol prism.4. If the incident convergent beam is making small angle with the front face, the ordinary makes less critical angle at the calcite and Canada balsam interface. Then the ordinary ray also emerges out m from the Nicol prism.9. Quarter wave plate .co A wave plate is an optical device, is used to produce path difference or phase differencebetween the ordinary and extraordinary rays. It is made from the doubly refracting uniaxial crystals.Its faces are cut parallel to the direction of the optic axis. bly When the polarized light is incident on the calcite crystal, the light splits up into O-ray and E-ray. The thickness of the device is cut in such way it can produce the path difference 4 or phasedifference 2 between the O-ray and E-ray is called quarter wave plate. ee Let be the thickness of the crystal. Let and be the refractive indices of E-ray and O-ray. Then the path difference between E-ray and O-ray is      For quarter wave plate path difference = 4   .w ics 4   4 ysFor negative crystals  For positive crystals ph 10. Half wav e plate A wave plate is an optical device, is used to produce path difference or phase difference gbetween the ordinary and extraordinary rays. It is made from the doubly refracting uniaxial crystals. rinIts faces are cut parallel to the direction of the optic axis.When the plane polarized light is incident on the calcite crystal, the light splits up into O-ray and E- eeray. When the refracting faces are cut parallel to the optic axis, O-ray and E-ray travels with differentvelocities along the same path.The thickness of the device is cut in such way it can produce the path difference 2 or path gindifference between the O-ray and E-ray is called half wave plate.Let be the thickness of the crystal. Let and be the refractive indices of E-ray and O-ray.Then the path difference between E-ray and O-ray is  en    For half wave plate path difference = 2   2   2For negative crystals  For positive crystals
  • 27. GLOBAL COLLEGE OF ENGINEERING &TECHNOLOGY: YSR DIST. Unit –I OPTICS - polarisation Engineering Physics 11. Theory of circular and elliptical polarized lights A beam of plane polarized light is obtained by passing the unpolarized light through the Nicolprism. Let be the maximum amplitude of the incident light which makes an angle with theoptic axis. When the plane polarized light is passes through a wave plate, it is split up into twoordinary and extraordinary rays. The amplitudes of ordinary and extraordinary rays are sin   and cos respectively. A sin m Q .co P A cos bly The ordinary and extraordinary rays travel with different velocities in the same direction, and eehence a phase difference is introduced between these two rays. Let be the phase difference betweenthe ordinary and extraordinary rays. The emergent beam from the wave plate can be represented as .w cos sin (1) sin sin (2) icsTaking cos and sin , the above equations can be written as  sin (3) ys  sin (4)Equations (3) and (4) can be written as ph  sin (5) g  sin (6) rinEquation (5) can be written as  sin cos    cos  sin (7) eeSubstituting the value of  sin and  cos from equation (6) in equation (7) gin  cos   1   sin (8)Squaring equation (8), we geten  cos 2  cos   1   sin  cos 2  cos     sin   sin  cos   sin 2  cos     sin 2  cos     sin (9)P.Sreenivasula Reddy M.Sc, (PhD) Page 7 of 9
  • 28. Equation (9) is a general expression for an ellipse.   when 0,   sin 0 and  cos 1Equation (9) becomes 2  0 0   mThe above expression represents a straight line. Therefore the light is plane polarized.   when ,   sin 1 and  cos 0 .coEquation (9) becomes  1 blyThis represents the equation of a ellipse. Thus the emergent light in this case will be ellipticallypolarized light.   when ,   sin 1,  cos 0 and ee  1 .wThis represents the equation of a circle. Thus the emergent light in this case will be circularly icspolarized light 12. Differences between polarized light and unpolarized light ysUnpolarized light Polarized light phWhen an unpolarized light is passed through the When a polarized light is passed through therotating tourmaline crystal, no variation intensity rotating tourmaline crystal, variation in theof light is observed. intensity of light is observed. gThe unpolarized light is symmetrical about the The polarized light is lack symmetry about thedirection of propagation of light. direction of propagation light. rinIn unpolarized light the vibrations are confined to In polarized light the vibrations are confined toany direction. one particular direction. eeThe unpolarized light is specified as The polarized light is specified as gin en Differences between O-ray and E-rayO-ray E-rayThe O-ray obeys the laws of refraction The E-ray does not obeys the laws of refractionThe O-ray has vibrations perpendicular to the The E-ray has vibrations parallel to the principalprincipal section. sectionIn negative crystal the velocity light for O- ray is In positive crystal the velocity light for E- ray isless than the E- ray less than the O- rayThe refractive index of O-ray is same for all The refractive index of E-ray is always variesangle of incidence with the angle of incidence
  • 29. GLOB BAL COLLE EGE OF EN NGINEERI ING &TEC CHNOLOGY YSR DIS Y: ST. Unit –I – OPTICS - polarisat tion Engineer E ring Phys sics13. A Analysis or detection of unknown p d f polarization n m .co bly ee .w ics ys g ph rin ee ginenP.Sr reenivasu Reddy ula M.Sc, (PhD) Page 9 of 9 e