interference & diffraction

basic engineering

1. UNIT I INTERFERENCE AND DIFFRACTION

2. Syllabus contents Interference and Diffraction Interference due to thin films of uniform thickness and non-uniform thickness (with derivation), Newton’s rings, Applications of interference. Fraunhoffer diffraction at a single slit; condition of maxima and minima, Plane Diffraction grating (diffraction at multiple slits).

3. Outline Introduction Definition of interference Young’s double slit experiment and principle of superposition Interference due to thin films of uniform thickness Interference due to thin films of non uniform thickness (wedge shaped film)

4. Techniques of Interference Division of wavefront Division of amplitude

5. THIN FILM INTERFERENCE Thin film : A film is said to be thin when its thickness is of the order of one wavelength of visible light ~ 5500 Å (0.55μm). ↑ Iridescence caused by interference : Colours as seen due to interference phenomena on oil layer, soap bubbles, beetle body, oil film on rock and morph butterfly wings, Colours on CD and DVD.

6. INTERFERENCE PRINCIPLE OF SUPERPOSITION : The resultant displacement of a particle of the medium acted upon by two or more waves simultaneously is the algebraic sum of the displacements of the same particle due to individual waves. y = y 1 + y 2 y’= y 1 - y 2

7. INTERFERENCE Definition: Interference is the physical effect caused by superimposition of two or more wave trains traveling in the same direction at the same time . When two or more waves superimpose, the resultant amplitude (intensity) in the region of superposition is different than the amplitude (or intensity) of individual waves. This modification in the distribution of intensity in the region of superposition is called interference.

8. INTERFERENCE OF WAVES The equation of two waves are The wave resulting from their superimposition is y=y 1 +y 2

9. INTERFERENCE OF WAVES The wave resulting from their superimposition is +

10. INTERFERENCE OF WAVES Constructive Interference: Sum of amplitudes due to two waves when they meet in phase Destructive Interference: Difference of amplitudes due to two waves when they meet at a point in opposite phase

11. INTERFERENCE OF WAVES Constructive Interference: Sum of amplitudes due to two waves Path difference:  =n=(2n) / 2 where n=0,1,2,3….  =0 , , 2, 3, …….. n  Phase difference:  =(2 / )  =( 2n  ) where n=0, 1, 2, 3….  =0 , , 2, 3 ,….n 

12. INTERFERENCE OF WAVES Destructive Interference : Difference of amplitudes due to the two waves Path difference:  =(2n+1) /2, where, n=0,1,2,3 ….. = /2 , 3 /2 , 5/2 …..( 2n+1)/2 Phase difference:  =(2/  )  = (2n+1)  where n=0,1,2,3….  = , 3  , 5  ,………

13. Techniques of obtaining interference These techniques can be divided into two broad classes: [1] Wave front splitting: A wave front of light emerging from a narrow slit can be divided it into two by passing it through two closely spaced slits. Young’s double slit expt., Fresenl’s expt. of double mirror, Fresenel’s biprism , Lloyd’s mirror, etc employ this technique. [2] Amplitude splitting: The amplitude (intensity) of a light wave is divided into two parts, namely reflected and transmitted components, by partial reflection at a surface. Interference in thin films (parallel, wedge shaped, Newton’s ring, etc), Michelson’s interferometer etc utilize this method. Also, optical elements such as beam splitters, mirrors are for achieving amplitude divison

14. INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS A transparent thin film of uniform thickness bounded by two parallel surfaces is known as a plane parallel thin film. ASSUMPTIONS : [1]According to Stokes, when a light wave is reflected at the surface of optically denser medium, it suffers a phase change of  i.e. a path difference of /2. [2]A distance ‘l’ traversed by light wave in a medium of refractive index ‘’ has it’s equivalent optical path ‘l’.

15. INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS When the film is observed in reflected light

16. INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS The effective path difference between waves BR and DR 1 is, =2tcosr + (/2 )

17. When the film is observed in reflected light Condition for darkness : The total path difference should be odd multiple of /2  = ( 2n + 1)/2 where n=0,1,2,3,….. 2 tcosr + (/2)= (2n + 1)/2 2tcosr = n Condition for brightness : The total path difference should be an integral multiple of   = n  where n=0,1,2,3,….. 2 tcosr + (/2)= n 2tcosr = (2n + 1)/2 INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS

18. When the film is observed in transmitted light INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS

19. When the film is observed in transmitted light Condition for brightness : The total path difference should be odd multiple of /2  = ( 2n + 1)/2 where n=0,1,2,3,….. 2 tcosr + (/2)= (2n + 1)/2 2tcosr = n Condition for darkness : The total path difference should be an integral multiple of   = n  where n=0,1,2,3,….. 2 tcosr + (/2)= n 2tcosr = (2n + 1)/2 INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS

20. INTERFERENCE DUE TO THIN FILMS OF UNIFORM THICKNESS Some important points [a] The conditions of interference depend on three parameters i.e. t,  and r. In case of parallel film, t and r remain constant. The conditions of interference solely depend on the wavelength ‘’. [b] When parallel beam of monochromatic light is incident normal to the film, the whole film will appear uniformly dark or uniformly bright. The film will appear bright in reflected light when the thickness of the film is /4, 3/4, 5/4….. The film will appear dark in reflected light when the thickness of the film is /2, /, 3/2….. [c] A change in the angle of incidence of the rays leads to a change in the path difference. Consequently, if the inclination of the film with respect to the light beam is changed gradually, the film appear dark and bright or bright and dark in succession. [d] If a parallel beam of white light is incident on the film, those wavelengths for which the path difference is ‘n ’, will be absent from the reflected light. The other colours will be reflected

21. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS ASSUMPTIONS : [1]According to Stokes, when a light waves is reflected at the surface of optically denser medium ( i.e ) surface backed by a denser medium), it suffers a phase change of  i.e. a path difference of /2. [2]A distance ‘l’ traversed by light wave in a medium of refractive index ‘’ has it’s equivalent optical path ‘l’.

22. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS

23. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS

24. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS From the complete geometry of above figure, BPQ and BPC have a common side BP, From BPQ, BQP= α QBP= 90 BPQ= 90-α From BPC, CBP=90- r CPB= 90-α Hence BCP= r+α PG is parallel to DP and ray BCP is assumed to be straight and intersecting at PQ and DP at C and P, hence BPD=r+α DS=SP=t, CS is common side of both triangle CSD and CSP, i . e. CP=CD, Hence, CDS = r+α

25. The optical path difference  is given by, =  (BC + CD) – BF =  (BE + EC + CD) –  (BE) =  (EC + CP) =  (EP) From EPD, = 2 t cos (r + ) Due to reflection from a surface backed by a denser medium suffers an abrupt phase change of  which is equivalent to a path difference of /2.  eff = = 2 t cos (r + )  /2 INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS

26. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF NON-UNIFORM THICKNESS Condition for brightness: The total path difference should be an integral multiple of   =n where n=0,1,2,3,….. 2tcos(r+) + (/2)= n 2 tcos(r+) = (2n + 1)/2 Condition for darkness: The total path difference should be odd multiple of /2 =(2n + 1)/2 where n=0,1,2,3,….. 2tcos(r+) + (/2)= (2n + 1)/2 2tcos(r+) = n

27. WEDGE-SHAPED FILMS Salient features of the Interference Pattern due to wedge shaped thin film Fringes are equidistant Consider two consecutive dark fringes at point A and C as shown in figure. The n th dark band be formed at point A at a distance x 1 from the edge of contact ‘O’ and t 1 be the thickness of film at A. The (n+1) th dark band is formed at point C at a distance x 2 from ‘O’ and t 2 is the thickness of film at point C.

28. WEDGE-SHAPED FILMS For destructive interference condition, 2 μ t cos r = n λ (normal incidence, cos r =1) or 2 μ t 1 = n λ ------- (1) For (n+1) th dark fringe, 2 μ t 2 = (n+1) λ --------- (2) Subtracting (1) from (2) we get, 2 μ (t 2 – t 1 ) = λ -------- (3) From figure, in right angled triangle AEC, tan θ = 𝐶𝐸/𝐴𝐸=t 2 − t 1 /x 2 −x 1 Since θ is small, tan θ ~ θ,t 2 − t 1 = (x 2 −x 1 ) θ ---- (4)

29. WEDGE-SHAPED FILMS Substituting value of t 2 -t 1 in equation (3) we get 2 μ ( x 2 -x 1 ) θ = λ Since x 2 -x 1 = β = Fringe width ( ie . distance between two consecutive dark fringes). Therefore, 2 μ β θ = λ or Fringe width β = 𝜆/2μ𝜃 For air film, refractive index μ =1, β = 𝜆/2𝜃 Since 𝜆, 𝜃 are constant, β is constant. Hence fringes are equidistant . Interference pattern

30. WEDGE-SHAPED FILMS [2] The fringe at apex is dark At the apex, the thickness of the wedge is very small compared to λ, i.e. 𝑡≪𝜆. Therefore thickness of film at apex is zero. The optical path difference becomes, Δ = 2μt−λ/2 = λ/2 For path difference of 𝜆/2, the interfering rays will always be 180° out of phase and interfere destructively . Therefore, the fringe at the apex of the wedge is always dark.

31. WEDGE-SHAPED FILMS [3] Fringes are straight and parallel The locus of points having the same thickness lie along lines parallel to the apex line of the wedge.Thus , the fringes are straight. Since the fringes are equidistant and straight, they are parallel. [4] Fringes of equal thickness Each maximum or minimum is a locus of constant film thickness, the fringes are called fringes of equal thickness.

32. WEDGE-SHAPED FILMS [5] Fringes are localized The fringes are located at the top surface of the wedge film. [6] Wedge Angle ( ) Experimentally, we can find the wedge angle ‘θ’ using a travelling microscope. As shown in the fig consider two dark fringes formed at points A and B at a distance x 1 and x 2 respectively from apex ‘O’.

33. WEDGE-SHAPED FILMS Let the thickness of the film be t 1 and t 2 at A and B respectively. At point A, 2 μ t 1 = n λ From the figure, t 1 = x 1 tan θ ~ x 1 θ (as θ is very small) ∴2μ x 1 θ = n λ ------(1) Similarly, at point B, 2μ x 2 θ = ( n+N ) λ ------- (2) where N is the number of fringes between A and B. Subtracting (1) from (2) we get, 2 μ( x 2 -x 1 ) θ = Nλ Hence 𝜃 = 𝑁 λ / 2 μ( x 2 -x 1 ) for air film, 𝜃 = 𝑁 λ / 2 ( x 2 -x 1 ) ----- (3)

34. WEDGE-SHAPED FILMS [ 7] Spacer thickness To find the thickness of spacer/sheet or diameter of wire ‘t’ From figure , t = 𝑙 tan θ where 𝑙 is the length of air wedge Substituting the value of θ from eqn.3 we get, 𝑡 = 𝑙 𝜃 =λ/2 μ (x 2 -x 1 ) -----(usingeqn.3)