Interference - Fresnel Biprism & Newton's Ring PDF

FRESNEL BIPRISM Fresnel Biprism is a device which is used to produce coherent sources and thus show the phenomenon of interference. In practice it consist of a thin glass plate with one of its face ground and polished till a prism is formed with one of its angle of about 179 0 and two side angles of...

FRESNEL BIPRISM Fresnel Biprism is a device which is used to produce coherent sources and thus show the phenomenon of interference. In practice it consist of a thin glass plate with one of its face ground and polished till a prism is formed with one of its angle of about 179 0 and two side angles of ½ o each. When a light is incident on an ordinary prism, the ray bent through an angle called angle of deviation. As a result the ray emerging out of the prism appears to have emanated from a source S’ located at a small distance above the real source. If we arrange two prisms placed base to base the bending of light takes place in such a way that light appears to come from two virtual sources S1 and S2. The refraction of light in biprism also produces two virtual sources S1 an S2. FRESEL BIPRISM Two ordinary prisms placed base to base forming two virtual sources Sand S2. A monochromatic light source illuminated the vertical slit which acts as a narrow linear monochromatic light source. The biprism is placed in such a way that the refracting edge is parallel to the length of the slit. A single cylindrical wave front impinges on the biprism. The top portion of the wave front is refracted downward and appears to have emanated from the virtual source S1. The lower segment falling on the lower part of the biprism is refracted upward and appears to have emanated from the virtual source S 2. The virtual source S1 and S2 are images of the same source S produced by the refraction due to biprism and hence are coherent. Hence the light source is in a position to interfere and produce fringes in a region beyond the biprism. Thus biprism creates two virtual images of the source (similar to the combination of two prisms placed base to base) The theory of interference and formation of fringe in case of Fresnel biprism is the same as described for Young’s Double Slit Experiment. At point O a central bright fringe is formed having alternate bright and dark fringes on either side of O. The fringe width i.e. the spacing between consecutive bright or dark fringes is given by:  𝜆𝐷 𝛽 = 𝑑 Where- D is the distance between source or slit and screen/eye-piece d is the distance between two virtual light sources (i.e. S1 and S2) λ is the wavelength of the monochromatic source Application of Fresnel Biprism Determination of wavelength of monochromatic light using Fresnel Biprism 𝛽𝑑 The wavelength of light can be determined using the equation𝜆 =. In order to use this 𝐷 relation the values of  β are to be measured experimentally  D  D Experimental Set-Up On an optical Bench; the Slit, biprism and the eyepiece are placed and adjusted at the same horizontal level. Light from a monochromatic source is allowed to be incident on the slit. The biprism and eyepiece is moved laterally on the optical bench till fringes are obtained in the focal plane of the eyepiece (due to the action of the biprism interference fringes are formed on the screen/eyepiece).  Determination of D (distance between the slit and the eyepiece) Once the distinct fringe pattern appears in the focal plane of the eyepiece, the uprights having the slit, the biprism and the eyepiece are fixed on the optical bench. The position of the slit and the position of the eyepiece are noted as X and Y in cm from the scale of the optical bench. D = Y-X = position of eyepiece – position of the slit (X any Y to be noted from scale of the optical bench)   Determination of fringe width β When the fringes are observed in the field of view of the eyepiece, the vertical cross wire is made to coincide with the centre of any nth bright fringes. The position of the crosswire is read as xn from the scale of the eyepiece in mm. The micrometer screw of the eyepiece is further moved slowly and the vertical cross wire is made to coincide with the centre of any other bright fringes for eg n+6th bright fringe, the position of the crosswire is again read as xn+6 from the scale of the eyepiece. The number of the bright fringes that pass across the crosswire between the two positions xn and xn+6 is counted as N (eg N=6).. The fringe width is given by: 𝑥𝑛+6 − 𝑥𝑛 𝛽 = 𝑁 Where xn = the position of the vertical crosswire coinciding with a bright fringe (in mm) xn+6 = the position of the vertical crosswire coinciding with a another bright fringe N= number of bright fringes counted between the positions xn and xn+6   Determination of d (distance between two virtual sources S1 and S2 ) 1. Displacement Method A convex lens of short focal length is placed between the slit and the eyepiece without disturbing their positions. The convex lens is moved back and forth near the biprism till a pair of sharp images of the slit is obtained in the field of view of the eyepiece. In order to obtain real images of the two virtual sources the distance between the slit (object) and eyepiece (screen) is kept >4f. The distance between the images is measured by coinciding the vertical crosswire of the eyepiece on the two images 𝑆 ′ and 𝑆 ′′. Let it be denoted by d1 and the position of the convex lens is L1. If u = is the distance of the slit and lens and v= is the distance of eyepiece from the lens (the distance u and v can be measured from the optical bench), Using magnification formula: 𝑣 𝑑1 = ……………. (1) 𝑢 𝑑 The lens is then moved to a position nearer to the eyepiece, where again a pair of images of the slit is seen. The distance between the two sharp images is again measured. Let it be d2 If 𝑢′ is the distance of the slit and lens and 𝑣 ′ is the distance of eyepiece from the lens (the distance 𝑢 ′ and 𝑣 ′ can be measured from the optical bench) the position of the convex lens be L2, the by magnification formula: 𝑣′ 𝑑2 = …………….. (2) 𝑢′ 𝑑 Since L1 and L2 are the two conjugated positions of the convex lens where the real image is obtained keeping the distance between the slit and the eyepiece same (>4f) Multiplying equation (1) and (2) 𝑑1𝑑2 = 1 𝑑2 𝑑 = √𝑑1𝑑2 d1 and d2= (to be obtained from scale of the eyepiece in mm by making the vertical cross wire overlap with the real images formed) d1 = is the distance between the two real images (of the virtual sources S 1 and S2) and d2 obtained when the convex lens is kept at position L1 d2= is the distance between the two real images (of the virtual sources S 1 and S2) and d2 obtained when the convex lens is kept at position L2 Note: Equation (1) was also sufficient to find the value of ‘d’ since u & v can be measured from the optical bench (cm). Also d1 can be measure from eyepiece scale (mm) [Now if we find the conjugate position of lens L2 and measure the image distance as d2. The final equation of ‘d’ depends now only on parameters d 1 and d2 which can be noted from eyepiece scale in mm, the measurement of u and v in this case is not required (which were in cm and not more precise as compared to mm scale)]  2. Deviation Method From the ray diagram given below S1 P S a 𝛼+𝛿 𝑠𝑖𝑛 ( ) By using the prim formula 𝜇 = 𝛼 2 ………………… (1) 𝑠𝑖𝑛 ( 2 ) Where α = are the angle of biprism and δ = angle of deviation. 𝛼+𝛿 𝛼+𝛿 Since α and δ are ~ small 𝑠𝑖𝑛 ( ) ~( ) 2 2 𝛼 𝛼 And 𝑠𝑖𝑛 ( ) ~ ( ) 2 2 Thus ∴ 𝛿 = (𝜇 − 1)𝛼 ……………………. (2) 𝑑⁄ 2 From the figure S1SP 𝑡𝑎𝑛 𝛿 = 𝑎 Since 𝛿~ 𝑠𝑚𝑎𝑙𝑙 tan 𝛿~ 𝛿 𝑑⁄ 2 ∴ 𝛿= ……………………. (3) 𝑎 Putting the value of δ in above equation (2) & 3) we have 𝑑 = 2 𝑎 (𝜇 − 1)𝛼 1 𝜋 Where α=½0 = 𝑥 radian 2 180 a= distance between the slit and the biprism to be noted from the optical bench (usually 10cm) μ = is the refractive index of the biprism (glass) = 1.5 Interference Due to Division of Amplitude: Thin Films In this case a film of refractive index ‘μ’ is bounded by two surfaces, The thickness of the film is ‘t’=10-100μm  The thickness of the films remains constant at all points  The thickness of the films is not same, it increases on moving away from the point of contact Interference in thin film When light is incident on the surface of the film. It divides into two components by partial reflection and partial refraction. Since each time the light strikes between two surfaces the intensity of incident light is divided into two pats (intensity is directly proportional to amplitude) thus it is called Division of Amplitude. The reflection Coefficient is given by Equation If μ1 (=1 air) and μ2 (=1.5 e.g.) be the refractive indices of the two surfaces. On striking the surfaces it divide into two components let the reflected light be R 1, R2, R3……and transmitted components are T1, T2, T3……..  The intensity of R1 , R2 & T1, T2 is appreciable , rest waves have negligible intensity- Only R1 and R2 contribute to form interference pattern  In order to obtain fringes over the entire surface of the film- It is illuminated by a Broad Source of light  The pattern formed due to R1 and R2 is more distinct as compared to T1 and T2 Interference in Thin Wedge Shape Film [Thing to remember: According to Stoke’s treatment, when a ray reflects from a denser medium then the reflected ray suffers a phase difference (∆φ) of π with incident 2π ray or a path difference (∆x) of λ/2 with incident ray [since ∆φ = ∆x therefore, π = λ 2π λ ∆x or ∆x = ]. λ 2 Here, incident ray S reflects from a denser medium (point P) (reflected ray is ray 1). Thus ray1 has phase difference of π with incident ray S. similarly, since the ray 2 reflects from a rarer medium (air) (point F) therefore it doesn’t suffers any phase difference with incident ray PF. Therefore, ray 2 emerged from point E, is in same phase as the phase of incident ray from S. Consequently, we can say that ray 1 and ray 2 has phase difference of π or path difference of λ/2( this path difference is associated with ray1)] A light ray incident from source S, reflects from upper surface (point P) and lower internal surface (point F).The light from F emerges from E, thus 1 and 2 light rays interfere(other rays reflected rays from inner surface don’t have much intensity to get interfere).  𝑃𝐾 sin 𝑖 𝑃𝐾 𝜇= = 𝑃𝐸 = sin 𝑟 𝑃𝑁 𝑃𝑁 𝑃𝐸 Or 𝑷𝑲 = 𝝁𝑷𝑵 … … … … The path difference between 1 and 2 ray: [since the ray 1 travels an optical path µ(air) PK in air than the ray 2 (with respect to upper surface). But the ray 2 travels an optical path µ (PF+FE) in medium of refractive index µ with respect to upper surface. Here, it is to mention that according to Stoke’s treatment, a path difference of λ/2 is already associated with ray 1. Therefore, the net optical path of ray 1 become µ(air) PK + λ/2. Thus, the net expression of path difference (∆x) between 1 and 2 interfering waves becomes: ∆x = (optical path travelled by ray 2 – optical path traveled by ray 1)] 𝝀 𝚫𝒙 = 𝝁 (𝑷𝑭 + 𝑭𝑬) − [𝜇𝑎𝑖𝑟 𝑷𝑲 + ] 𝟐 𝜆 Δ𝑥 = 𝜇 (𝑃𝐹 + 𝐹𝐸 ) − [𝑃𝐾 + ] 2 𝝀 Δ𝑥 = 𝜇 [(𝑃𝑁 + 𝑁𝐹) + 𝐹𝐸 ] − 𝜇𝑃𝑁 − 𝑓𝑟𝑜𝑚 𝑓𝑖𝑔. 𝑃𝐹 = 𝑃𝑁 + 𝟐 𝑃𝐹 𝑎𝑛𝑑 𝑃𝐾 = 𝜇𝑃𝑁 (𝑒𝑞. 1) 𝝀 Δ𝑥 = 𝜇𝑃𝑁 + 𝜇 (𝑁𝐹 + 𝐹𝐸 ) − 𝜇𝑃𝑁 − 𝑓𝑟𝑜𝑚 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝐹𝐸 = 𝐹𝐿 𝟐 𝝀 Δ𝑥 = 𝜇 (𝑁𝐹 + 𝐹𝐿) − 𝟐 𝝀 Δ𝑥 = 𝜇 (𝑁𝐿) − … … … … ….. 𝟐 𝑁𝐿 From right triangle ENL cos(r + θ) = 𝐸𝐿 Or 𝑁𝐿 = cos(r + θ) 𝐸𝐿 Or 𝑁𝐿 = 2𝑡 cos(r + θ) 𝑓𝑟𝑜𝑚 𝑓𝑖𝑔. 𝐸𝐿 = 𝐸𝑀 + 𝑀𝐿 = 2𝑡 Substituting value of NL into eq.  𝝀 Δ𝑥 = 2𝜇𝑡 cos(r + θ) − 𝟐 This is the expression of path difference between ray1 and ray2. It depends upon the refractive index (µ) of wedge medium, thickness of the film (t) as well as refractive angle (r) and angle of wedge (θ).  If at thickness t, there is any bright fringe then applying the condition of a bright fringe Δ𝑥 = 𝑛𝜆 𝝀 or Δ𝑥 = 𝑛𝜆 = 2𝜇𝑡 cos(r + θ) − 𝟐 𝝀 or 𝟐𝝁𝒕 𝐜𝐨𝐬(𝐫 + 𝛉) = (𝟐𝒏 + 𝟏) 𝟐 This expression shows that the condition of bright fringe is that the term 2µt cos (r+θ) must be equal to odd multiple of λ/2.  If at thickness t, there is any dark fringe then applying the condition of a dark fringe 𝝀 Δ𝑥 = (2𝑛 + 1) 𝟐 𝝀 𝝀 or Δ𝑥 = (2𝑛 + 1) = 2𝜇𝑡 cos(r + θ) − 𝟐 𝟐 𝟐𝝁𝒕 𝐜𝐨𝐬(𝐫 + 𝛉) = 𝑛𝝀 This expression shows that the condition of dark fringe is that the term 2µt cos (r+θ) must be equal to multiple of nλ Fringe Width β in a WEDGE FILM B A C t1 t2 Let at point A on the film a nth dark fringe is formed where the thickness of the film is t1 Let at point B on the film a n+1th dark fringe is formed where the thickness of the film is B t2 Then the fringe width is the distance between two consecutive bright or dark fringes β= AB The condition for nth dark fringe/ minima at point A= 2𝜇𝑡1 cos(r + θ) = 𝑛𝜆 …. (1) The condition for (n+1)th dark fringe/ minima at point B= 2𝜇𝑡2 cos(r + θ) = (𝑛 + 1)𝝀 In the above equations if we consider the case of Normal incidence r=0 θ~small ∴ cos(r+θ) ~ 1 Eq(1) and (2) can now be written as At point A 2𝜇𝑡1 = 𝑛𝜆 …………… (3) At point B 2𝜇𝑡2 = (𝑛 + 1)𝝀 …………… (4) Subtract Eq (3) from (4) λ t 2 − t1 = ……….…… (5) 2μ 𝐵𝐶 𝑡2 −𝑡1 From ΔABC 𝑡𝑎𝑛𝜃 = = ……………... (6) 𝐴𝐶 𝐴𝐶~𝐴𝐵=𝛽 Since angle of wedge θ is small tanθ~θ  𝑡2 −𝑡1 ∴ 𝜃= ……..…… (7) 𝛽 Substitute Eq. (7) in (5) 𝝀 𝜷= 𝟐𝝁𝜽 NEWTON’S RING Newton’s ring is an example of Interference due to Division of Amplitude in a Wedge Shape Film. The Wedge shape film is created by placing a plano-convex lens of large radius of curvature with its convex surface in contact with a plane glass plate, an air film is formed between the two surfaces (glass plate & curved surface of plano-convex lens). The thickness of the air film gradually increases on moving away from the point of contact. If monochromatic light (sodium lamp) is allowed to fall normally on the palno-convex lens and viewed as shown in figure given below, then alternate dark and bright circular fringes are observed. These rings are known as Newton’s rings.. Point of contact Newton’s rings are formed because of the interference between the waves reflected from the top and bottom surfaces of the air film formed between the palno-convex lens and glass plate. These rings are concentric circles which are seen by using a microscope given in Figures. 𝜆 The path difference between R1 and R2 is given by: = 2 𝜇 𝑡 𝑐𝑜𝑠(𝑟 + 𝜃 ) - …… (1) 2 The wedge shape film enclosed between the plano-convex lens and plate is air μ=1 Light is incident normally on the wedge shape air film r = 0 The palno-convex lens used is of large radius of curvature (R=150cm), thus the angle of wedge shape air film formed is Ɵ ~ 0 hence cos(r + Ɵ) = 1 𝜆 Hence path difference between R1 and R2 is given by: = 2𝑡- …………….. (2) 2 Here the path difference depends on the thickness of the air film which is variable (increases on moving away from the point of contact).  For Bright fringe/ ring path difference = n λ 𝜆 i.e. 2𝑡 - = nλ 2 𝜆 𝑡 = (2𝑛 + 1) ………………. (3) 2 𝜆 For dark fringe/ ring path difference = (2𝑛 + 1) 2 𝜆 𝜆 i. e. 2𝑡 - = (2𝑛 + 1) 2 2 𝑡=nλ..…………….. (4) From Eq. 3 and 4 it is clear that at any point on the film the condition of Bright (maxima) or dark (Minima) depends only on ‘t’-   Air film is symmetrical. The locus of point of constant thickness is circular. Thus the rings formed are also circular. Let R= radius of curvature of plano-convex lens Let rn= the radius of nth ring formed at point B on the film Let t=thickness of the air film formed between points AB Thus- CO=CB=R AB=OD=t CD=R-t AO=BD=rn From the Figure in triangle CBD: 𝑅 2 = (𝑅 − 𝑡)2 + 𝑟𝑛2 𝐫𝐧𝟐 Let 𝑡 2 = negligible 𝐭= ……………. (5) 𝟐𝐑 Dn (diameter) D2n D2n since [rn (radius) = ] → [rn2 = ] t = 2 4 8R On putting the value of equation (5) in equation (3)  (2𝑛+1)𝜆𝑅 For bright rings 𝑟𝑛2 = 2 𝐷𝑛2 = 2(2𝑛 + 1)𝜆𝑅 ……………. (6) 𝑫𝒏 𝒃𝒓𝒊𝒈𝒉𝒕 𝜶 √(𝟐𝒏 + 𝟏) It show that diameter of bright ring is proportional to square root of odd integer number. 𝑫𝒏=𝟏 𝒃𝒓𝒊𝒈𝒉𝒕 𝜶 √𝟑 = 1.73 0.52 Decrement 𝑫𝒏=𝟐 𝒃𝒓𝒊𝒈𝒉𝒕 𝜶 √𝟓 = 2.25 0.39 𝑫𝒏=𝟑 𝒃𝒓𝒊𝒈𝒉𝒕 𝜶 √𝟕 = 2.64 It shows that spacing between two successive bright rings decreases as we move away from the point of contact. On putting the value of equation (5) in equation (3) For Dark Rings 𝑟𝑛2 = 𝑛𝜆𝑅 𝐷𝑛2 = 4𝑛𝜆𝑅 …………… (7) 𝐷𝑛 ∝ √𝑛 It show that diameter of bright ring is proportional to square root of an integer number. 𝑫𝒏=𝟏 𝒅𝒂𝒓𝒌 𝜶 √𝟏 = 1 0.41 Decrement 𝑫𝒏=𝟐 𝒅𝒂𝒓𝒌 𝜶 √𝟐 = 1.41 0.32 𝑫𝒏=𝟑 𝒅𝒂𝒓𝒌 𝜶 √𝟑 = 1.73 q It shows that spacing between two successive dark rings decreases as we move away from the point of contact.  Determination of wavelength on Monochromatic light Using Newton’s Ring Experiment The diameter of the nth dark fringe Dn2=4n𝜆𝑅 ……………. (8) The diameter for the (n+p)th dark fringe D2n+p= 4 (n+p)th λ R …………….. (9) Where p= integer of your choice in order to move to any higher order ring On subtracting equation (9) and (8) D2n+p-D2n= 4 p R λ ….…………… (10) 2 − 𝐷2 𝐷𝑛+𝑝 𝑛 𝜆= ….…………… (11) 4𝑝𝜆𝑅 In order to calculate the wavelength of monochromatic light following parameters will be required Dn+p is the diameter of (n+p)th order ring It is measured by making a tangent on that particular ring using micrometer eyepiece Dn the diameter of nth order ring p= known integer R= Radius of curvature of the plan-convex lens (R=150cm calculated by using instrument known as spherometer)  Determination of refractive index (µ) of a transparent liquid Using Newton’s Ring Experiment Once the experiment is performed in air wedge film then The diameter of the nth dark fringe in air Dn2=4n𝜆𝑅 ………. (1) The diameter for the (n+p)th dark fringe in air D2n+p= 4 (n+p)th λ R ……….. (2) From Eq (1) & (2) we have ………………….. (3) (Dn+p2 - Dn2 ) (air) = 4pλR The same steps are repeated / the experiment is performed again by introducing the transparent liquid (refractive index μ) between the wedge shaped film (between the lens and plane glass plate). Again the diameter of nth dark (say D’n2) and n+pth dark ring (say D’n+p2) is determined. Then Then for any nth dark ring D’n liq2 = 4nλR/µ ………..………. (4) And for any n+pth dark ring D’n+p liq2 = 4(n + p) λ R/µ …………….….. (5) Then Eq. (4) – (5) gives us the relation (D’n+p2 – D’n2 ) (liquid) = 4pλR/µ …………………..(6) From above two expressions by dividing Eq (5) & (6) we have (𝑫𝟐𝒏+𝒑 −𝐃𝟐𝐧 ) (𝐚𝐢𝐫) µ= (𝑫𝟐𝒏+𝒑 −𝐃𝟐𝐧 ) (𝐥𝐢𝐪𝐮𝐢𝐝) Thus by determining the value (Dn+p2 - Dn2) experimentally in air and in liquid wedge film, we can determine the refractive index. On Substituting Eq (1) in (4)  𝐷𝑛2 𝑎𝑖𝑟 𝐷𝑛2 𝑙𝑖𝑞 = 𝜇 𝐷𝑛 𝑎𝑖𝑟 𝐷𝑛 𝑙𝑖𝑞 = √𝜇 Thus the diameter on nth ring in liquid gets contracted by a factor of 1/√μ times the diameter of the same ring in air. Experimental Set Up