Page 1 : Engineenng, , PII, , 4-20, , ilarities and, , 2, , Dissimilarities between Newton's, , Films, , Interferance in Thin Fikm, , Rings and, , Wedge-shaped, , milaritles, , Fringes are formed due to enclosed thin film, , Both can, , be, , t h Can be, , ained only by the concept of division, , explaine, , of, , ampitude, , used for determination of optical flatness, , pissimilarities, , Newton's Rings, , Sr. No., , Wedge-shaped fiim, , We get alternate dark and bright, , We, , rings, , bright fringes, , Air gap has its thickness linearly, , Air gap, , straight, , get, , is, , alternate, , dark, , and, , non-linearly inereased, , increased, , Popularly used for, , Popularly used for, , determination of, , As, , we, , go, , thickness of, , for, , rings, , of very, , small thickness, , unknown wavelength, 4., , determination, , Fringe width, , orders, , higher, , remains, , constant, , reduces, , 4.13 Solved Problems, Problems on Thin, , Film, , soap film of, , 4.13.1:, , White light falls at, , an, , the fiim will it appear, , on a parallel, angle of 45°, of wavelength, bright yellow, , 5896 A, , in, , refractive, , index 1.33. At what, , the reflected, , minimum, , light?, , Soin., Ven:i=, , ormula:, , 45",, , H, , 1.33, 2, , =, , 2 ut cos, , nmum, , r=, , A, , 5896 A 5896 x 10 m, =, , =, , (2n, , )5,, , -, , thíckness,, , n, , is, , n, , =, , For bright fringe, , 1, 2, 3,, , n, minimum i.e., , t, , =, , 1, , 2x 24 cos r, 5896 x 10 1, 2x2x 1.33 x cosr, , Now, , in, , r, 13305316, , 33, , cos, , 8469, VI-s n ' r -, , thickness of

Page 2 : Engineering Physics -I(MU-FE), , 5896 x, , t, A, , light, , of wavelength 5500 Aincident, , 10-10, , x 2x1.33 x, , 2, , Ex. 4.13.2:, , interference in Thin Fim, , 4-21, , on, , 0.8469, , .Ans, , 1304.5A, thin, , transparent, , denser, , of refraction is 45°, the thickness of thin medium if the angle, , medium, , (Consider n, , =, , has refractive, , or, , stermine, , 1)., , Soln., side, Here, it is not specified whether reflected, , index 1.45., , be, transmitted side is to, , considered., , Let us assume reflected side., Also it is not specified whether dark fringe or bright fringe condition 1s Satisnea, , Let us assume dark fringe., , Condition for dark fringe on reflected side,, , 2ut, , cosr, , nl, , =, , 2u, '., , t, , nA, , cos r, , 1 x5500x 10-3, 2x 1.45 x cos 45, Ans., , 2.68x 10-5 cm, , Ex. 4.13.3: Light ofwavelength5880A is incidenton a thinfim ofglass of=1.5such that the angle of refraction in te, , plateis 60° Calculate thesmallestthickness ofthe platewhich willmakeit dark by reflection., MU -Dec. 14, 3 Marks, Soln., a, , 5880 x 10- cm, u = 1.5, r= 60, t=?, , Condition for film to appear dark is,, 2 ut cosr, , = nA, , The smallest thickness will be for n = 1., , 2x 1.5 x t xcos 60, , = 1x 5880x 10-8, , 5880x10-8, , t2x 1.5 x 0.5, t, , Ex. 4.13.4:, , x, , 10-8 cm, , A soap film of refractive index 1.43 is illuminated by white, examined by a spectroscope in which dark band, , For the, , thin film, refracted, , transmitted system is given by,, , A, , light incident, , corresponding, , Calculate the thickness of the film., , Soln., , 3920, , to, , at an, , t, , s, , angle 30°. The refracted light, , wavelength, , 6, , x, , 10, , light forms transmitted system and, the condition, 2ut cos r, , n, , m, , is observe, , o f minima ", , = (2n -1) / 2, , (2n-1), 4u cos T, , Tech, , Knoul, , 1 0 n 5, , UDIICatv

Page 3 : Engineering, Given:, , A =6x 10-" m,, , 1.43,, , Interference in Thin Film, , 4-22, , Physics-I (MU - FE), , i, , =, , 30°, , Using SnelW's law,, Sin 1, Sin r, , sin i, , sinr, , cos r = V1-sin2r, , sini, u2, , 1-, , sin 30, , --, , 1.43, , = 0.9369, , 2n-1)x 6x10-7, So,, , t, , 4x1.43x0.9369, , =, , n, , For minimum, , of film,, , thickness, , n, , =, , 1, 2, 3,, , . ., , .Ans., , 1.12, , =, , x, , 10-m, , on the s u r t a c e, , Ex. 4.13.5:, , 10-"m, , x, , 1, t, , Hence,, , 1.12, , (2n - 1) x, , =, , of, of a tank of water, , c.C. is dropped, is, volume 0.2, which is i n i d e n t normally, and white light, surface, the, with wavelength 5.5, band coinciding, uniformly over, one dark, , An oil, , is, , seen, , spreads, 1 sq. meter. The film, The, a spectrometer., , observed through, , drop of, , anectrum, , area, , x, , 10, , cm, , in air. Find the refractive, , to contain, , index of oil., , Soln., 0.2, of volume, The oil drop, , c.c., , spreads unitormly, , given by,, , The film, , dark, , by, , reflected, , thickness of the film, 1 m; hence the, , so, , formed iis, , 0.2, (100)22x, 10-5cm., , t, , appears, , over, , light., , 2ut, , cos r, , na, , =, , Hence,, For normal incidence, , r, , =, , 0, , ' . cos r, , n, , =, , =, , 1 and a, , 1, =, , 5.5, , x, , 10-5, , cm., , Refractive index of oil is, , 2t, , nA, , 1x5.5 x 10-5, , cos r 2 x 2x 10-6x1, , 1.376, , Ans., TechKnesledge, , PuDIIC attons

Page 4 : Engineering Physics - I(MU- FE), Ex. 4.13.6:, , A drop of oil of =, , Interference in Thin Film, , 4-23, , 1.20 floats on water with, , 1.33 surface and is observed from above by reflected light r, , The, the, , thickness of the oil drop at the edge is very smal-almost zero and gradually increases towards the middle ofh, , drop. Answerthe following, )Will the thinnest outer region of the drop correspond to a bright or a dark region? Give reason., , (i) What will be the thickness of oil drop where wavelength of 4800 A° is intensified in reflected light for the thi, , third, , order?, , Soln., ), , The thinnest region of the drop corresponds to a bright region because both the reflected, rays, one fom, the boundary between air and oil and another from the boundary between oil and water are in, Hence the condition for brightness is 2 ut cos r =na, , phass, , At the, , edge, t, , =, , 0 and satisfies the condition for maximum, , intensity., , ii) The condition for maximum intensity in reflected light is 2 ut cos r, , na, , For normal incidence,, 2 ut, , = nl, , t24, The thickness when n = 3,, , =, , 4800 Å and u = 1.2 is, , tt, Ex. 4.13.7:, , Light, , of, , wavelength, , 4800A, 2Xx 1.2 =, , 6000 Å, , 5893, , Ais reflected at nearly normal incidence from, What is the least thickness of the film that will appear, () Black, , Ans., , a, , soap film of refractive index 1.42., , (i) Bright, , Soln., i), , In reflected system, the condition of dark is, na, , 2utcos r, For normal incidence, r = 0, cos r = 1, , So,, , 2ut, , =, , nA,, , t=, , nA, , 2u, , For minimum thickness of the film, n = 1, , Hence,, , 5893 x 10-8, , t242x142, t, , CIm, , 2075 A, , Ans., , (ii) In reflected system, the condition of bright is, 2ut cos r, , = (2n - 1) / 2, , Tech, , Knowledge, , PuDlitatio

Page 5 : I (MU, Cogineering Physics, , normal, , incidence,, , r, , =, , 0,, , -, , FE, , 4-24, , cos r, , =, , interferencein ThinFim, , 1, , for, , thickness of the film, n = 1, f o m, r, i n i m u m, , 1.42, , Cm, , t1037 A, soap film of thickness 5x 10, 4.13.8: Light falls normally onbea refiected, ?, most, , Ans, , and of refracsve index 1.33 Which waelength, MU-Dec. 12.5 Marks, , cm, , strongly, , the visible region will, , Soln., maxima is given, The condition of, , by,, 2 u t cosr, , where n = 1. 2.3. ., , = (2n - 1) 2, , =, , t, , Given, , 1.33, , 5 10 cm, Le. cos r =, , ., , =0, , 4 ut cosr, 2 n -1), , NoW,, , 4x1.33, 5x10, (2n 1), By substituting, , the values ofn, , =1, 2,, , . . ., , we, , series, , a, , get, , of wavelengths, , which shall be predominantly, , reflected by the film., , Forn 1,, , 1335 105, -1, , =, , Similarly, For n, , 26.66, , 5.32, , h3, , For n= 3,, , x, , 10 cm, , x 10*em, , 3.8, , For n = 4,, Out of these wavelengths, ence, , 5320 Å i s the, , Ex.4.13.9, , most, , Whie ligtt, , is, , pectroscope, An, , the, , 5.320, , strongly, , is, , lies, , n, , vIsible, , the, , region (4000 Á to, , 7500, , ÅA, Ans, , reflected, , a, incident on, , cm, , 10, , cm, , x 10 em, , 8.866, , 2,, , 10, , x, , soap, , wavelength, n, , ar, , ), , aiye, , an, , the retfected, , COtespond to waveiengths, , DS, wo conbecuuv, tound thal, cakulate he thickness, is 4/3,, fuin, the, index of, , d, , and, , lignt, 61, , is odserved with a, , to, , and 6.0, , 10, , retractive, , Soln. :, , We have, IVe, , the, , condition, , bad, for dark, , 2 t, , system,, i n rellected, , cosr, , nAA, , Tec, , nenledge, , D I C a t i g o y

Page 6 : Ifn and (n, , +, , 1), , are, , Interferencein Thin Film, , 4-25, , Engineering Physics I(MU FE), the orders of, , consecutive dark, , bands for, , wavelengths A1 and z respectivelv, then, en,, 2 ut cos, , 2 ut cosr, , = nA1,, , 2 ut cos r, , = nAj = (n + 1) 2, , n, , r, , =, , (n+ 1), , ..1), , = (n + 1) A2, , n, , -2, , Put the value of n in equation (1), we have,, , 2utcos T, t, , (-), (a, , g) 24 cos, , -, , r, , 2, , (A-Ag)24V1Given, , = 4/3, sin i = 4/5, , sin, , As, , and, , sin r, , cos r =V1 -sin2r, , cosr 1/1-)Given:, , =6.1 x 105 cm, g = 6.0 x 105 cm, u = 4/3, , Put all these values in equation (2),, , 6.1x 10-5x 6.0x 10-5, , (6.1- 6) x 10-5 x 2x 4/3 x 4/5, t, , Ex. 4.13.10: A soap fim of refractive, , index, , = 0.0017 cm, , and thickrness 1.5, , x, , ..Ans, , 10 om, , is illuminated, , by white light incident at an anged, , 45". The light reflected by it is examined by a spectroscope in which is found a dark band corresponding t, wavelength of 5 x 10 cn. Calculate the order of interference band., , Soln., , Given, , t = 1.5x 10, , em., i = 45°, À =5 x 10, , em, , Formula: For dark band,, , 2 ut cosr, , nA;, , 2ut cos r =na, , Now, , sin 46, 4/3, a, , k, , a, , s, , w, , l, , e, , d

Page 7 : E, , n, , g, , i, , n, , e, , e, , r, , n, , i, , n, , g, , P, , h, , y, , s, , I, , C, , -1Sin45, , r, , = sin-1, , 4 /3, , sin, , cos, Order of, , 32.02, , =0.84778, , r, , dark band is, , 2utcosSr, , =, , n, , 0.8478, 2x1.33 x 1.5 x 10x, , 5x 10-3, , or, , n, , = 6.7, Ans, , 6., Order of dark band is, white, , Ex.4.13.11:, , index u is illuminated by, A film of refractive, and, fringes of wavelength A,, consecutive, , bright, , Az, , light, , are, , at, , an, , angle of, , found overlapping., , incidence, , i. in reflected, , light, , two, , thickness of, Obtain expression for, , film., , Soln., , Say, nth bright fringe, , with (n, of A1 overlaps, , +, , 1)th fringe of A2, , For maxima of A, 2ut cos r, , = (2n -, , ...(1), , 1) 2, , And for 2, 24t, , (2n, , So, , cos r, , 1), , { 2(n + 1 ) - 1}, , =, , =, , 2n (A-Ag), , ., , (2, , {20n + 1)-11, , = A1 -Az, , 2n, , ThickneBs of the, , film t,, , we, , equation (1)), get from, , r-A-1, 2u cos r (ÀL - Àu), Ex.4.13.12: Wtite g t is ncidert al an angie of 45" on a soap tim 4, , 10, , Ans, , em thick Find the waveiengtn of light in the, , vinible spectrum whith will be absentin the rolectod ight ( - 1 2 ) ., , Soln,, , Tecemedga, tr, ttan

Page 8 : Hore whil, , hs, , Interferencein Thin Film, , 4-27, , FngiiesingPiyaivs | (MU FE), , light is mads ineident, and on reflected side it, is expected to find the absent wavelength ia, , whioh will Halisly the condition for dark., , one, , 2 Jut c o s r = n A, , Here,, , =4, , i, , X 10" em, , = 45, , =, , 1.2, , sin45, , 12 s i n r, , r, , sin, , 46, , 1.2, , = 36,104, cOsr, , = 0,8079, , Now find A for various order n such that it remains between 4000 A to 8000 A i.e. visible spectrum., , For n, , 1, condition for dark fringe, , 1xa = 2 x 1.2 x 4 x 10- x 0.8079, , =, , 7755x 10- cm, , This is in visible range and it will remain absent, Similarly for n, , = 2, , 2x1.2x 4x10-6 x 0.8079, 2, = 3877 x 10- cm, , T'his is not in visible range,, , 7755 A will remain absent,, , Ans., , Ex, 4,13.13: A plane Wave of monochromatio lightfalls normally on a uniform thin film of oil, which covers a glass plate. The, , wavelength of the source can be varled continuously, Complete destructive interference, , Wavelengths 5000 A and 7000 A Find the thickness of the oil layer, Given R.I ofol= 1.3 and R.I of glass=1.5, , obtained only for, , MU-May 13,7 Markss, , oln., lere the path is trom air to oil and oil to glas8, , (1 orair to oil, P'or destruetive interferenee tlhe condition is, (for normal incidence), , 244uil oil, , ..1, , (2n + 1), , A the arrangement remaina NAme for both the wavelengths, Hoil and to will be the same or constant., const., , = (2n + 1D5, Tech, P, , Knowledg, u, , D, , I, , I, , r, , a, , t, , i, , g, , n, , '

Page 9 : Order of, , destructive interference is inversely, , .=7000 Å take order n, For, ForA, , =, , proportional to wavelength., and for, next wavelength Ag = 5000 Åtake next order ie., , n, , +1, , 7000 A, , 2 i toil, , (2n + 1) 7000, 2 A, , .(2), , For 5000 Å, 2 Hoil toil = (2 (n + 1) + 1) x, , (2n +1)x, , 5000, , ..3), , 5000, (2 (n + 1) +1)x 2, , 7000, , 5000, , (2n+1), , (2n+3) 7000, On solving, n = 2, , Substitutein equation (2), , (2(2) +1)x7000, 2x 1.3 x 2, , Toil, , ..Ans., , = 6730.769 Å, Problems on, , Wedge-shaped Film, , is, , of 40 seconds, an angle, , illuminated by, , having, The distance measured, microscope., vertically through a, used., , shaped airfilm, Ex.4, A 4.13.14: A wedge, , monochromatic, , light and fringes, , are, , bright fringes is 0.12, between consecutive, , MU- Dec. 17, May 18, 5 Markss, , observed, wavelength of light, cm. Calculatethe, , 40degrees, , Soln., , degrees, seconds =3600, , 40, , Given, , 40, 3600, , B, , ormula :, , Spacine b e t w e e n, , the, , consecutive, , 180 radians,, , 0.12 cm., , bright fringes, , 1s,, , (For air film), , B 26, 40 x TT, a, , 28 0=2x0.1z x 3600x180, , =, , The wavelength isS,, =, , 4654, , x, , 4654, , A, , 10-8, , cm, , Ans., , A, , Ex. 4.13.15: Light, h a t are, , avelength, , 2.5, , mm, , =, , on, normally on, falls, falls normally, 5500 A°, , apat., , Find the, , angle of, , aa, , film of refractive, thin wedge-shaped, , wedge in, , index 1.4 forming fringes, , seconds., , TechuDICatlons, Kaewledge

Page 10 : Interference in Thin Fim, , 4-29, , Engineering Physics-1 (MU FE), , Soln, We have fringe width,, , 2u, , 2uf, 5500, , Given, , 10 em, ju 14,B 0.25 cm, , 5600x 10, , *, , 2 14x0.26 7 86, 7.86x, -, , 180, , 10, , 0.0045, , 3600, , x, , 10 radian, , 0.0046°, , 16.2 see., , Ans, , Problems on Newton's Rings, Ex. 4.13.16: Newton's nings are obtained with reftected light of wavelength 5500 A. The diameter of 10" dark ning s, 5 mm. Now the space between the lens and the plate is filed with a liquid of retractive index 125. What s he, , diameter of the 10 ring now ?, , Soln., , Given:, , 5500 A, , Formula: Diameter, , =, , 5500x 10, , of nh, , " cm, Dio, , 5 mm, , =, , dark ring is given, , =, , 0.5 cm,, , u, , =, , 1.25, , by,, , D, , 4nRA, , For the air film, u = 1, , Hence diameter of 10th dark ring is,, , (0.5, For the, , liquid film,, , 4x 10xRx 5500x 10", , D, , 1, , the diameter of the 10h dark ring is,, , D, , D, , D, , 4x 10 x Rx 5500x 10, 1.25, , 1, , 126, , D, D 125, D, , D, , 26" 11180, , 0.6, 18, , 0 4472 cm, Diameter of 10h dark ring tur the, liquid filu, , D, , 4.474, , Ans.

Page 11 : inering Physics-I, , Engineeni, , (MU FE), -, , -30, fnterference in Thin Fim, , ewton's rings formed with sodium light between a, ihat will be the order of the dark ring which will have flat glass plate and a convex lens are viewed normally, double the diameter of that of the 40 dark ring ?, , 413.17, , Ne, , MU-May 19,3Marks, ln., l e the, t, , diameter, , ofthe nth dark ring be double the diameter of 40h dark ring, D, 2 D, , Nor the diameter of n"h dark ring is given by the expression, , D, , ..1, , =4n R, , Where, R, , Radius of curvature of lens., , =, , =Wavelength oflight., Hence for the 40th dark ring., , D, From equations (1) and (2), , we, , 4, , =, , 40, , x, , x, , 2, , R, , have, , D, , = 4x D, , 4nRA, , x4x 40 x, , 4, , =, , R, , .Ans, n, , 4.13.18:, , The diameter of 5th dark ring, , = 160, , ring experiment, , in Newton's, , was found, , to be 0.42, , cm., , Detemine the diameter of, , MU-May 16,4 Markss, , 10 dark ring., , Soln., , D=4 nRA, AS, diameter, , Now,, , of 5th dark ring, , diameter, , =0.42 cmn, , ring, of 10th dark, , ?, , =, , 45)R, , 4(10) R, , D0, , 2 (D) D, =, , =, , Dio, , 2, , =, , (D,), , =, , V2 (042), , 0.594 cm, , Ans, Diameter, , Ex. 4.13.19:, , ring, of 10th dark, , =, , is, , A, , N, , fing, , 4.5, the, , 10, , 0.594, , usod, , arrangement, , om., , e, that ihe, t is fourd, , is, curved s u r t a c e, , 00, , cm,, , cmn, , wth, , a, , source, , emitting wo wavelengths à, coincikkes with, , A, concikes, duo oto A,, ring due, dask tug, ndark, 3 dark nng tor A,, , f d the, , diameler, , of, , (, n.", ah, 1) ddark, (n., , 6, , rng tor, , ns, , 10, , Ag, , cm, , and, , If the radius

Page 12 : Interferencein Thin Fim, , 4-31, , Engineering Physics I (MU -FE), Soln., , Given:, , A, , =, , 6, , 10, , x, , cm,, , Aa, , =, , 4.5, , x, , 10, , cm,, , R, , 90 cm, , (D, , D.., , Formula: Diameter of nth dark ring is,, , D= 4n Rà, , (A= 1), , For the nth dark ring A1, D, , and for the (n, , = 4 n R1, , 1), , 1)th dark ring h2, , +, , D., ., , 4(n +1) R, , 4nRA, , =, , n, , 4 (n +, , ..2), , 1) R 2, , =(n+1)2, n, , -m, , 0.45 x105, , (6-0.45) x 10-5, n, , Using equation (1), the diameter of, , = 3, , 3rd dark ring for A^ is, =, , D, , 4x3x90 x6x 10-5, , = V4x3x90 x 6 x 105, = 0.2545 cm., , Diameter of third dark ring for, Ex., , 4.13.20, , Light, , glass plate. Now, dark ring, due, , containing, , if the, , two, , wavelengths, , n" dark ring due, , to, , R, , to , is N-, , is, A, , 0.2545, and, , A2, , cm., falls, , normally, , on, , a, , convex, , à, coincides with (n, +1) dark ring due, , lens, , to, , 2,, , of, , radius, , of, , curvature, , R,, , resti, A, ng n s, , then prove that the radius of the, , ", , Soln., The, , diamneter, , of, , For nh dark, , n", , dark ring is, , ring of A., , given by a relation, D, 4n RA, , D, , 4, , n RA1, , ..1), , Technouled, Pubiiratiou

Page 13 : Engineeil, , a n d, , (n, , +, , 1)th, , dark, , ring of Ag,, D., Now, , 4 (n + 1) RA, , D, , = D., , 4nRA, ., , (2), , as per data, , = 4 (n + 1) RA2, , n, , a -m, ring of A1 is,, , the diameter of nth dark, As per equation (1),, , D, = 21/n RA1, Radius of nth dark, , ring of A is,, , D-nR, , Tn, , .R, , or, , A, , =, , Hence proved., separation, , Ex.4.13.21:, , goes, , Soin.:, , that, with clear examples,, number of, a s the serial, , Show,, , on, , consecutive similar, between two, , ring, , According to theory, , diameter, , ring the, , Da, , n,, , D,, Or, , wheren, , example, first, , 1ence, , the, , is the serial number of, , calculate, , separation, , the separation, , between, , V5, , D, , 4, , 5th, , and 4h, , the, , separation, , and, , between, , We can, , rings, , experiment, , dark, , ring is given by,, , the ring., =, , =, , ring,, , 2.236 and, , 2, , ring is, , «0.236in SI, , 79th and 80h, , unit., , ring is, 8.9442, , Da0, , V80, , D, , V79 =8.8881, , Ds0- Dr, , Hence,, , in Newton's, , V4n RA, , =, , Ds-D, , rly,, , of the nh, , dark, of 5th and 4th, , Ds, , rings,, , increases., , reducing, , of Newton's, , R, , -, , =, , 0.0660, , in SI unit., , conclude that,, DO-D, , Ds, , D,hence proved., , Tec, , ed

Page 14 : 4.14, , Interference in Thin Film, , 4-33, , Engineering Physics I (MU -FE), , Determination of Thickness of Very Thin Wire or Foil, , There are many applications wherein we need to know the thickness of a very thin wire or an, equivalent for example "contact lens". As mentioned in introduetion, concept of inference can lead us to, an experimental arrangement which gives us accurate measure up to one tenth of a micrometer., Measurement of thickness of very thin wire or foil:, , As seen in the case of "spacing between two consecutive bright bands"', we have derived fringe width B=, 20, , The same concept is used for determination of thickness of thin wire. We take two glass slides (optically, flat) and put them in touch at one end and at the other we put the wire or foil whose thickness is to be, determined. Hence we prepare a wedge with very small angle 0., Here we make the use of same setup which is used for Newton's rings. In place of plano convex lens we, take the above mentioned wedge., On viewing through microscope when illuminated by a monochromatic light of wavelength A at normal, , incidence, we get alternate dark and bright lines., We take readings of dark lines at the spacing of some interval say P number of lines., We draw, , a, , graph for Reading- order of dark line and find the fringe width B'., , From experimental set up measure ", , tan 6 -=, , ., , (as, , is very smal), , 6, , Fig. 4.14.1, , as, , .t, Since A, l and ß, , EX.4.14.1, , are, , known we, , can, , 20, , 23B, , calculate thickness ofa foil, , ..(4.14.1), or, , thin wire., , wo, , plane glass surfaces in, contact along one edge are separated at the, opposite edge by a thin wire. n, interterence fringes are, observed between these edges in sodium, light at normal incidence, what is e, thickness, , of the wire? (Given, , =, , a, , 5893 A), , Tech, , Knowledye, , ubIications

Page 15 : Physics, , I (MU, , FE), , 4-34, , Engineering, , Fringe width B =, , For, , airfilm, , 2u6, , Thickness of wire t, , (For normal incidence), , B=2, , 1,, , =, , Interferencein Thin Film, , = X, tan 8, , (for small 6,, , Xn, , tan 0-0), , 20.B.0, As there, , are, , 20 fringes between two edge points, Wire, , n, , Fig. P. 4.14.1, , Putting the value, , of ß,, , we, , get,, , 10A, , 20 9, , t, , 10, , =, , x, , 5.893, , =, , x, , 5893, , 10-10, , 10-, , x, , m, , Ans., , = 5.893 um, M, , Determination, , .15 Deterr, 4.15, Lens by, , With, , the, , Rings, , the radiusof, , 15.1, , o, , c, , h, , r, , a, , t, , ic Light, i, , necessary, , expressions,, , lens, , expressions, , explain how, , Newtor, ring experimentis, Newton's, , cleaned, , of a, , lens, , u, , or, , large, , radius, , (Dec.14,8 Marks), , placed, of c u r v a t u r e is, , G, Another glass plate, I, , at, is held, , normal incidence on the film., , on a, , plano, , convex, , plane glass plate P. (Fig. 4.15.1), , of, , (May 17, 5 Marks, , refractiveindex of liquid?, determine the, to, how, explain, , lens., , surface, , Curvature, , useful to detemine the, , necessary, , experiment, , of, , 13, Dec. 18, 5 Marks), (May, , ow, explain how, , mental Arrangement, , carefully, , Radius, o r Radius, or, , c, , 17, Dec. 18, , convex, , A, , m, , MU- May 13, Dec. 14, May, isuseful, experi, ring experiment, Newton's, ton'sring, , convex, ofa plano, , and, , o, , of, , curvature, , plano-convex, of c u r v a t u r e of, , Newton's ring, .Wth Newton, , n, , Method, , and, proper diagram, , p e r diagram, With prope, , radius, , Wavelength, , Newton's, , help of, , L m i n e, , of, , o, , suitable, , distan, ance, , above at, , an, , angle 45° with the, , vertical, , (to make the, , a, , TechKneledge, Publications

Page 16 : Interference in Thin, , 4-35, , Engineering Physics- I (MU - FE), , lamp), , (sodium, , is allowed to fall, , condensing lens, maximum light from, the lens L and plate P., plate G reflects light onto the air film between, With, , source, , a, , G. The inels, ncli, , on, , Microscope, , 45, Sodium, , lamp, , P, , Fig. 4.15.1, , Newton's rings are formed, bottom faces of the air film., , They are seen through a, , as a, , result of interference between the rays reflected from the, top, , low power, , microscope focussed on the, , air film where the, , rings, , are, , an, , formed., , Theory, The effective, , path difference between the interfering rays is,, , 2utcos(r+0)+5, Where, k = R.I. of film., , If, , D, , below, , is the, , Angle of film at any point, =, Wavelength of light, , diameter of nth dark ring then, , as, , per the, , theory, , of, , Newton's rings described, , in, , equatio, , D= 4 n Rà, where R= Radius of, curvature of lower surface of, lens., Let x, , D+D, , D.p -Da, Alternately if wavelength of incident, , Ri.e. the radius of, curvature., For a more accurate, approach, , we, , =, , =, , 4 (n+ p) R, , for (n+p)h ring), , .(4.15.1), , 4 pRA, 4 pR, ...(4.15.2), , monochromatic ray is known,, , plot graph, a, , of D, , vs n as, , shown, , using equation, (4.15.2),, , in, , we cam, , find, , Fig. 4.15.2., , TechKnouledge, P u b l i c a t i o n e

Page 17 : Physics I (MU FE), , 4-36, , -, , Interference in Thin Film, , gheering, , asiring the diameter of nth and (n + p)" dark rings and the radius of curvature R, the, Thus., by., , carelength A, , can, , be calculated., , n v e l, , of the rings is, , ameter, h ediame, , measured with the, , equation, , lens, termined by using, jeterm, , or, , by, , a, , travelling microscope, , and the radius of curvature, , can, , be, , spherometer., , D, Dn+P, , n, , n+p, , Fig. 4.15.2, , Lens, of Radius of Curvature of, , Determination, , 15.2, This, , can, , and the formula, of spherometer, be done easily with the help, , ...(4.15.3), , Rh, Where,, , l, , two, Distance between, , =, , h, , =Difference, , legs of spherometer, , the diameter, Newton's rings experiment, , in, , . 5 9 0 cm., , Soln., Given, , Find the radius of, , Ds=0.336 cm,, , Di5, , 5890 Å, , 5890, , =, , Formula D.,- D, :, , =, , of 5, , ring, , 10e, , lens a s well a s, , when placed, , on, , of 15, used is 5890 A, if the wavelength of light, , was, , lens, curvature of plano convex, , 0.590, x, , on, , when placed, in the reading, , 0.336, , cm, , and the, , diameter, , surface, ring, , was, , MU-May 15,Dec. 18, 5 Marks, , cm,, em, , 4pRA, R, , D, D, 4, p a, (0.590)-(0.336), , 4 x 10, , x, , 5890, , x, , 10-*, , = 99.91 cm, , R, , 4.16 Dete, , ermination, , ., , 99.91 cm, , Index of a Liquld, of Refractive, , With, , Ans., , by Newton's Rinas, , the refractive, how to determine, , index of liquid, , explain, NewMon', On'ss ring, expe, How is Newt, indexofliquid, won's ring experiment used to determine refractive, experiment, , medium?, , MU-May 14, May 16, (May 14, 4 Marks), , (May16,4 Marks)

Page 18 : Interferencein Thin Flm, , 4-37, , Engineering Physics -1 (MU -FE), , There is a popular branch of engineering known as "hydraulics" in which liquid is in motion and, transmits energy, viscosity and refractive index are some of the parameters which can be used to decide, , the suitability of the oil (called grade of the oil). In such applications, the correct measurement of the, R.I. is thus an important issue., Consider that a transparent liquid whose refractive index is to be determined is placed between the lens, L and plate P of the Newton's rings arrangement., Tf the liquid is rarer than glass, a phase change of r will occur at the reflection from the lower surface of, , liquid film., If the, , liquid, , is denser than, , glass, then, , a, , phase change of r, , will, , occur, , due to reflection at the upper, , surface of the film., Hence in either case,, , a, , path difference, , of /2 will be introduced between the, , interfering, , rays in the, , reflected system, and hence the effective path difference between them will be, , 2ut cos (r+8) +, Now, , r, , = 0 for normal incidence, , 0, , for large R, , Path difference 2 ut+, , For nh dark ring, we have,, , 4. D, R n, , +2 =(2n +1) /2, D, , nR2, , Similarly for the (n + p)th dark ring we have,, , 4 (n+p)R2, , Dp, , ..4.16.1), , D., -D 4pR-A, , Dn+p, , Diameter of (n, , +, , p)h dark ring, then, , we have, , 4pRA, , (D.-D, =, , For, , liquid, , 4 pR, D . - D ] =1(* u=1 for a, air, , more, , accurate, , approach we plot a graph of D, , vs n as, , shown below., , ..04.16.2), , ...4.16.3)

Page 19 : E n g i n e e r i n g, , P h y s i c s s, , similar, , to Fig., , 4.15.2., , (4.16.3), 4.16.2) and, , tis, e, , q, , u, , a, , t, , i, , o, , n, , have,, , we, , s, , D,-D., air, , From, , p.-D., liquid, , Tf we, , obtain, , if, , slowly, , we, , using, , structure, , ring, , D, , (liquid), , diameter, , then, , medium, , the, , as, , R.I. u then, , liquid with, , insert a, , air, , is given by, , ratio, "4ni Take, Take ratio, , =, , Newton's, , measured, , formed, , are, , rings, , film. The, with air, , u,, film., for air, , are, , The, , between, , and, lens L, , diameters, , transparent, , liquid is, and, , of nth, , diameters, , measured, , formed, Newton's, , in, , reflected, , 6, , of the, , If the, , light of, , wavelength, , mm, , is 3.1, bright ring, , are, , rings, , diameter, , surfaces., , with a, 6000 A°, , be, , can, , X . 4.16.1:, , with, , liquid, , index, , of the, , p)*h, , +, , film., , between, calculated., , (4.16.4), k, , (n, , rings, , dark, , using, , Hence, , are, , dark rings, , p), , +, , (n, , ofnth and, , plate P., , introduced, , equation, , these two., ofthese, , D (air), , find, Thus to, , the, , 4nRA. Now, , =, , ...(4.16.4), , D(liq), , Then, , (air), , D, , and the, , radius, , the plane, , and, , liquid, of, , curvature, , of the, , curved, , curved, , surface, , is, , liquid., , refractive, , the, calculate, , 100 cm,, , Soln., The, , diameter, , of n, , th, , bright, , 2(2n-1)AR, , ring is,, , D, R=, , n, , Given, , =6,, , a, , 6000 x, , 10-*, , 2(2n-1) AR, D, , 100 cm,, , cm,, , D6, , =, , cm, , 0.31, , 22 x6-1), , 6000 x 10-8 x, , 100, , (0.31), , =, , 2 x11x6x10-3, (0.31), , Ans., , = 1.373, , Interference, , of, , Applications, , 4.17, , one, , fact, , that, , understand, , if a, , has its, thin film, , easily, Reader, , a, , can, , =wavelength, , of monochromatic, , to, ray used, , illuminate,, , then total, , thickness, , path, , altered, , difference, , is, , T, , reflected ray., , where, , by, , e, , c, , h, , K, , n, , o, , w, , PubiiC, , l, , e, , d, , g, , e

Page 20 : Engineering Physics I (MU FE), Now, , is the path, , This is, , Interference in Thin Film, , 4-39, , interference, difference which converts, , easily observable. Hence if we, , fringes, , from dark to, , bright, , which is illuminated, consider any thin film, , or, , do., bright to dark, , by Na lamp (5Rac, , then a charge of, , x, , 5896, , =, , 1474, , A, , can be detected by change in interference dark to bright (or bright to dark). It is important to note that, , 1474 A 0.1474 micrometer., Now coming to the area of surface finish, it is essential to know that surface finish of parts an, , significantly affect their friction, wear, fatigue, corrosion, tightness of contact joints, position accuracy, and so on. Surface finish has always been considered an important factor for manufacturing process, , monitoring and quality control inspection., Surface finish is, , a, , representation, , of the vertical deviations of, , a measured from its ideal form. If the, deviations are substantial then the surface is, rough and if these deviations are minor the surface is, smooth. For many engineering, applications, the finish on the surface can have a big effect or the, or, performance durability of parts. Hence, a quick measurement of flatness is, an, important area and, optics through interference is considered useful. As, mentioned above, interference is, capable o, detecting a variation of 0.1474 micrometres using Na, lamp., , The smoothness of surface, , can, , be detected, , by using, , surface usually within a few tens of, nanometres used, unknown surface may be, , an, as, , compared., , An, is, , optical flat which is precisely, polished flat, a reference, against which the flatness of an, , optical flat is usually placed upon a surface under, investigation., , used, , to, , illuminate the workpiece, a series, interference fringes determine the flatness, accuracy of few micrometres, formed., , as, , of dark and, bright, of the work, , discussed above. By, using, , monochromatic light (Na lamp, interference fringes are formed. These, a, , piece relative to the, optical flat, a, wedge as shown in Fig., 4.17.1,, , Optical flat, , Work piece, , Fig. 4.17.1, , If, , up to the, , fringes, , are

Page 21 : workpiece, , is perfectly flat then, , straight parallel interference fringes will form., , the, , Flg. 4.17.2, , More a n d, , indicate, , thinner fringes, , indicate, wider fringes, steeper wedge while fewer but, , a, , wedge., , ce, , fthe workpiece, , is, , concave o r convex,, , the, , fringes will be, , as, , shown in, , Fig., , smaller, , 4.17.3., , Fig. 4.17.3 (b), , Fig. 4.17.3 (a), , f, , towards, , the fringes, , are, , curved, , fringes, , are, , curved away, , the, , 4.17.1, , Testing, , the Optical, , the, , edge, the, , contact, , surface, , the work piece, , Flatness, , of, , workpiece, , is, , surface, , Fig. 4.17.3, , convex, , is, , concave, , (Fig., , 4.17.3, , (a)) and, , (b)., , Surfaces, , Fringes of equal, thickness, , B, , Fig. 4.17.4, , testing, , T, Eomenon, , of, , interference, , OB (Fig., surfaces OA and, , Iftwo surfaces, , in thickness from ), ickness, , If the, inges, , To test the, ne, , used in, is also, perfectly, are, , 4.17.4), , fringes, to A. The, , of the, , flatness, , as, , each, , are observed, , them g r a d u a l l y, , varies, , the points, fringe is the locus of, , at, , value., , a, , t h a t the, , it, , are, , urfaces, , sur, , not, , plane., , m e a n s, , surtace, , of a, , surfaces., , nim between, the a r, , constant, , film has, , ptically plane surface (OA)., , "The giri e s, , plane,, , of the, , ofequal, , not ofequal, , optical, , plainness, , thickness, , are, , thickness,, , are, , the, , surface,, , the, , (OB), to be tested, , is, , nlacod, , an, , specimen, , they, fIf they, of view., field, in the, , of equal, , a r e or e q u a l, , t, th, hi, ic, ck, kn, ne, es, ss, s, , urtace, surface, the s, , OR, OB :is, , plane. If, , hen surface OB is not plane., Tech Kneledge, PudIIcations

Page 22 : The surface OB is, polished and the process is, means the surface OB is, , plane., , terrerence in hin Film, , repeated. When fringes observed, , The accuracy in this, method is far more, accuracy level is of the order of, , fraction, , 4.18, , Concept of Anti-reflecting, , Q., , superior compared, of micrometer., , a, , other, , equal width, it, , technique adopted. The, , Coating (Non-reflecting Films), , Describe in detail the, concept of, film, the material should satisfy to act anti-reflecting, as, anti, , What do you understand, by, , to any, , are of, , MU-May 12, Dec. 13, Dec. 17, Dec. 18, , with a proper ray diagram of thin film interference., , reflecting film?, , Which condition, , (May 12, 8 Marks), (Dec. 13, 8 Marks, (Dec. 17, 5 Marks), , anti-reflecting coating? Derive the conditions with proper diagram., anti-reflecting film with a proper ray diagram., What is antireflection, coating? What should be the refractive index and, minimum thickness of the coatin9?, , Describe in detail the concept of, , We, , are aware, , When the, , n, , =, , that, , (Dec.18,3 Marks), compound microscope, telescope, camera lenses,, , ete., , use a, , combination of lenses., , light enters the optical instrument at the, glass-air interface, around, and glass with n2, 1.5) that too at single reflection is lost, , 1, , =, , undesirable. For advanced telescopes the total loss, if working under low intensity, applications., In order to reduce the reflection, This film is known as, , loss,, , a, , 4%, , of light (for air with, , by reflection which, , comes, , out to be, , nearly 30%, , is, , highly, , and cannot be tolerated, , transparent film of proper thickness is deposited on the surface., , "non-reflecting film"., , Popular material used is MgF2 because, , its refractive index is, , (n = 1.36) is also used., , Thickness of the film, , may be, , obtained for given purpose, R, , as, , 1.38 (i.e., , shown in, , air n, t, , between air and glass). Cryolite, , Fig. 4.18.1., , = 1, , Thin film, R. I., , =, , 4, , glass n2, Fig. 4.18.1 Thin film coating, Let, as, , a, , ray I be incident upon thin film of MgF2 coated, , Ri and from lower surface, , incident ray enters from, , as, , rarer, , on, , glass. This, , ray is, , reflected from upper surface, , R2. The optical path difference between these, , to denser twice i.e. at air to film and film to, , two rays is n,, , (2t),, , the, , as, , glass., , If both the rays Ri and R2 interfere with each other and path difference is (2n + 1) /2 (for n = 0, 1, 2,, ....) then destructive interference will take place., , Tech, U, , Knouied, B, , I, , c, , a, , t, , g, , ", , '

Page 23 : no, , n2, , -, , -, , ---*, , R, R2, Fig. 4.18.2, (for n = 00), , 2, , n t:, , nt44, have, It means, in order to, , destructive, , interference a, , layer of n1t, , =, , is coated, , on, , glass plate., , 1 and ray 2, of reflected rays, ray, amplitudes, the, that, condition requires, , Amplitude condition, The amplitude, , are, , equal. That, , E=E2, , is,, intensities, , of two, , reflected, , beams should be equal, , condition,, , For complete, It, , destructive, , requires that,, , LHg +H, , LH+HaJ, , where, , a, , ., , 1 , the, , and e, above, , Take square, , are, , the, , expression, , root, , on, , refractive, , may be, , both s1des, , to, , indices, , rewritten, , compare, , -1) (Hg +, , air,, , thin film, , and, , glass, , substrate respectivelv., , As, , as, , their, , E+1, , or, , amplitudes,, , Pg+HE, , (g-H) (Hf+, , 1), , Teck, , nenlelge, , Publiationns