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Subject: PHYSICS-I Subject Code: PHY-101F L T P 3 1 0 Internal (Sessional + Class performance) 50 Marks Exam 100 Marks Total 150 Marks Duration of Exam 3 Hrs. SYLLABUS (Ist Semester) 1)Interference 2)Diffraction 3)Polarization 4)Laser 5)Fiber Optics 6)Dielectrics 7)Special Theory of Relativity 8)Superconductivity Section A Interference: Coherent sources, conditions for sustained interference. Division of Wave-Front - Fresnel’s Biprism, Division of Amplitude- Wedge-shaped film, Newton’s Rings, Michelson Interferometer, applications (Resolution of closely spaced spectral lines, determination of wavelengths). Diffraction: Difference between interference and diffraction Fraunhofer and Fresnel diffraction. Fraunhofer diffraction through a single slit, Plane transmission diffraction grating, absent spectra, dispersive power, resolving power and Rayleigh criterion of resolution. Section B POLARISATION Polarised and unpolarised light, Uniaxial crystals double refraction, Nicol prism, quarter and half wave plates, Detection and Production of different types of polarized light, Polarimetry; Optical and specific rotation, Biquartz and Laurent’s half shade polarimeter. LASER Spontaneous and Stimulated emission, Laser action, characteristics of laser beam-concept of coherence , spatial and temporal coherence, He-Ne and semiconductor lasers (simple ideas), applications Section C FIBRE OPTICS Propagation of light in optical fibres, numerical aperture, V-number, single and multimode fibres, attenuation, dispersion, applications. DIELECTRICS Molecular theory, polarization, displacement vector, electric susceptibility, dielectric coefficient, permittivity & various relations between these, Gauss’s law in the presence of a dielectric, Energy stored in a uniform electric field, concept of local molecular fields and Claussius Mossotti relation. Section D SPECIAL THEORY OF RELATIVITY Michelson’s Morley Experiment, Postulates of Special Theory of Relativity, Lorentz transformations, Consequences of LT (length contraction and time dilation), addition of velocities, variation of mass with velocity, mass energy equivalence. SUPERCONDUCTIVITY Introduction (Experimental survey), Meissner effect, London equations, Hard and Soft superconductors, Elements of BCS Theory. Text Books: Fundamentals of Engineering Physics Modern Physics for Engineers M. S. Khurana, MR Pub, Delhi S P Taneja; R Chand Publication Engineering Physics Satya Prakash, Pragati Prakashan Engineering Physics R. K. Gaur & S. L. Gupta Reference Books: Concepts of Modern Physics Beiser Optics, A. Ghatak, Optics Eugene Hecht Fundamentals of Optics Jenkins & White Lectures on Physics Feynman Practical Books: Fundamentals of Engineering Physics – M. S. Khurana, MR Pub, Delhi Advanced Practical Physics – B.L. Worshnop and H. T. Flint Practical Physics – S. L. Gupta & V. Kumar (Pragati Prakashan). Advanced Practical Physics Vol.I & II – Chauhan & Singh. Course Objective: The objective of this course is to apply the concepts of physics (i) to different optical phenomena (i) devices based on these phenomena INTERFERENCE Interference? Why we need to study? There are two possibilities: 1. 2. Maximum intensity position Minimum intensity position Young’s Experiment: crest A Trough S (Coherent Source) B (Slit) (Screen) Coherent sources ? Conditions for sustained interference: Two light sources must be coherent. Two coherent sources must be narrow, otherwise a single source will act as a multi sources. The amplitude of two waves should be equal so that we can get good contrast between bright and dark fringes. The distance between two coherent sources must be small. The distance between two coherent sources and screen should be reasonable. The large distances of screen reduce to intensity. Theory of Interference P A O d S (Coherent Source) B D Slit Screen The path difference is an even multiple of 2 , the waves arrive at a point in the same phase and produce maximum intensity. If the path difference is an odd multiple of 2 , the two waves arrive at a point in the opposite phase, they cancel out and the resultant intensity should be minimum. How to calculate the fringe width: To determine the spacing between the bands/ fringes and the intensity at point P. P x S1 d 2 S (Coherent Source) M O d d 2 N S2 D Slit Screen The path difference can be calculated from fig. Path difference (Δ) = S2P-S1P To calculate S2P, consider the ∆S2NP S 2 P 2 S 2 N 2 NP 2 d S2 P2 D2 x 2 1 2 2 d x 2 S 2 P D 1 or D2 Expending by binomial theorem and neglecting higher power term of D as D>> x d 2 2 We have 2 d x 1 2 S 2 P D 1 2 2 D d D is very very large in comparison of x 2 Therefore, higher power term of D can be neglected. Then we get 2 d x 2 S 2 P D 1 2D 2 or 2 d x 2 S 2 P D 2D Similarly, we can calculate S1P, consider the ∆S1MP 2 d x 2 S1 P D 2D Then the path difference is (Δ) = S2P-S1P putting the value of S2P and S1P then we have Path difference For the nth xd D xn d fringe the path difference = D (a) Bright Fringes: We have already pointed out that for bright fringe, the path difference should be equal to n . xn d n D D xn n d where n = 0, 1, 2, 3, 4, ---------------- x 0 Therefore, the distance between two consecutive fringes is also known as fringe width. D n 1 d D xn n d xn 1 x n 1 x n D n 1 D n d d D D D n n d d d x n 1 x n D d D d (b) Dark Fringes: Similarly, for dark fringes, the path difference should be equal to 2n 1 2 Point P to be dark; xn d 2n 1 D 2 D x n 2n 1 2d where n = 0, 1, 2, 3, 4, -------------- x n 1 x n 2 D D 2d d D d From the above equations, it is clear that fringe width depends upon following factors: It is directly proportional to the distance between two coherent sources and screen D It is directly proportional to the wavelength of light It is inversely. proportional to the spacing between two coherent sources 1 d Type of interference: Interference by division of wave front Interference by division of amplitude Fresnel’s Biprism: Fresnel’s biprism is a device to produce two coherent sources by division of wave front. P S1 G Overlap region S S2 O H a b D Q Construction: A biprism consists of a combination of two acute angled prisms placed base to base. The obtuse angle of the biprism is 179º and other two acute angles are 30’. 30 179° 30 Applications of Fresnel’s Biprism: Determination of thickness of thin sheet of transparent material like glass or mica. or How to calculate the displacement of fringes when a mica sheet is introduced in the path of interfering rays? t m x S1 Po d S2 P O D Let the point P is the center of the nth bright fringe if the path difference is equal to nλ The path difference between S2P and S1P is S 2 P S1 P t mt n or S 2 P S1 P m 1 t n We have already calculated that S 2 P S1 P xn d D xn d m 1t n D or xn D m 1t n d (1) Where xnis the distance of the nth bright fringe from the central fringe in the absence of mica. The position of the central bright fringe when the mica sheet is placed in the path S1P is obtained by putting n=0 in equation (1) we get x0 Since m D m 1 t d (2) >1 so that x0 is positive. The fringe width is x n 1 x n D d Using equation (1) It means the fringe width is not affected by introduce of mica sheet. D d Put these values in equation (2) we get, t x0 d Dm 1 or t x0 m 1 Thus we can calculate the thickness of mica sheet. (b) Determination of the distance between two virtual sources: Displacement method is one of the methods to calculate the distance between two virtual coherent sources: m n n m According to the linear magnification produced by the lens: size of the im ageI dis tance between im ageand lens size of the object O dis tance between object and lens d1 n d m Further the lens moves towards the eyepiece and sets a potion of image of virtual sources S1 and S2 in eyepiece again. This time the image separation of S1 and S2 should be appear different (d2) so that: d2 m d n From equation (1) and (2), we get or d2 d d d1 d 2 d1d 2 d d1d 2 Interference due to Reflection: (Reflected rays) R1 R2 Source i iN M A r r B C t D T1 T2 (Transmitted rays) Path ABC in film path AN in air m AB BC AN (1) BM cos r AB BM = t AB and also BC t cos r t cos r AB BC Now, for AN AM tan r BM AM BM tan r or AM t tan r 2t tan r AC = AM + CM (because AM = CM) Now, AN sin i AC or AN AC sin i or AN 2t tan r sin i 2t sin r sin r sin i cos r sin r 2 mt sin r sin r cos r or sin i m sin r sin 2 r AN 2mt cosr So that, t sin 2 r t m 2mt cosr cosr cosr 2mt sin 2 r 2mt cosr cosr 2mt 1 sin 2 r cos r 2 mt cos 2 r cos r 2mt cosr So that the actual path difference: 2 mt cos r 2 Production of colors in thin films: When a thin film of oil on water, or a soap bubble, exposed to white light (such as sunlight) is observed under the reflected light. The brilliant colors are seen due to following reasons: 2mt cosr The path difference depend µ or the wavelength. It means the path difference will be different for different colors, so that with the white light the film shows various colors from violet to red. The path difference also varies with the thickness of film so that it passes through various colors for the same angle of incidence when seen in white light. The path difference changes with the angle r and angle r with change with angle i. So that the films assumes various colors when viewed from different directions with white light. Non uniform thickness (Wedge shaped film): X X’ E M’ i i D r i µ t B r O θ N r+θ M C I I’ The path difference (Δ) = path [BC+CD] in film – path BE in air Δ = µ (BC+CD)-BE or Δ = µ(BC+CI)-BE because CD=CI = µBI-BE = µ (BN+NI)-BE In right angle triangle BED, sin i In right angle triangle BND, sin r BE BD BN BD …………………(1) …………………(2) …………………(3) From equations (2) and (3) we have sin i BE sin r BN or m BE BN or Putting the value of BE in equation (1), we have m BN NI mBN or mNI In right angle triangle DNI, NI cos( r ) DI We know DI DH HI t t 2t , so that NI 2t cos(r ) Therefore, mNI 2mt cos(r ) BE mBN Applying Stoke’s law, Therefore, the net path difference is 2mt cos( r ) 2 is very small, so can be neglected then the path difference is 2 mt cos r 2 Now, the condition for brightness is 2mt cos r 2 n n = 0, 1, 2, 3, …………… or 2mt cos r 2n 1 t 2n 1 4m cos r 2 Putting n = 0, 1, 2, 3,…………… t 4 m cos r , 3 5 , , …………………..etc. 4 m cos r 4 m cos r The film will appear dark if 2mt cos r 2 2n 1 2 or 2mt cosr n 1 n = 0, 1, 2, 3, ………… or 2mt cos r n t n = 1, 2, 3, ………… n 2 m cos r It is observed that in the direction of increasing thickness of the film there will be alternate bright and dark fringes parallel to the edge of the film. Fringe width: t θ XN Let us consider XN is the position of Nth bright fringe, where the thickness of the film is t then tan t XN Since θ is very small, so that t XN or Putting the value of t then we have (for bright fringe) t .X N 2n 1 4m cos r .X N For the small angle of incidence cos r 1 because angle r will also be small then XN 2n 1 and 4m X N 1 Then the fringe width X N 1 X N 2m 2m 2n 3 4m Similarly, we can calculate the fringe width for dark fringes. We have already calculated that t .X N Now we calculate the fringe width for dark fringe put the value of t form equation (2), we have, n .X N 2m cos r Similarly for the small angle of incidence cos r 1 n .X N 2m and XN or X N 1 n 2m n 1 2m Then the fringe width is X N 1 X N or 2m 2m Newton’s Rings: Source Actually the path difference between the interfering rays is 2mt cos r 2 Therefore, the effective path difference 2mt cos r 2 For normal incidence cosr =1, then the path difference For maxima 2 mt or 2 2 mt 2n 1 For minima 2 mt or or 2mt 2n 1 2 2 n 2 2 n 1 2mt n 2 where n = 0, 1, 2, 3, 4, ------- 2 2n 1 2mt where n = 0, 1, 2, 3, 4, ----- where n = 1, 2, 3, 4, ------- How to calculate the radius or diameter of the nth fringe: Let n be the radius of nth bright ring and t is the thickness of air film. Let R be the radius of plano-convex lens. By the geometry of the circle D R 2 n2 R t 2 R R t 2Rt 2 or But 2 n 2 R 2 O n2 2 Rt t 2 R R-t t 2 R A B so that the higher power terms are neglected. Therefore, we have n2 2 Rt or t n2 2R ρn C For constructive interference: We have, 2 mt 2n 1 or 2m 2R 2n 1 n2 2n 1 or or n2 2 2 R 2m R2n 1 2m n R2n 1 2m Diameter Dn 2 n 2 Dn2 4 R2n 1 2 R2n 1 2m m Now, for the air film the refractive index m 1 Therefore, Dn2 2R2n 1 n p th Now the diameter of bright ring, n p 2R2n p 1 D2 Therefore, D2 n p Dn2 2R2n p 1 2R2n 1 4nR 4 pR 2R 4nR 2R or D2 n p Dn2 4 pR D2 n p Dn2 4 pR Another important application of Newton’s ring: With the help of Newton’s ring experiment we can determine the refractive index of a liquid. For this purpose, first of all the experiment is performed with air film and diameter of nth and (n+p)th rings. Liquid We have already calculated that for air, D2n p Dn2 4 pR air -----(1) Now the liquid is introduced between plano-convex lens and glass plate and the same experiment is repeated. Let m be the refractive index of the liquid used, then 4 pR D 2 Dn2 -----(2) n p liquid m From equation (1) and (2), we have D 2 Dn2 n p liquid D 2 Dn2 n p air m D 2 Dn2 n p liquid D 2 Dn2 n p air m Michelson Interferometer: M1 M 2' M2 Monochromatic Source (S) B Telescope (T) C Let the distance between M1 and M2’ is d, which is also known the thickness of the air film and θ is the angle which reflected beam makes with the normal. As the ray after reflection from M2 suffers reflection at the silvered surface of the glass plate B and the additional path difference of 2 is introduced between the two interfering rays.So that the path difference is 2 mt cos r 2d cos 2 2 Now condition for bright and dark rings For dark rings For bright rings Here or t = d, r is θ and µ = 1 (for air) so that Forms of the fringes: (i) Circular fringes: When M1 and M 2' are exactly parallel to each other. The fringes produce concentric circles. (ii) ' Curved and straight fringes: When M1 and M 2 are inclined so that air film is wedge shaped. M 2' M1 M 2' M1 M 2' Fig: Various types of fringes. M1 APPLICATIONS OF MICHELSON INERFEROMETER: (i) Determination of wavelength of monochromatic light When the mirror move through a distance , one fringe shift and there 2 is a corresponding change of in the path difference. If N fringes shift across the cross wire, when the mirror moved through a known distance x, then xN 2 2x N By knowing x and N, the wavelength can be determined. (ii) Determination of refractive index or thickness of transparent sheet: When a thin transparent sheet of thickness ‘t’ and refractive index ' m ' is introduced in between glass plate and mirror M2 and adjustment are made so that the light pass through it normally. In this position the path difference will be m 1 t As the beam travels the plate twice, an extra path difference m 1 t is added. So that the total path difference = 2m 1 t Therefore, the fringes must shift when the movable mirror is moved till the fringes are back to their initial position. If the displacement of the movable mirror is x, then 2x 2m 1 t or t x m 1 (iii) Determination of difference in wavelengths: we can write, 2 x n1 2 x n 12 and or or n 2x 1 n 1 2x 2 Subtracting equation (1) from (2), we get n 1 n 2x 2 2x 1 or 1 2 1 2 2x Since 1 and 2 are almost equal, thus, we have 12 2avg . So that 1 2 2avg . 2x ------(1) ------(2)