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Subject: PHYSICS-I Subject Code: PHY-101F

Subject: PHYSICS-I Subject Code: PHY-101F L T P Internal (Sessional + Class performance ) 50 Marks 3 1 0 Exam 100 Marks Total 150 Marks Duration of Exam 3 Hrs. SYLLABUS (I st Semester) Interference Diffraction

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Subject: PHYSICS-I Subject Code: PHY-101F

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  1. Subject: PHYSICS-I Subject Code: PHY-101F L T P Internal (Sessional+ Class performance) 50 Marks 3 1 0 Exam 100 Marks Total 150 Marks Duration of Exam 3 Hrs.

  2. SYLLABUS • (Ist Semester) • Interference • Diffraction • Polarization • Laser • Fiber Optics • Dielectrics • Special Theory of Relativity • Superconductivity

  3. 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.

  4. 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

  5. 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.

  6. 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.

  7. Text Books: Fundamentals of Engineering Physics M. S. Khurana, MR Pub, Delhi Modern Physics for Engineers S P Taneja; R Chand Publication Engineering Physics Satya Prakash, Pragati Prakashan Engineering Physics R. K. Gaur & S. L. Gupta

  8. 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.

  9. Course Objective: • The objective of this course is to apply the concepts of physics • to different optical phenomena • devices based on these phenomena

  10. INTERFERENCE Interference? Why we need to study? There are two possibilities: 1. Maximum intensity position 2. Minimum intensity position

  11. Young’s Experiment: crest A B Trough S (Coherent Source) (Screen) (Slit)

  12. Coherent sources ?

  13. 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.

  14. D Theory of Interference P A d B O S (Coherent Source) Slit Screen

  15. The path difference is an even multiple of , the waves arrive at a point in the same phase and produce maximum intensity. • If the path difference is an odd multiple of , the two waves arrive at a point in the opposite phase, they cancel out and the resultant intensity should be minimum.

  16. x How to calculate the fringe width: To determine the spacing between the bands/ fringes and the intensity at point P. P S1 d S2 M O S (Coherent Source) N D Slit Screen

  17. The path difference can be calculated from fig. Path difference (Δ) = S2P-S1P To calculate S2P, consider the ∆S2NP or Expending by binomial theorem and neglecting higher power term of D as D>>

  18. We have D is very very large in comparison of Therefore, higher power term of D can be neglected. Then we get or

  19. Similarly, we can calculate S1P, consider the ∆S1MP Then the path difference is (Δ) = S2P-S1P putting the value of S2P and S1P thenwe have Path difference For the nth fringe the path difference =

  20. Bright Fringes: • We have already pointed out that for bright fringe, the path difference should be equal to where n = 0, 1, 2, 3, 4, ---------------- . Therefore, the distance between two consecutive fringes is also known as fringe width.

  21. (b) Dark Fringes: Similarly, for dark fringes, the path difference should be equal to Point P to be dark; where n = 0, 1, 2, 3, 4, --------------

  22. 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 • It is directly proportional to the wavelength of light • It is inversely proportional to the spacing between two coherent sources .

  23. Type of interference: • Interference by division of wave front • Interference by division of amplitude

  24. Overlap region a b D Fresnel’s Biprism: Fresnel’s biprism is a device to produce two coherent sources by division of wave front. P S1 G S O S2 H Q

  25. 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’. 179°

  26. 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 P S1 d S2 x Po O D

  27. 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 or We have already calculated that or (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

  28. (2) >1 so that Since is positive. The fringe width is Using equation (1) It means the fringe width is not affected by introduce of mica sheet. Put these values in equation (2) we get, or Thus we can calculate the thickness of mica sheet.

  29. m n m n (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: According to the linear magnification produced by the lens: 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:

  30. From equation (1) and (2), we get or

  31. Interference due to Reflection: (Reflected rays) R1 Source R2 N i i M A C t r r D B T1 T2 (Transmitted rays)

  32. (1) BM = t and also Now, for AN

  33. or AC = AM + CM (because AM = CM) Now, or or or

  34. So that, So that the actual path difference:

  35. 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: • 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.

  36. X X’ E M’ i D r i µ i t B r r+θ N θ O M C I I’ Non uniform thickness (Wedge shaped film):

  37. 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, …………………(1) …………………(2) In right angle triangle BND, …………………(3)

  38. From equations (2) and (3) we have or or Putting the value of BE in equation (1), we have or In right angle triangle DNI, We know , so that Therefore,

  39. Applying Stoke’s law, Therefore, the net path difference is is very small, so can be neglected then the path difference is Now, the condition for brightness is n = 0, 1, 2, 3, …………… or

  40. Putting n = 0, 1, 2, 3,…………… , , , …………………..etc. The film will appear dark if or n = 0, 1, 2, 3, ………… or n = 1, 2, 3, ………… 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.

  41. XN Fringe width: t θ Let us consider XNis the position of Nth bright fringe, where the thickness of the film is t then Since θ is very small, so that or Putting the value of t then we have (for bright fringe)

  42. For the small angle of incidence because angle r will also be small then and Then the fringe width

  43. Similarly, we can calculate the fringe width for dark fringes. We have already calculated that Now we calculate the fringe width for dark fringe put the value of t form equation (2), we have, Similarly for the small angle of incidence or and Then the fringe width is or

  44. Newton’s Rings: Source Actually the path difference between the interfering rays is Therefore, the effective path difference

  45. For normal incidence cosr =1, then the path difference For maxima or where n = 0, 1, 2, 3, 4, ------- For minima or where n = 0, 1, 2, 3, 4, ----- or where n = 1, 2, 3, 4, -------

  46. ρn How to calculate the radius or diameter of the nth fringe: 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 Let D R O R or R-t C A But B so that the higher power terms are neglected. Therefore, we have or

  47. For constructive interference: We have, or or or Diameter

  48. Now the diameter of bright ring, Now, for the air film the refractive index Therefore, Therefore, or

  49. 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, -----(1) Now the liquid is introduced between plano-convex lens and glass plate and the same experiment is repeated. Let be the refractive index of the liquid used, then -----(2) From equation (1) and (2), we have

  50. Michelson Interferometer: M1 M2 Monochromatic Source (S) B C Telescope (T)

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