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- 1. Welcome to Wave optics
- 2. Does Light always exhibit Rectilinear Propagation?
- 3. Explanation for bending of light Wave theory of light Christiaan Huygens Wavefront Huygensβ Principle Huygens Wave theory of light
- 4. ο All points on a particular circle will oscillate with the same phase. ο Wavefront is the locus of all points at which the wave disturbance is in the same phase. Wavefront ο In 3π·, for a point source, the wavefronts are spherically symmetric. ο Point source creates wave traveling in all directions according to equation. π¦ = π΄sin(ππ‘ Β± ππ₯) ο Points π΄, π΅, πΆ have same phase ππ‘ Β± ππ₯1.
- 5. The direction of propagation of light wave is perpendicular to the wavefront at any given point. Direction of Propagation of Light Spherical Wavefront
- 6. ο Time taken by each ray to propagate from one wavefront to the next wavefront in any medium remains same. ο Hence, distance covered by the light in each time interval also changes. βΉ Speed of light changes. ο Refractive index, π = π π£ When medium changes, π1 π2 = π£2 π£1 Properties of Wavefront π ο Distance between two consecutive wavefronts = π.
- 7. ο Spherical wavefront is observed for a point source. β’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Spherical Wavefront
- 8. β’ Planar wavefronts are observed when the source is at infinity. β’ Direction of propagation of light: Perpendicular to the plane. Planar Wavefront
- 9. β’ Cylindrical wavefront is observed for a line source. β’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Cylindrical Wavefront
- 10. S If the distance between the wavefronts in the medium with refractive index π2 is π2, then what will be the distance π1 between the wavefronts in the medium with refractive index π1? π = π π£ We known that, ΰ΅ π2 π1 = ΰ΅ π£1 π£2 π = ΰ΅ π£ π β β ΰ΅ π1 π2 = ΰ΅ π£1 π£2 = ΰ΅ π2 π1 Here, π1 = π1, and π2 = π2 Hence, π1 π2 = π2 π1 β π1 = π2 π2 π1
- 11. ο Every point on the wavefront acts as a point source called secondary wave source and generates secondary wavelets. ο The common tangent to the secondary wavelets in the forward direction gives the secondary wavefront. ο Intensity is maximum in forward direction and zero in backward direction. Secondary Wavelets Secondary Wavefront Primary Wavefront Primary Source Huygensβ Principle
- 12. Shape of Wavefront Object at infinity for a convex lens Object at infinity for a concave lens
- 13. Object at infinity for a concave mirror Shape of Wavefront Object at infinity for a prism
- 14. Reflection of a plane wave by a plane surface Huygensβ Principle - Law of Reflection Reflecting plane surface π΄ π΅ π΅πΆ = π£π‘ πΆ π· π·π΄ = π£π‘ π‘ = 0 π π‘ Incident Wavefront π΄π΅ at π‘ = 0 Reflected Wavefront πΆπ· at time = π‘ Secondary Wavelet From π΄
- 15. Reflection of a plane wave by a plane surface Huygensβ Principle - Law of Reflection β π·πΆπ΄ = π β Angle of reflection β π΅π΄πΆ = π β Angle of incidence ο Angle of incidence is the angle between the incident wavefront and the reflecting surface. ο Angle of reflection is the angle between the reflected wavefront and the reflecting surface. π΄ πΆ π· π π΅ πΆ π΄ π β΄ βπ΄π·πΆ β βπΆπ΅π΄ βΉ β π = β π From βπ΄π·πΆ and βπΆπ΅π΄: β π΄π·πΆ = β π΄π΅πΆ = 90Β° π΄π· = π΅πΆ = π£π‘ π΄πΆ = π΄πΆ [π . π». π Congruency]
- 16. From βπ΄π·πΆ: From βπ΄π΅πΆ: sin π = π΅πΆ π΄πΆ = π£1 π‘ π΄πΆ sin π = π΄π· π΄πΆ = π£2 π‘ π΄πΆ sin π sin π = π£1 π£2 = π2 π1 = Constant Huygensβ Principle - Law of Refraction Medium π2 π΄ π΅ π£1π‘ πΆ π· π π π‘ = 0 π π Incident Wavefront π΄π΅ at π‘ = 0 Refracted Wavefront πΆπ· at π‘ = π‘ Secondary Wavelet From A π£2π‘ π Medium π1 π1 < π2
- 17. B A D C Solution: The direction of propagation β ΖΈ π βΉ The direction of wavefront β β₯ π‘π ΖΈ π In the given options, the plane β₯ to ΖΈ π is represented by π₯ = π. π₯ = π π¦ = π π§ = π π₯ + π¦ + π§ = π Light waves travel in vacuum, along the π₯ β axis. Which of the following may represent the wavefronts?
- 18. Hole Screen Hole Screen Light behaves like a ray (i.e., exhibits rectilinear propagation) for: Obstacle dimensions β«ππππβπ‘ Obstacle dimensions β ππππβπ‘ Light behaves like a wave for: Does Light always exhibit Rectilinear Propagation?
- 19. OPTICS Study of the behavior and properties of light GEOMETRICAL OPTICS Explains refraction and reflection, but not interference, diffraction and polarization WAVE OPTICS Explains reflection and refraction, as well as interference, diffraction and polarization
- 20. Interference is the phenomenon in which two waves superpose to form the resultant wave of lower, higher or same amplitude. Interference
- 21. βWhen two or more waves cross at a point, the displacement at that point is equal to the vector sum of the displacements of individual wavesβ Τ¦ π¦πππ‘ = Τ¦ π¦1 + Τ¦ π¦2 + Τ¦ π¦3 β¦ β¦ + Τ¦ π¦π π¦ π₯ π π¦1 π¦2 π¦πππ‘ Superposition Principle
- 22. Case 1: When two crest meet Case 2: When crest and trough meet Resultant of Waves
- 23. Phase Difference and Path Difference β Path difference (βπ₯): Difference in the path traversed by the two waves. β Phase difference (πΏ): Difference in the phase angle of the two waves. We know that, For a path difference π, phase difference = 2π So, for path difference βπ₯, phase difference = 2π π βπ₯. πΏ = 2π π βπ₯ Source π1 Source π2 π¦1 = π΄1 sin ππ‘ + ππ₯ π¦2 = π΄2 sin ππ‘ + ππ₯ + πΏ π¦πππ‘ = π¦1 + π¦2 π
- 24. Combination of Waves π¦πππ‘ = π΄1 sin ππ‘ + ππ₯ + π΄2 sin ππ‘ + ππ₯ + πΏ π΄πππ‘ = π΄1 2 + π΄2 2 + 2π΄1π΄2 cos πΏ tan πΌ = π΄2 sin πΏ π΄1 + π΄2 cos πΏ π¦πππ‘ = π΄πππ‘ sin ππ‘ + ππ₯ + πΌ If π΄1 = π΄2 = π΄ π΄πππ‘ = 2π΄ cos πΏ 2
- 25. Constructive Interference Interference that produces maximum possible amplitude (or maximum intensity) is called constructive interference. π 2π 3π β Path Difference = 3π β 2π = π π΄πππ‘ = π΄1 2 + π΄2 2 + 2π΄1π΄2 cos πΏ For maximum amplitude: cos πΏ = 1 β πΏ = 2ππ π΄ = π΄πππ₯ = π΄1 + π΄2 2ππ = 2π π Γ (βπ₯) Path difference = βπ₯ = ππ πΏ = 2π π βπ₯ Constructive Interference
- 26. β‘ π₯ π¦ π¦πππ‘ = 2π΄ sin(ππ‘ + ππ₯) + Constructive interference occurs when the crest and trough of one wave overlaps with the crest and trough of another wave. Constructive Interference π₯ π¦ π¦1 = π΄ sin(ππ‘ + ππ₯) π π₯ π¦ π¦2 = π΄ sin(ππ‘ + ππ₯) π
- 27. Destructive Interference Interference that produces minimum possible amplitude (or minimum intensity) is called destructive interference. π΄πππ‘ = π΄1 2 + π΄2 2 + 2π΄1π΄2 cos πΏ π 7.25π 9.75π For minimum amplitude: cos πΏ = β1 β πΏ = (2π + 1)π π΄ = π΄πππ = π΄1 β π΄2 (2π + 1)π = 2π π (βπ₯) Path difference = πΏ = 2π π βπ₯ βπ₯ = 2π + 1 π 2 β Path Difference = 9.75π β 7.25π = 2.5π Destructive Interference
- 28. π₯ π¦ π¦1 = π΄ sin(ππ‘ + ππ₯) Destructive interference is resulted when the crest of one wave overlaps with the trough of another wave. β‘ π₯ π¦ π π¦ = 0 + π₯ π¦ π¦2 = π΄ sin(ππ‘ + ππ₯ + π) Destructive Interference π π
- 29. T Two-point light sources π1 and π2 are separated by a distance of 4.2π. If an observer standing at the centre πΆ of the two sources starts moving towards π2, then find the minimum distance travelled by the observer to meet the first maxima. π1 π2 2.1π πΆ π βπ₯ = 0 π₯ 2.1π βπ₯ = π Solution: For maxima at π: βπ₯ = π β π1π β π2π = π β (2.1π + π₯) β (2.1π β π₯) = π β 2π₯ = π π₯ = 0.5π
- 30. Coherent and Incoherent Sources Coherent Waves Incoherent Waves Coherent Sources β Same wavelength β Same frequency β Constant phase difference Incoherent Sources β Different wavelength β Different frequency β Varying phase difference Note: We will always consider coherent sources in our discussion.
- 31. Youngβs Double Slit Experiment Particle nature of light Wave nature of light
- 32. According to Huygensβ principle, the sources π1 and π2 will behave as independent sources. Youngβs Double Slit Experiment π· β Light source must be monochromatic. β Sources π1 and π2 must be coherent. β Width of the slit is comparable to the wavelength of light. β Waves coming from sources π1 and π2 will interference and obtain different interference pattern on the screen.
- 33. Youngβs Double Slit Experiment β’ ππ1 = ππ2 π· (β΄ No path difference till slits.) π β’ β²πβ² is distance between slits. β’ ππ΅ is the central line. π π΅ β’ β²π·β² is distance between screen and slits plane. π π π¦ β’ Consider point π at a distance π¦ from central line ππ΅. β’ β πππ΅ = π. β’ π2π > π1π β π2π β π1π = βπ₯ β’ βπ₯ is the path difference.
- 34. Youngβs Double Slit Experiment π· π π π΅ π π π¦ π/2 π/2 β Path difference at any general point π, βπ₯ = π2π β π1π βπ₯ = π¦ + π 2 2 + π·2 β π¦ β π 2 2 + π·2 Approximation 1: π· β« π Approximation 2: π is verysmall βπ₯ β π¦π π·
- 35. Youngβs Double Slit Experiment βπ₯ = π2π΄ = π sin π π is verysmall β sin π β tan π βπ₯ β π tan π = ππ¦ π· βπ₯ β π¦π π·
- 36. β For Maxima (Constructive interference): βπ₯ = ππ π¦ = πππ· π Where, π = 0, Β±1, Β±2, Β±3, β¦ β ππ¦ π· = ππ π = 0 correspondstothecentralmaxima. π = Β±1 correspondtothe1π π‘ maxima. π = Β±2 correspondtothe2ππ maxima. π π΅ π π π¦ π1 π2 π· βπ₯ Condition for Constructive Interference β π¦ = 0 β π¦ = Β± ππ· π β π¦ = Β± 2ππ· π
- 37. β For Minima (Destructive interference): βπ₯ = π + 1 2 π β ππ¦ π· = π + 1 2 π π¦ = π + 1 2 ππ· π π = 0,β1 correspondtothe1π π‘ minima. π = 1,β2 correspondtothe2ππ minima. π π΅ π π π¦ π1 π2 π· βπ₯ Condition for Destructive Interference Where, π = 0, Β±1, Β±2, Β±3, β¦ β π¦ = Β± ππ· 2π β π¦ = Β± 3ππ· 2π
- 38. In YDSE, white light is passed through the double slit and interference pattern is observed on a screen 2.5 π away. The separation between the slits is 0.5 ππ. The first violet and red maxima are formed at distances of 2 ππ and 3.5 ππ away from the central white maxima, respectively. The wavelengths of red and violet light, respectively, are: Given: To find: Solution: π· = 2.5 π, π = 0.5 ππ, π¦π£πππππ‘ = 2 ππ, π¦πππ = 3.5 ππ ππ£πππππ‘, ππππ Distance of first maxima from the central maxima is given by: For violet light, π¦ = 2 ππ β 2 Γ 10β3 = 2.5 Γ ππ£πππππ‘ 0.5 Γ 10β3 β ππ£πππππ‘ = 2 5 Γ 10β6 π π¦ = π·π π For red light, π¦ = 3.5 ππ β 3.5 Γ 10β3 = 2.5 Γ ππππ 0.5 Γ 10β3 β ππππ= 0.7 Γ 10β6 π ππππ = 700 ππ ππ£πππππ‘ = 400 ππ
- 39. Resultant Amplitude of the Wave πΈ1 = πΈ01 sin ππ₯ β ππ‘ πΈ2 = πΈ02 sin ππ₯ β ππ‘ + πΏ πΈ01, πΈ02 ππ π΄01, π΄02 β Electric Field Amplitude. Electric field of the wave from π1: Electric field of the wave from π2: πΈ0 2 = πΈ01 2 + πΈ02 2 + 2πΈ01πΈ02 cos πΏ πΈπππ‘ = πΈ1 + πΈ2 = πΈ0 sin ππ₯ β ππ‘ + πΏ Net electric field after the interference: tan π = πΈ02 sin πΏ πΈ01 + πΈ02 cos πΏ
- 40. Intensity of Waves πΌ = 2π2 π2 ππ£πΈ0 2 πΌ β πΈ0 2 πΌπππ‘ = πΌ1 + πΌ2 + 2 πΌ1 πΌ2 cos πΏ πΈ0 2 = πΈ01 2 + πΈ02 2 + 2πΈ01πΈ02 cos πΏ π π΅ π π π¦ π1 π2 π· πΈ2 πΈ1
- 41. Intensity for Identical Waves πΌπππ‘ = 4πΌ cos2 πΏ 2 πΌπππ‘ = πΌ1 + πΌ2 + 2 πΌ1 πΌ2 cos πΏ β For identical slits: πΌ1 = πΌ2 = πΌ πΌπππ‘ = πΌ + πΌ + 2 πΌ πΌ cos πΏ πΌπππ‘ = 2πΌ + 2πΌ cos πΏ = 2πΌ(1 + cos πΏ) πΌπππ‘ = 2πΌ Γ 2 cos2 πΏ 2 π π΅ π π π¦ π1 π2 π· πΌ2 πΌ1
- 42. β For constructive interference: (πππ₯πππ) Intensity of Waves β For destructive interference: (ππππππ) πΈ0 πππ₯ = πΈ01 + πΈ02 πΈ0 πππ = πΈ01 β πΈ02 πΌ β πΈ0 2 β πΌπππ₯ πΌπππ = πΈ0 πππ₯ πΈ0 πππ 2 πΌπππ₯ πΌπππ = πΈ01 + πΈ02 πΈ01 β πΈ02 2 β When cos πΏ = 1, πΏ = 2ππ (πππ₯πππ) β When cos πΏ = β1, πΏ = 2π + 1 π (ππππππ) πΌπππ‘ = πΌ1 + πΌ2 + 2 πΌ1 πΌ2 β πΌ1 + πΌ2 2 πΌπππ‘ = πΌ1 + πΌ2 β 2 πΌ1 πΌ2 β πΌ1 β πΌ2 2 πΌ = πΌ1 + πΌ2 2 , constructive interference πΌ1 β πΌ2 2 , destructive interference
- 43. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: πΌ1 πΌ2 = 25 9 Solution: Given: To find: πΌπππ₯ πΌπππ = 16 πΌ1 πΌ2 πΌπππ₯ πΌπππ = 16 β πΌ1 + πΌ2 πΌ1 β πΌ2 2 = 16 1 β πΌ1 + πΌ2 = 4 πΌ1 β 4 πΌ2 β 3 πΌ1 = 5 πΌ2 β πΌ1 πΌ2 = 5 3 Ratio of intensities: πΌ1 πΌ2 = 5 3 2
- 44. Intensity Variation π¦ = πππ· π π¦ = π + 1 2 ππ· π β Constructive Interference: β Destructive Interference: π¦ 1π π‘ order maxima 2nd order maxima Central maxima 1π π‘ order minima 2π π‘ order minima π¦ = Β± ππ· π , Β± 2ππ· π , Β± 3ππ· π . . . . . . . π¦ = Β± ππ· 2π , Β± 3ππ· 2π , Β± 5ππ· 2π . . . . . . . π· π π1 π2 π πΌ0 πΌ0
- 45. Intensity Variation πΌ = 4πΌ0 cos2 πΏ 2 β Intensity at any point: β πΌπππ₯ = 4πΌ0, πΌπππ = 0 πΏ = Phase difference between the two waves from π1 and π2. πΌ0 πΌ0 4πΌ0 πΌ = 0 πΌ = 0 πΌ = 0 πΌ = 0 π· π π1 π2 π 4πΌ0 4πΌ0 4πΌ0 4πΌ0
- 46. Shape of Fringes on Screen If we replace slits by pin holes in YDSE, then we will see a Hyperbolic Fringe pattern. Pointsourcesplacedon theperpendicularaxis tothescreencreateconcentriccircularfringes.
- 47. Fringe Width π· π π1 π2 π ππ« π Fringe width is the distance between two consecutive maxima/minima. π½ = 3ππ· 2π β ππ· 2π π½ = ππ· π
- 48. Fringe Width when the setup is inside a Medium If experimental setup is dipped in liquid π = πππ£πππ’π’π πππππππ’π β ππππππ’π = ππ£πππ’π’π π π½ = ππ· π Fringe Width Refractive index π is speed of light in vacuum π£ is speed of light in liquid π½ = ππππππ’ππ· π β π½ = ππ£πππ’π’ππ· ππ Fringe Width inside liquid, π = π π£
- 49. Angular Fringe Width π π΅ π π π π¦ = π½ π1 π2 π· Central Maxima First Maxima tan π β π = π½ π· We know π½ = ππ· π After putting π½ value in π = π½ π· π = π π
- 50. Position of Maxima/Minima π¦ = πππ· ππ π¦ = π + 1 2 ππ· ππ β Constructive Interference: β Destructive Interference: π¦ = Β± ππ· ππ , Β± 2ππ· ππ , Β± 3ππ· ππ . . . . . . . π¦ = Β± ππ· 2ππ , Β± 3ππ· 2ππ , Β± 5ππ· 2ππ . . . . . . . 2ππ· ππ 3ππ· 2ππ
- 51. Incoherent Light Sources Incoherent Light Sources β However, the human eye cannot the capture the rapidly changing bright and dark fringes. β In ultra slow motion, the fringes on the screen flicker. β So, the eyes see a continuous band of light. β Net Intensity on the screen is the sum of intensity from two sources.
- 52. In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance π· from the slits. The slits are separated by a distance π and are illuminated by monochromatic light of wavelength π. Find the distance from the central point π΅ where the intensity falls to half the maximum. πΌ = 4πΌ0 cos2 πΏ 2 β cos πΏ 2 = 1 2 β πΏ = π 2 Phase difference πΏ = 2π π βπ₯ = π¦π π· π 4 = π¦π π· β β βπ₯ = π 4 π¦ = ππ· 4π ππ· 2π π· π π1 π2 π πΌπππ₯ πΌπππ₯ 2 ππ· 4π π΅ Path difference βπ₯ 4πΌ0 cos2 πΏ 2 = 1 2 4πΌ0 Intensity is half the maximum, Solution: 2π π βπ₯ = π 2
- 53. Two coherent point sources π1 and π2 vibrating in phase emit light of wavelength π. The separation between the sources is 2π. Consider a line passing through π2 and perpendicular to the line π1π2. What is the smallest distance from π2 where the intensity is minimum? T Solution: Given: To find: ππ1π2 = 2π π₯ 2π 2 + π₯2 β π₯ = π + 1 2 π When π₯ > 0, 16π β 2π + 1 2 π > 0 So, 2π + 1 < 4 π < 3 2 β΄ π = 1 π₯ = 16π β 2π + 1 2 π 4 2π + 1 β 16π β 9π 12 π₯ = 7π 12 π₯ = 16π β 2π + 1 2 π 4 2π + 1
- 54. Solution: T Optical path difference between π΅π0 and π΄π0 : βπ₯ = π sin π β π tan π βπ₯ = π π 2π· = π2 2π· βπ₯ = π 3 = π2 2π· π = 2ππ· 3 Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let π΅π0 β π΄π0 = π/3 and π· β« π. Show that in this case π = 2ππ·/3.
- 55. Path Difference between the Two Waves
- 56. Optical Path length πΈ = πΈ0 sin ππ₯0 β ππ‘ Original wave equation: πΈ = πΈ0 sin π π₯0 + πΏ β ππ‘ Equation of a wave when it is ahead by a length πΏ: πΈ = πΈ0 sin ππ₯0 β ππ‘ + ππΏ Phase difference βπ π = 2π π So, βπ = 2π π πΏ β΄ Two points on a wave separated by a path length of πΏ will have a phase difference of 2π π πΏ. πΈ π₯ πΏ π₯0 π₯0 + πΏ
- 57. Optical Path length 2π ππ πΏπππ = 2π ππ ππΏπππ Phase Difference between points π΄ and π΅ on the wave, travelling in air: βππππ = 2π ππππ πΏπππ Phase Difference between points π΄ and π΅0 on the wave, travelling in medium: ππππ = ππ π = 2π ππ π πΏπππ βππππ = 2π ππ ππΏπππ βππππ = βππππ If, βππππ = 2π ππ πΏπππ πΏπππ = ππΏπππ β πΏπππ π πππ πΏπππ π΄ π΅ π΄ π΅0 Optical Path Length in a medium is the corresponding path that light travels in vacuum to undergo the same phase difference.
- 58. πππ· = πππΏπΌπΌ β πππΏπΌ = ππππ β 1 πΏ Optical Path Difference Optical Path Length in air: πππΏπΌ = π΄π΅ + πππππΏ + πΆπ· = π΄π΅ + πΏ + πΆπ· Optical Path Length in medium: πππΏπΌπΌ = π΄π΅ + πππππΏ + πΆπ· Optical Path Difference: Phase Difference: πΏ = βππΌπΌ β βππΌ = 2π ππ ππππ β 1 πΏ πΏ = 2π ππ πππ· πππ· = πππΏπΌπΌ β πππΏπΌ = π2 β π1 πΏ and Phase Difference: πΏ = βππΌπΌ β βππΌ βππΌπΌ = 2π ππ π2πΏ πΏ = 2π ππ π2 β π1 πΏ Phase Difference in medium 1 and 2: Optical Path Difference: βππΌ = 2π ππ π1πΏ
- 59. Thin Transparent Film in YDSE Optical Path Difference: πππ· = πΌ β πΌπΌ = ππ‘ + π1π΅ β π‘ β π2π΅ Optical Path Travelled by wave πΌ: ππ‘ + π1π΅ β π‘ π2π΅ Optical Path Travelled by wave πΌπΌ: π1π΅ = π2π΅ πππ· = ππ‘ β π‘ = π β 1 π‘ β’ Introduction of slab causes change in πππ· by π β 1 π‘. β’ Path length of the ray π1π΅ increases when it encounters the thin film. β’ Now the central maxima does not lie at π΅.
- 60. Thin Transparent Film in YDSE Optical Path Difference: βπ₯ = π2π β π1π βπ₯ = π¦π π· βπ₯ = πΌπΌ β πΌ Total Optical Path Difference at point P: βπ₯ = πΌπΌ β πΌ βπ₯ = π¦π π· β π‘ π β 1 β’ After inserting thin film, light ray π1π has travelled an extra path of π β 1 π‘. βπ₯ = π2π β π1π + π‘ π β 1 β’ After inserting thin film, light ray π2π has travelled an extra path of π β 1 π‘. Total Optical Path Difference at point P: βπ₯ = πΌπΌ β πΌ βπ₯ = π¦π π· + π‘ π β 1 βπ₯ = π2π + π‘ π β 1 β π1π
- 61. Shift in Central Maxima π π΅ π π π π¦ π₯ π₯ + Ξπ₯ π· π‘, π π β 1 π‘π· π At central maxima, π1 π2 πΌ πΌπΌ Screen πππ· = 0 π¦π π· β π β 1 π‘ = 0 π¦ = π β 1 π‘π· π Shift in central maxima =
- 62. Number of Fringes Shifted π β 1 π‘π· π Shift in central maxima: Number of Fringes shifted π = π¦ π½ = π β 1 π‘π· π Γ π ππ· π = π β 1 π‘ π = π βπππ‘ ππ ππππ‘πππ πππ₯πππ ππππππ π€πππ‘β π = π β 1 π‘π· π Shift in central maxima: Number of fringes shifted: π = π β 1 π‘ π
- 63. T The Youngβs double slit experiment is done in a medium of refractive index 4/3. A light of 600 ππ wavelength is falling on the slits having 0.45 ππ separation. The lower slit π2 is covered by a thin glass sheet of thickness 10.2 ππ and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 π from the slits. Find the location of the central maximum on the π¦ βaxis. 10.2ππ 7.5 ππ 1.25 ππ 4.25 ππ B A D C π0 π π = 4/3 π1 π2 Screen πΌ πΌπΌ
- 64. Given: Solution: To find: ππππ = 4 3 , π = 0.45 ππ, π· = 1.5 π, π = 600 ππ, π‘ = 10.2 ππ, ππ‘ = 1.5 Position of central maxima πππ· = πΌ β πΌπΌ πππ· = π1π0 ππππ β πππ‘ + π2π0 ππππ β πππππ‘ πππ· = π1π0 β π2π0 ππππ β π‘ ππ β ππππ πππ· = π¦π π· ππππ β π‘ ππ β ππππ At central maxima πππ· = 0 π¦π π· ππππ β π‘ ππ β ππππ = 0 π¦ = π‘π· πππππ ππ β ππππ Substituting all the values we get, π¦ = 4.25 ππ π π = 4/3 π1 π2 Screen π0 πΌ πΌπΌ π¦
- 65. Given: Solution: To find: π = 5000 αΆ π΄, π = 1.5, π‘ π¦ = π β 1 π‘π· π π¦5π΅ = 5π·π π π¦ππ΅ = ππ·π π Shift in central maxima: π β 1 π‘π· π = 5π·π π π‘ = 5π π β 1 = 5 Γ 5000 Γ 10β10 1.5 β 1 π‘ = 5 ππ T In YDSE, find the thickness of a glass slab (π = 1.5) which should be kept in front of upper slit π1 so that the central maxima is formed at a place where 5π‘β bright fringe was lying earlier (before inserting slab). π = 5000 αΆ π΄
- 66. Given: Solution: To find: Wavelength = π, π‘1 = π‘2 = π‘ =? Path difference due to both the slabs at π0: For π‘ to be minimum, put π = 0, β π1 β π2 π‘ = π + 1 2 π Ξπ₯ = π + 1 2 π For minima at π0, we know, π1 β 1 π‘ β π2 β 1 π‘ = π1 β π2 π‘ Refractive index = π1 & π2 Screen π· π0 π π π1 π2 π‘ π1 π2 T π‘ = π 2 π1 β π2 Two transparent slabs, having equal thickness but different refractive indices π1 and π2 π1 > π2 , are pasted side by side to form a composite slab. This slab is placed just after the double slits in a Youngβs experiment so that the light from one slit goes through one material and light through other slit goes through other material. What should be the minimum thickness of the slab so that there is a minimum at point π0 which is equidistant from the slits?
- 67. Phase change Yes No Reflected from Denser Medium Rarer Medium π π Air Air π No phase change No phase change π Phase change in Reflection & Refraction
- 68. A narrow slit π transmitting light of wavelength π is placed at a distance π above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance π· from the slit. (π) What will be the intensity at a point just above π. Given: To find: Solution: Wavelength = π, Distance of screen = π·, ππ β πππ β path difference = 0 (when π is just above π) Screen π π· π π π π Ξ΄ = π (due to reflection of light from a denser medium) Distance of source from the mirror = π Intensity just above π T πΌπππ‘ = 0 If phase difference is odd integral multiple of π, destructive interference will take place.
- 69. A narrow slit π transmitting light of wavelength π is placed at a distance π above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance π· from the slit. Given: To find: Solution: T (π) At what distance from π does the first maximum occur? Screen π π· π1 π π π π2 π π π¦ Wavelength = π, π¦1π π‘ πππ₯πππ Distance of screen = π·, Distance of source from the mirror = π βπ₯ = 2π sin π = 2π tan π = 2ππ¦ π· βπ₯πππ‘ = 2ππ¦ π· + π 2 = ππ = 1 π β 2ππ¦ π· = π 2 π¦ = ππ· 4π
- 70. Given: To find: Solution: π = 700 ππ, π· = 1 π, π½ = 2 Γ ππ· 4π = ππ· 2π π½ = 700 Γ 10β9 Γ 1 2 Γ 10β3 = 3.5 Γ 10β4 π π = 1 ππ Fringe width = π½ T A narrow slit π transmitting light of wavelength π is placed at a distance π above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance π· from the slit. (π) If π = 1 ππ, π· = 1 π and π = 700 ππ, then find the fringe width. π π· π1 π π π π2 π π ππ· 4π ππ· 4π π½ = 0.35 ππ
- 71. A narrow slit π transmitting light of wavelength π is placed at a distance π above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance π· from the slit. (π) If the mirror reflects only 64 % of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen? Given: To find: Solution: πΌπππ₯ = πΌ1 + πΌ2 2 , πΌπππ₯ πΌπππ T πΌπππ₯ πΌπππ = 81 π π· π1 π π π2 π πΌ1 = πΌ0 πΌ2 = 0.64 πΌ0 Wavelength = π, Distance of screen = π·, Distance of source from the mirror = π πΌ2 = 0.64 πΌ0 πΌπππ = πΌ1 β πΌ2 2 πΌπππ₯ πΌπππ = πΌ1 + πΌ2 2 πΌ1 β πΌ2 2 β πΌ0 + 0.64 πΌ0 2 πΌ0 β 0.64 πΌ0 2 = 1 + 0.8 2 1 β 0.8 2 = 1.8 2 0.2 2
- 72. Condition for Thin Film Interference π π π air air π π β’ The film thickness should be comparable to the wavelength of light π β π . β’ The incident light should be white (non - monochromatic). β’ The angle of incidence should be small π β 0.
- 73. Thin Film Interference β’ Inside the film, when a particular colourβs path difference is even integral multiple of the wavelength, it undergoes constructive interference, so these colours appear bright. β’ Interference of light wave being reflected off two surfaces that are at a distance comparable to its wavelength is known as thin film βThin film interferenceβ. β’ When a particular colourβs wavelength is odd integral multiple of the path difference inside the film, it undergoes destructive interference, so these colours donβt appear at all. π π Air Air π No phase change No phase change π
- 74. βπ₯ = 2ππ Interference due to Thin Film from Transmitted Light angles are very small π π π air air π Path difference of 1π π‘ and 2ππ transmitted light wave π π π cos π = 2 π cos π π When angle is very small, π β 0 cos π β 1 Path difference π 1π π‘ 2ππ
- 75. For constructive interference: Or For destructive interference: Or 2ππ = ππ 2ππ = π + 1 2 π πΏ = 2ππ πΏ = 2π + 1 π angles are very small π π π air air π Interference due to Thin Film from Transmitted Light
- 76. βπ₯ = 2ππ β π 2 Path difference of 1π π‘ and 2ππ reflected light wave π π π cos π = 2 π cos π π β 2ππ Total path difference π π π air air π angles are very small 1π π‘ 2ππ Interference due to Thin Film from Reflected Light Because phase change of π after reflection of 1π π‘ , path difference = β π 2
- 77. For constructive interference: For destructive interference: 2ππ = ππ 2ππ = π + 1 2 π π π π air air π π angles are very small Interference due to Thin Film from Reflected Light 2ππ β π 2 = ππ 2ππ β π 2 = π + 1 2 π β 2ππ = π + 1 π π is an integer.
- 78. Interference due to Thin Film For constructive interference: β’ Colours will be strongly reflected/transmitted. Destructive interference: β’ Colours will be poorly reflected/transmitted. This gives coloured appearance of the film.
- 79. A soap film of refractive index 1.33 is illuminated by the light of wavelength 400 ππ at an angle of 45Β°. If there is complete destructive interference then, find the thickness of the film. Given: To find: Solution: T π = 1.33, π = 45Β°, π = 400 ππ π‘ π = sin π sin π β 1.33 = sin 45 sin π sin π = 3 4 2 cos π = 1 β sin2 π For Destructive interference: 2ππ‘ cos π = ππ β 2 1.33 π‘ 0.85 = 1 400 Γ 10β9 β π‘ = 1.27 Γ 10β7 π β cos π = 0.85 45Β° soap film
- 80. Diffraction β’ Bending of a wave or its deviation from its original direction of propagation while passing through a small obstruction is known as diffraction. Original direction β’ Diffraction is explained by wave nature of light.β β’ Condition for bending: size of obstacle β π β’ Every point on the wavefront makes a secondary wave according to Huygens Principle.
- 81. β’ Source and screen are at infinite distance from diffraction element. β’ Source and screen are at finite distance from diffraction element. β’ High intense interference pattern is observed on screen. β’ Diffraction was first demonstrated by this experiment. ππ ππ Diffraction of Light Waves Fraunhofer Diffraction Fresnel Diffraction
- 82. ππ π π π Screen π 2 π 2 sin π π β’ Path difference between the waves, βx = π 2 sin π Fraunhofer Diffraction β Path Difference
- 83. ππ π π π Screen β’ The condition for 1π π‘ dark fringe formed at point π: Fraunhofer Diffraction β first Dark Fringe π 2 sin π = π 2 π sin π = π β’ The condition for ππ‘β dark fringe: π sin π = ππ β’ Central Maxima is at π0, where π = 0
- 84. ππ π π π Screen β’ Electric Field Amplitude at π: Fraunhofer Diffraction β Intensity at general point πΈβ² = πΈ0 sin π½ π½ β’ When π = 0, π½ = 0, πΈβ² = πΈ0 πΌ = πΌπ sin2 π½ π½2 Where πΌ0 is the intensity at Central Maxima Where π½ = 2π π π 2 sin π π½ = phase difference β’ Intensity β πΈ2 β’ Intensity at a general point:
- 85. Fraunhofer Diffraction β Graph of Intensity πΌ = πΌπ sin2 π½ π½2 π½ = ππ sin π π πΌπ β 2π π sin π πΌ 0 β π π π π 2π π πΌπ 22 πΌπ 62.5 β’ sin π = 0 ; Central Maxima β’ sin π = ; Minima ππ π β’ sin π = Β± π π , Β± 2π π , Β± 3π π β¦ β¦ β¦ β¦ β’ Maximum intensity is distributed between β π π and π π . β’ If π is large, π π β 0, then Single fringe is observed and no diffraction is seen.
- 86. Fraunhofer Diffraction β Graph of Intensity π β’ Fraunhofer Diffraction βΆ Intensity decreases away from the centre. β’ πΌ0 β Intensity at the central bright fringe β’ YDSE βΆ Intensity is same for all fringes. β’ πΌ0 β Intensity from a single slit. πΌ = πΌπ sin2 π½ π½2 πΌ = 4πΌπ cos2 π 2
- 87. Diffraction Difference between Interference and Diffraction Interference It is the phenomenon of superposition of two waves coming from two different coherent sources. It is the phenomenon of superposition of two waves coming from two different parts of the same wavefront. In interference pattern, all bright fringes are equally bright and equally spaced. All bright fringes are not equally bright and equally wide. Brightness and width decreases with the angle of diffraction. All dark fringes are perfectly dark. All dark fringes are perfectly dark, but their contrast with bright fringes and width decreases with angle of diffraction. In interference, bright and dark fringes are large in number for a given field of view. In diffraction, bright and dark fringes are fewer for a given field of view.
- 88. A beam of light of wavelength 600 ππ from a distant source falls on a single slit 1 ππ wide and a resulting diffraction pattern is observed on a screen 2 π away. The distance between the first dark fringes on either side of central bright fringe is 1ππ πΌ π π· = 2 π π sin π β π Solution: π = π π πΌ = 2π = 2π π πΌ = π¦ π· = 2π π βΉ π¦ = 2ππ· π π¦ = 2 Γ 600 Γ 10β9 Γ 2 10β3 = 24 Γ 10β4 π π¦ = 2.4 ππ First maxima is formed at π π distance away on both side of central maxima. π¦ (π is small)
- 89. π΄ π΅ π 3π 2 π π sin π = 3π 2 π1π΅ = 3π 2π π¦1π΅ = 3π·π 2π β’ Condition for 1π π‘ secondary maxima (bright): β’ The angle of diffraction (π1B) for 1st maximum (bright) is: β’ The position of 1st maximum (bright) from the centre of the screen is: Single Slit Diffraction β πππ Secondary Maxima
- 90. Screen First secondary maxima Second secondary maxima Central maxima First secondary maxima Second secondary maxima Where, π = 1, 2, 3, 4 β¦ β¦ β¦ β’ The angle of diffraction ππ for ππ‘β maxima (bright) is: β’ The position of ππ‘β maxima (bright) from the centre of the screen is: πππ΅ = 2π + 1 π 2π π¦ππ΅ = 2π + 1 π·π 2π Single Slit Diffraction β πππ Secondary Maxima
- 91. β’ Angular position for minima + π π β π π + 2π π β 2π π β’ Angular position for maxima sin π = Β± π π , Β± 2π π , β¦ β¦ , Β± ππ π sin π = Β± 3π 2π , Β± 5π 2π , β¦ β¦ , Β± (2π + 1)π 2π Single Slit Diffraction + 3π 2π β 3π 2π + 5π 2π β 5π 2π 0
- 92. Given: To find: Solution: Angular position of first minima is given by, sin π1π· = π π βΉ 1 2 = π π Angular position of first secondary maxima is given by, sin π1π΅ = 3π 2π βΉ sin π1π΅ = 3 4 π1π΅ = sinβ1 3 4 T π1π· = 30Β°; π = 5000 αΆ π΄ Angular position of first secondary maximum (π1π΅) In a diffraction pattern due to a single slit of width π, the first minima is observed at an angle 30Β° when the light of wavelength 5000 αΆ π΄ is incident on the slit. The first secondary maxima is observed at an angle of:
- 93. β’ When a monochromatic light is incident on the hole we see concentric circular bright and dark spots on the screen. β’ The size of hole should be comparable to the wavelength of incident light. β’ Central bright spot contains the most energy. β’ Brightness of the rings decreases as we move away from the centre. β’ For the first dark ring, sin π = 1.22π π Fraunhofer Diffraction for Hole π π· π π π = 1.22ππ· π
- 94. A convex lens of diameter 8.0 ππ is used to focus a parallel beam of light of wavelength 6200 Γ . If the light be focused at a distance of 20 ππ from the lens, what would be the radius of the central bright spot? 6200 Γ π = 8 ππ π 20 ππ Given: To Find: Solution: π· = 20 ππ, π = 8 ππ and π = 6200 Γ π Radius of central bright spot, π = 1.22ππ· π β π = 1.22 Γ 6200 Γ 10β10 Γ 20 Γ 10β2 8 Γ 10β2 π = 1.89 Γ 10β6 π π
- 95. Binary Star
- 96. Limit of Resolution Just resolved: Well resolved: Unresolved: Diffraction discs from both sources overlap. π1 π2 π1 π2 π1 π2 The periphery of the diffraction disc of one object touches the centre of the diffraction disc of the other object. The diffraction discs formed by two objects are well separated from each other.
- 97. Rayleigh criterion The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. Unresolved Just resolved Well resolved (Rayleigh criterion)
- 98. Limit of Resolution of a Telescope β’ Clear image is formed when the diffraction discs from two sources is just resolved. ππ‘ππ1 ππ‘ππ2 Just resolved π βπ Telescope lens β’ Distance between diffraction disc > π β Well resolved β’ Distance between diffraction disc < π β Unresolved
- 99. Resolving Power of Telescope β’ Angular limit of resolution of telescope: βπ = 1.22π π βπ = Limit of resolution β’ Angle subtended by the first dark fringe > βπ β Well resolved β’ Angle subtended by the first dark fringe < βπ β Unresolved π π π βπ
- 100. Radius of the Central Bright Spot β’ Radius of central bright region is: β’ Resolving power of a telescope: π = 1.22ππ π π . π. = π 1.22π π = πβπ π . π. = 1 βπ β’ Bigger lens β larger π β smaller Limit of Resolution βπ β higher Resolving Power π . π. π π π βπ
- 101. Disadvantages of using Lens 1. Difficult and expensive to build large lenses. 2.Providing mechanical support to large lenses require large and complex machinery. 3. Chromatic aberration ( light rays passing through a lens focus at different points, depending on their wavelength). The largest lens objective in use has diameter of 40 inch (~1.02 π). It is at the Yerkes Observatory in Wisconsin, USA. Fact: Chromatic aberration
- 102. Advantages of using Mirror as the Objective in Telescope 1. No chromatic aberration 2. Parabolic mirror used to counter spherical aberration. 3. Large mirrors can be supported from the back. The viewer sits near the focal point of the mirror, in a small cage. Fact: Eyepiece Objective mirror
- 103. Given: To find: Solution: Limit of resolution of telescope is given by: βπ = 1.22 Γ 500 Γ 10β9 200 Γ 10β2 T π = 200 ππ, π = 500 ππ Limit of resolution of telescope 305 Γ 10β9 ππππππ βπ = 1.22π π Calculate the limit of resolution of a telescope objective having a diameter of 200 ππ, if it has to detect light of wavelength 500 ππ coming from a star. JEE Main 2019
- 104. Limit of Resolution of a Microscope Angular limit of resolution of microscope: sin π β π = 1.22π π π = π£π = π£ Γ 1.22π π π = π π β π = π π Magnification of convex lens: π = π£ Γ 1.22π ππ β’ Clear images can be seen in the microscope if the diffraction discs are just resolved. Lens Formula: 1 π£ β 1 π’ = 1 π 1 β π£ π’ = π£ π π = 1 β π£ π β 1 β π = π£ π π = π£ π’ π β β π£ π π£ β« π β π£ π β« 1 π = π£ Γ 1.22π ππ We know that: π = 1.22π ππ Γ βππ = 1.22ππ π tan π½ = π 2π β π = 2π tan π½ π½ is small, π = 2π sin π½ ππππ = 1.22ππ π = 1.22ππ 2π sin π½ ππππ = 1.22π 2 sin π½
- 105. Microscope immersed in Oil When the setup is immersed in oil, ππππ changes to π π ππππ = 1.22ππππ 2 sin π½ βΆ ππππ = 1.22π 2π sin π½ Note: The product πsinπ½ is called the numerical aperture and is sometimesmarked on the objective. Resolving power of a microscope: β’ Resolving power of the microscope increases when it is immersed in a medium. β π . π. = 2π sin π½ 1.22π π . π. = 1 ππππ
- 106. Validity of Ray Optics β’ Consider a single slit of width βπβ.Diffraction pattern will be observed. β’ There will be central maxima due to diffraction. β’ Angular size of central maximum, π = π π If π¦ β π β’ The width of diffracted beam after it has travelled by π§, π¦ = π§ Γ π = ππ§ π π§πΉ is Fresnel distance. β’ If π§ < π§πΉ, the ray optics is valid. β’ If π§ > π§πΉ, spreading due to diffraction dominates. π§ β π2 π = π§πΉ
- 107. Given: π§πΉ = 2 Γ 10β3 2 600 Γ 10β9 = 6.7 π T π = 2 ππ; π = 600 ππ Solution: Ray optics is a good approximation up to Fresnel distance only. π§ β π2 π = π§πΉ β’ If π < 6.7 π βΆ Ray optics holds. β’ If π > 6.7 π βΆ Spreading due to diffraction dominates. For what distance is ray optics a good approximation when a plane light wave is incident on a circular aperture of width 2 ππ having wavelength 600 ππ?
- 108. When a parallel beam of light passes through a medium, a part of it appears in directions other than the incident direction. This phenomenon is called scattering of light. Aparticle Scattering β’ Light is an EM wave, it oscillates the charged particles in a medium because of its oscillating electric field. β’ Oscillating charged particles emit EM waves. β’ If the oscillating electric field of incident light has frequency π, the frequency of scattered wave will also have frequency π. Incidentlight
- 109. Rayleighβs Law of Scattering β’ When size of particles < π β’ Intensity of scattering depends on Intensity of scattered wave β 1 π4 1. Wavelength of light 2. Size of particles causing scattering π πΌ π΅ π π π πΊ π β ππππππ’π π β πππ₯πππ’π Scatters most Scatters least π increases and Intensity of scattering decreases
- 110. Unpolarized Light β The light having electric field oscillations in all directions in the plane perpendicular to the direction of propagation. Note: Light wave is coming out of the screen, and arrows show direction of oscillation of electric field. β Examples of source of unpolarized light: Candle, Bulb, Sun etc.
- 111. Plane Polarized Light Plane Polarized light β When electric field at a point always remains parallel to a fixed direction as time passes. Plane of polarization β Plane containing electric field and direction of propagation. π₯ β ππ₯ππ β Direction of propagation. π¦ β ππ₯ππ β Oscillation of electric field. π₯π¦ πππππ β Plane of polarization. When the polarizer is placed in the path of unpolarized light, the direction of oscillation of the electric field becomes parallel to the transmission axis. Note: If any unpolarized light of intensity πΌ0 is incident on a polarizer, we get a polarized light of intensity . πΌ0 2
- 112. Law of Malus β πΌ = Intensity of transmitted light β πΌ0 = Intensity of incident light Intensity of the wave after crossing the polarizer: Electric field amplitude of the wave after crossing the polarizer: πΈ = πΈ0 cos π πΌ β πΈ2 πΌ = πΌ0 cos2 π π πΈ0 πΈ0 cos π πΌ0 πΌ Polarized light Unpolarized light 2πΌ0
- 113. Law of Malus πΌ = πΌ0 cos2 0Β° = πΌ0 β Case 1 β when π = 0Β° : β Case 2 β when π = 90Β° : πΌ = πΌ0 cos2 90Β° = 0 πΌ0 πΌ0 πΌ0
- 114. T Two βcrossedβ polaroids π΄ and π΅ are placed in the path of a light beam. In between these, a third polaroid πΆ is placed whose polarization axis makes an angle π with the polarization axis of the polaroid π΄. If the intensity of light emerging from the polaroid π΄ is πΌπ, then the intensity of light emerging from polaroid π΅ will be Solution: Intensity of light emerging from polaroid πΆ: Intensity of light emerging from polaroid π΅: πΌπ΅ = πΌπΆ cos2 90Β° β π πΌπ΅ = πΌ0 cos2 π . cos2 90Β° β π πΌπ΅ = πΌ0 cos2 π . sin2 π {Applying the law of Malus.} = πΌ0 4 2 sin π cos π 2 πΌπΆ = πΌ0 cos2 π πΌπ΅ = πΌ0 4 sin2 (2π)
- 115. Unpolarized light with amplitude π΄0 passes through two polarizers. The first one has an angle of 30Β° clockwise to vertical and second one has an angle of 15Β° counter-clockwise to the vertical. What is the amplitude of the light emitted from the second polarizer? Given: To find: π΄ = π΄0 π΄2 Solution: We know that: πΌ β π΄2 . β΄ π΄ β πΌ β π΄1 π΄2 = πΌ1 πΌ2 π΄1 π΄0 = πΌ1 πΌ0 = πΌ0 2πΌ0 = 1 2 π΄1 = π΄0 2 πΌ1 = πΌ0 2 T π΄1 = π΄0 2 πΌ2 = πΌ1cos2 π πΌ2 = πΌ0 2 cos2 45Β° = πΌ0 2 1 2 2 = πΌ0 4 π΄2 π΄0 = πΌ2 πΌ0 = πΌ0 4πΌ0 = 1 4 π΄2 = π΄0 2 π΄1 π΄2 = πΌ1 πΌ2 Again, π = 30Β° β β15Β° = 45Β°
- 116. Medium(π) π Oscillations β₯ to the incidence plane Oscillations in the incidence plane Polarization by Reflection β The reflected light has more vibrations perpendicular to plane of incidence. β The refracted light has more vibration parallel to the plane of incidence. β The percentage of polarization in reflected light changes as we change the angle of incidence π.
- 117. β For a particular angle of incidence ππ΅ , the reflected light becomes completely plane polarized. β The required condition for this purpose is: ππ΅ + π = 90Β° β Brewsterβs Law tan ππ΅ = π ππ΅ = Brewster angle/polarising angle Brewsterβs Law tan ππ΅ = π2 π1 β If the light ray travels from one medium to another with refractive π1 and π2 respectively, then Brewsterβs law becomes, Medium(π) ππ΅ π
- 118. Relation between Critical Angle and Brewster Angle β΄ tan ππ΅ = 1 sin ππΆ β Brewsterβs Law tan ππ΅ = π ππ΅ = Brewster angle/polarising angle sin ππΆ = 1 π β Critical Angle: ππ΅ = tanβ1 1 sin ππΆ And ππΆ = sinβ1 1 tan ππ΅ Medium(π) ππ΅ π
- 119. T The polarizing angle of diamond is 67Β°. The critical angle of diamond is nearest to: [Given tan 67Β° = 2.36 ] To find: Solution: Critical angle Given: ππ΅ = 67Β° Critical angle is given by: ππΆ = sinβ1 1 2.36 1 2.36 < 1 2 β sinβ1 1 2.36 < sinβ1 1 2 β ππΆ < 30Β° β΄ Out of the four option only 22Β° is less than 30Β° ππΆ = sinβ1 1 tan ππ΅ ππΆ = sinβ1 1 tan 67Β°
- 120. Scattering by Polarization β If an unpolarized light gets scattered from air molecule, light perpendicular to original ray is plane polarized. β If we draw a plane perpendicular to the incident light, then from every viewpoint on the plane we can see the plane polarized light. Air molecule Incident Light Observer Observer
- 121. Given: To find: Solution: ππ = 60Β° Velocity of light in glass (π£π) As reflected light is plane polarized, ππ = 60β According to Brewsterβs law, π = tan ππ = tan 60β = 3 As, π = π£π π£π β π£π = π£π π T π£π = 3 Γ 108 3 = 3 Γ 108 π/π A ray of light, travelling in air, is incident on a glass slab with angle of incidence 60Β°. It is found that the reflected ray is plane polarized. The velocity of light in the glass is: Air Plane polarized Glass (π) 90Β° 60Β°