phy_1.pdf

PH113: Physics (B.Tech. I) Instructors: Dr. S K Mukherjee and Dr. Anupam Roy Dept of Physics, BIT Mesra. Drop an e-mail: royanupam [AT] bitmesra [DOT] ac [DOT] in. Section L, O & P ...

PH113: Physics (B.Tech. I) Instructors: Dr. S K Mukherjee and Dr. Anupam Roy Dept of Physics, BIT Mesra. Drop an e-mail: royanupam [AT] bitmesra [DOT] ac [DOT] in. Section L, O & P  PH113: Physics (B.Tech. I) Syllabus Module-1: Physical Optics Polarization, Malus’ Law, Brewster’s Law, Double Refraction, Interference in thin films (Parallel films), Interference in wedge-shaped layers, Newton’s rings, Fraunhofer diffraction by single slit, Double slit. Elementary ideas of fibre optics and application of fibre optic cables. Module-2: Electromagnetic Theory Gradient, Divergence and Curl, Statement of Gauss theorem & Stokes theorem, Gauss’s law, Applications, Concept of electric potential, Relationship between E and V, Polarization of dielectrics, dielectric constant, Boundary conditions for E & D, Gauss’s law in magnetostatics, Ampere’s circuital law, Boundary conditions for B & H, Equation of continuity, Displacement current, Maxwell’s equations. Module-3: Special Theory of Relativity Introduction, Inertial frame of reference, Galilean transformations, Postulates, Lorentz transformations and its conclusions, Length contraction, time dilation, velocity addition, Mass change, Einstein's mass energy relation. Module-4: Quantum Mechanics Planck's theory of black-body radiation, Compton effect, Wave particle duality, De Broglie waves, Davisson and Germer's experiment, Uncertainty principle, Brief idea of Wave Packet, Wave Function and its physical interpretation, Schrodinger equation in one-dimension, free particle, particle in an infinite square well. Module-5: Modern Physics Laser – Spontaneous and stimulated emission, Einstein's A and B coefficients, Population inversion, Light amplification, Basic laser action, Ruby and He-Ne lasers, Properties and applications of laser radiation, Nuclear Physics- Binding Energy Curve, Nuclear Force, Liquid drop model, Introduction to Shell model, Applications of Nuclear Physics, Concept of Plasma Physics, and its applications. Text books: 1: A. Ghatak, Optics, 4th Edition, Tata Mcgraw Hill, 2009 2: Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press ( 2001) 3: Arthur Beiser, Concept of Modern Physics, 6th edition 2009, Tata McGraw- Hill 4. F. F. Chen, Introduction to Plasma Physics and controlled Fusion, Springer, Edition 2016. Reference books: 1: Fundamentals of Physics, Halliday, Walker and Resnick Dr. Anupam Roy 2  PH113: Physics (B.Tech. I) Module 1 Module – 1 : Physical Optics Polarization, Malus’ Law, Brewster’s Law, Double Refraction, Interference in thin films (Parallel films), Interference in wedge-shaped layers, Newton’s rings, Fraunhofer diffraction by single slit, Double slit. Elementary ideas of fibre optics and application of fibre optic cables. Text book: T1: A. Ghatak, Optics, 6th Edition, 2017, McGraw-Hill Education (India) Pvt. Ltd. References: (1) Optics – E Hecht (2) Fundamentals of Optics – Jenkins & White Class structure: 4 Lectures including 1 Tutorial per week. (8 hours ~ 2 weeks for this module!)​ Dr. Anupam Roy 3 PH113: Physics (B.Tech. I) Module 1 Date: 05.09.2023 Lecture: 1 Dr. Anupam Roy 4  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Optics ❑Geometrical Optics and Physical Optics ❑Geometrical Optics is useful when the objects are very large compared to wavelength of light, e.g., lenses and mirrors. ❑Physical Optics deals with situations where the objects have dimensions comparable to the wavelength. ❑ Several new aspects of light-matter interaction – interference, diffraction, polarization (can only be explained by Physical Optics). Dr. Anupam Roy 5  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Optics ❑ Two types of waves: Longitudinal and Transverse ❑ Longitudinal wave: particles of the medium moves in the direction of propagation of the wave. Example: Sound wave, expansion & compression of spring. ❑ Transverse wave: particles movement is orthogonal/perpendicular to the direction of propagation. Example: EM wave (light wave). ❑ Light is an electromagnetic (EM) wave – consists of electric and magnetic field vectors in perpendicular directions, and propagates in a direction perpendicular to both the fields. Dr. Anupam Roy 6  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Direction of polarization is determined only by the oscillating direction/plane of the electric field, 𝑬 (not by the magnetic field, 𝑩) (𝑬 =𝑐𝑩 ≫ 𝑩) ❑ Oscillate one end of a string up and down – a transverse wave is generated. ❑ Each point of the string executes a sinusoidal oscillation in a straight line (along the x-axis) and the wave is, therefore, known as a linearly polarized wave. It is also known as a plane polarized wave because the string is always confined to the 𝑥-𝑧 plane. ❑ Polarization can be defined only in case of transverse waves and not in the longitudinal wave. Dr. Anupam Roy 7  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑Polarization: When the electric (or magnetic) field variations in a light wave are in a fixed plane, we say that the light is polarized. Dr. Anupam Roy 8  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ Light emitted from a common source like Sun or Electric bulb, the planes of vibration are random (each ray in the beam can vibrate in an arbitrary direction) – Unpolarized Light. ❑ Alternative way to represent unpolarized light: Each electric field vectors can be decomposed into x- and y-components. ❑ Amplitudes of vibrations along 𝑥-direction is 𝐸𝑥 and along 𝑦-direction is 𝐸𝑦. ❑ Polarizer is a device that make a unpolarized/randomly polarized light into a polarized light. Dr. Anupam Roy 9  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ Types of polarization: ❑ Plane or linear polarization: The electric field vector oscillate in one fixed plane ❑ Circular polarization: The electric field vector oscillate in a circle. Left & Right CP ❑ Elliptical polarization: The electric field vector oscillate in elliptical path (a) Linear polarization (b) circular polarization, and (c) elliptical polarization Dr. Anupam Roy 10  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Mathematically, the plane/linearly polarized light along 𝑥-axis is given by 𝐸1 (𝑧, 𝑡) = 𝑥ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) (1) Plane/linearly polarized light along 𝑦-axis is given by 𝐸2 (𝑧, 𝑡) = 𝑦ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡 + 𝜑) (2) 𝜑 is the phase difference compared to wave in (1) Positive phase shift Negative phase shift Dr. Anupam Roy 11  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Superposition of two linearly polarized waves with 𝝋 = 0 𝐸(𝑧, 𝑡) = 𝐸1 (𝑧, 𝑡) + 𝐸2 (𝑧, 𝑡) = 𝑥ො 𝐸01 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) + 𝑦ො 𝐸02 𝑐𝑜𝑠 𝑘𝑧 − 𝜔𝑡 = 𝑥ො 𝐸01 + 𝑦ො 𝐸02 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) Amplitude of the resultant wave 2 2 𝐸01 + 𝐸02 = 2𝐸0 (when 𝐸01 = 𝐸02 = 𝐸0 ) 𝐸02 𝐸0 Polarization (direction of field) of the resultant wave 𝜃= 𝑡𝑎𝑛−1 = 𝑡𝑎𝑛 −1 = 45𝑜 𝐸01 𝐸0 ❑ When two in-phase linearly polarized light waves are superposed, resultant wave has fixed orientation as well as amplitude. ❑ In other words, any linearly polarized light can be resolved into two components: along 𝒙-axis and along 𝒚-axis. Dr. Anupam Roy 12  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Consider uniform plane waves of same amplitude propagating along 𝑧Ƹ but polarized along 𝑥ො and 𝑦. ො Their superposition: 𝐸(𝑧, 𝑡) = 𝑥ො 𝐸01 + 𝑦ො 𝐸02 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) Note the relative phase! 45° linearly polarized light ! Dr. Anupam Roy 13  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ What is the amplitude and direction of resultant field when two linearly polarized light with 𝜑 = 𝜋 superpose with each other? ❑ Solve Dr. Anupam Roy 14  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization 𝜋 Superposition of two linearly polarized waves with 𝐸01 = 𝐸02 = 𝐸0 and 𝜑 = ± = 90𝑜 2 𝐸1 (𝑧, 𝑡) = 𝑥ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) 𝐸2 (𝑧, 𝑡) = 𝑦ො 𝐸0 𝑠𝑖𝑛(𝑘𝑧 − 𝜔𝑡) Superposed/resultant field: 𝐸(𝑧, 𝑡) = 𝐸1 𝑧, 𝑡 + 𝐸2 𝑧, 𝑡 = 𝐸0 [𝑥ො 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) + 𝑦ො s𝑖𝑛 𝑘𝑧 − 𝜔𝑡 ] Amplitude of the resultant field 1Τ2 𝐸 = 𝐸1 ∙ 𝐸2 = 𝑥𝐸 ො 0 𝑐𝑜𝑠 𝑘𝑧 − 𝜔𝑡 + 𝑦𝐸 ො 0 𝑠𝑖𝑛 𝑘𝑧 − 𝜔𝑡 ∙ 𝑥𝐸 ො 0 𝑐𝑜𝑠 𝑘𝑧 − 𝜔𝑡 + 𝑦𝐸 ො 0 𝑠𝑖𝑛 𝑘𝑧 − 𝜔𝑡 1Τ2 = 𝐸0 Resultant amplitude is constant and equal to the amplitude of the initial wave Dr. Anupam Roy 15  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ Direction or polarization of the field From scalar part of two initial waves 𝐸1 (𝑧, 𝑡) = 𝑥ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) 2 2 𝐸1 𝐸2 + =1 (Equation of circle) 𝐸2 (𝑧, 𝑡) = 𝑦ො 𝐸0 𝑠𝑖𝑛(𝑘𝑧 − 𝜔𝑡) 𝐸0 𝐸0 The resultant field equation rotate in a circle. The polarization is called Circular Polarization 𝝅 Hence, superposition of two plane/linearly polarized wave with equal amplitude and phase difference = ± 𝟐 leads to circular polarization. Problem: What is the difference between polarization direction when phase difference is /2 and −/2? ‘Or’ Define left handed polarized (LHP) and Right handed polarized (RHP) light. Dr. Anupam Roy 16  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ Direction or polarization of the field 𝐸1 (𝑧, 𝑡) = 𝑥ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) 𝐸1 (𝑧, 𝑡) = 𝑥ො 𝐸0 𝑐𝑜𝑠(𝑘𝑧 − 𝜔𝑡) 𝐸2 (𝑧, 𝑡) = 𝑦ො 𝐸0 𝑠𝑖𝑛(𝑘𝑧 − 𝜔𝑡) 𝐸2 (𝑧, 𝑡) = −𝑦ො 𝐸0 𝑠𝑖𝑛(𝑘𝑧 − 𝜔𝑡) Left Circular Linear Right Circular Dr. Anupam Roy 17  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ Use of Polarization ❑ Astronomy, Communication and RADAR applications (all radio transmitting and receiving antennas are intrinsically polarized, special use in radar), 3-D movies and polarized 3-D glasses ❑ Polarization is used in infrared spectroscopy, use in ophthalmic instruments ❑ In sunglasses, polarization is used to reduce glare. Dr. Anupam Roy 18  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization ❑ If an unpolarized wave falls on a slit 𝑆1 then the displacement associated with the transmitted wave is along the slit length. ❑ A rotation of the slit will not affect the amplitude of the transmitted wave although the plane of the polarization of transmitted wave depend on the orientation of the slit. Thus, the transmitted wave will be linearly polarized and slit 𝑆1 is acting as a polarizer. ❑ If this polarized beam falls on another slit 𝑆2, then by rotating 𝑆2 we can vary the amplitude of transmitted wave. This second slit 𝑆2 is acting as an analyzer. This slit can be used to detect the direction of polarization of the linearly polarized wave. Dr. Anupam Roy 19  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Malus’ law ❑ This law tells us how the intensity transmitted by the analyzer varies with the angle that its plane of transmission makes with that of the polarizer (ideal polarizer). ❑ In the case of two piles of plates, the plane of transmission is the plane of incidence, and for the law of Malus to hold we must assume that the transmitted light is completely plane-polarized. ❑ The law of Malus states that the transmitted intensity varies as the square of the cosine of the angle between the two planes of transmission. Dr. Anupam Roy 20  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Malus’ law (Contd..) ❑ If an 𝑥-polarized beam is passed through a polaroid 𝑃2 whose pass axis makes an angle 𝜃 with the 𝑥-axis, then the intensity of the emerging beam will be given by, 𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃 ❑ where 𝐼0 represents the intensity of the emergent beam when the pass axis of 𝑃2 is also along the 𝑥-axis (i.e., when 𝜃 = 0). This equation (𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃) represents the Malus’ law. An unpolarized light beam gets 𝑥 -polarized after passing through the polaroid 𝑃1 , the pass axis of the second polaroid 𝑃2 makes an angle 𝜃 with the 𝑥-axis. The intensity of the emerging beam will vary as 𝑐𝑜𝑠 2 𝜃. Dr. Anupam Roy 21  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Malus’ law (Contd..) ❑ 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 = 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 2 ❑ If incident wave has amplitude 𝐸0 , then the outgoing wave has an amplitude of 𝐸 = 𝐸0 𝑐𝑜𝑠𝜃. ❑ 𝐼 = 𝐸 2 = 𝐸0 2 𝑐𝑜𝑠 2 𝜃 = 𝐼0 𝑐𝑜𝑠 2 𝜃 ❑ 𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃 Dr. Anupam Roy 22  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) Malus’ law (Contd..) 2 ❑ 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 = 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 ❑ For any 𝜃, 𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃 𝐼0 ❑ But 𝜃 is random => 𝐼 = 𝐼0 𝑐𝑜𝑠 2 𝜃 = 2 1 ❑ Average value of 𝑐𝑜𝑠 2 𝜃 is 2 (Using Malus law: incident light is polarized) Dr. Anupam Roy 23  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization Malus’ law (Contd..) Example: A beam of vertically polarized light is incident on a linear polarizer. It is observed that the intensity of the light emerging out of the polarizer is 25% of the incident intensity when the angle between the transmission axis of the polarizer and the vertical is 60o. Solution According to Malus law, 𝐼 = 𝐼0 𝐶𝑜𝑠 2 𝜃 When the transmission axis of polarizer is 60o, the intensity of transmitted light is, 2 𝐼0 𝐼 60 = 𝐼0 𝐶𝑜𝑠 60 = = 0.25 𝐼0 4 Dr. Anupam Roy 24  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization So far… ▪ Polarization of light is due to its transverse nature. ▪ Types of polarization: (1) Linear (2) circular (RHP & LHP) (3) Elliptical ▪ Superposition of two linearly polarized waves with phase difference 0, ±𝜋 leads to another linearly polarized wave. ▪ A linearly polarized light can be resolved into two components perpendicular to each other (along 𝑥 and 𝑦). 𝜋 ▪ Superposition of two linearly polarized wave with equal amplitude and phase difference = ± leads to 2 circular polarization. 𝜋 ▪ Superposition of two linearly polarized wave with phase difference other than 0, ±𝜋 and ± leads to 2 formation of elliptical polarized wave. ▪ Malus’ law provides a condition, 𝐼 = 𝐼0 𝐶𝑜𝑠 2 𝜃 for an ideal polarizer. Dr. Anupam Roy 25  PH113: Physics (B.Tech. I) Module 1 Question? Dr. Anupam Roy 26 PH113: Physics (B.Tech. I) Module 1 Date: 08.09.2023 Lecture: 2 Dr. Anupam Roy 27  PH113: Physics (B.Tech. I) Module 1 Module – 1 : Physical Optics Polarization, Malus’ Law, Brewster’s Law, Double Refraction, Interference in thin films (Parallel films), Interference in wedge-shaped layers, Newton’s rings, Fraunhofer diffraction by single slit, Double slit. Elementary ideas of fibre optics and application of fibre optic cables. Text book: T1: A. Ghatak, Optics, 6th Edition, 2017, McGraw-Hill Education (India) Pvt. Ltd. References: (1) Optics – E Hecht (2) Fundamentals of Optics – Jenkins & White Class structure: 4 Lectures including 1 Tutorial per week. (8 hours ~ 2 weeks for this module!)​ Dr. Anupam Roy 28  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) Polarization of unpolarized light: ❑Several strategies ❑By scattering ❑By reflection ❑By double refraction ❑By absorption Dr. Anupam Roy 29  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 1) Polarization by Scattering ❑ If an unpolarized light is allowed to fall on a gas, then the beam scattered at 90° to the incident beam is linearly polarized. ❑ It follows from the fact that the waves propagating in the y-direction are produced by the 𝑥 -component of the dipole oscillations. The 𝑦 - component of dipole oscillations will produce no field in the 𝑦-direction. ✓ Unscattered part of the incident beam remains unpolarized. ✓ The waves scattered by the gas molecules in perpendicular direction is linearly polarized. ✓ Waves scattered in other directions are partially polarized. Dr. Anupam Roy 30  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 2) Wire Grid Polarizer and Polaroid ❑ Consists of a regular array of parallel metallic wires (e.g., copper), placed in a plane perpendicular to the incident beam. Wires are placed closed to each other such that the spacing ≲ 𝜆 in order to reduce diffraction effects. ❑ Suppose an unpolarized EM wave incidents on the grid. The 𝑦-component of the incoming field (parallel to wires) moves the electrons of the wire and generate currents in the wire. ❑ This leads to Joule heating and the 𝑥-component gets converted into the heat energy. The 𝑥-component of field in not absorbed and passes through. ❑ Thus, it converts an unpolarized light into a linearly polarized with the electric vector along the 𝑥-axis. Dr. Anupam Roy 31  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 2) Wire Grid Polarizer and Polaroid ❑ Note: For this system to be effective, spacing between the wires should be ≲ 𝜆. Remember, light waves are associated with a very small wavelength. So, fabrication of a polarizer with such a small spacings is extremely difficult. ❑ Instead of thin wires, one prepares sheets of long polymers chains. These atoms provide high conductivity. ❑ When a light beam is incident on such a polaroid, the molecules (aligned parallel to each other) absorb the component of electric field (which is parallel to the alignment). It, thus, acts like the wire grid polarizer. ❑ Now the spacing between two adjacent long chain molecules is small compared to optical wavelength. So, it is very effective in producing linearly polarized light and can be used for wavelengths in visible range. Dr. Anupam Roy 32  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 3) Polarization by reflection: Brewster’s law 𝜃 𝜃 ❑ Consider unpolarized light to be incident at an angle 𝜃 on a dielectric like glass, as shown in Figure. There will always be a reflected ray and a refracted ray. The reflected ray is partially plane-polarized. ❑ Only at a certain angle, about 57° for ordinary glass (air-glass interface), it is completely plane-polarized. At this polarizing angle 𝜃𝐵 , the reflected and refracted rays are just 90° apart. ❑ This is Brewster's law, which shows that the angle of incidence for maximum polarization depends only on the refractive index. This remarkable discovery enables one to correlate polarization with the refractive index. Dr. Anupam Roy 33  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 3) Polarization by reflection: Brewster’s law ❑ When light wave travels from low density (refractive index) medium (e.g., air) to high density (refractive index) medium (e.g., water or glass), at a particular angle of incidence, the reflected light is polarized. The angle at which polarization occurs is known as Brewster angle (𝜃𝐵 ). ❑ At Brewster angle, the angle between reflected and transmitted (refracted) wave is 90°. Unpolarized light incident (a) at angle I and (b) at angle B  Wave polarized parallel to the plane of incidence Wave polarized perpendicular to the plane of incidence Dr. Anupam Roy 34  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 3) Polarization by reflection: Brewster’s law ❑ Consider unpolarized light to be incident at an angle 𝜃 on a dielectric like glass, Air (n1) as shown in Figure. When the reflected and refracted rays become perpendicular, the reflected ray becomes completely plane-polarized. The Glass (n2) corresponding angle of incidence is called the Brewster’s angle (𝜃𝐵 ). 𝑛2 ❑ 𝜃𝐵 = tan−1 𝑛1 ❑ Proof: 𝜃𝐵 + 𝜃𝑟 = 90° Tangent of Brewster angle is equal to the ratio of the refractive indices of ❑ Snell’s law: 𝑛1 sin 𝜃𝐵 = 𝑛2 sin 𝜃𝑟 = 𝑛2 sin(90° − 𝜃𝐵 ) = 𝑛2 cos 𝜃𝐵 the media at whose interface incident 𝑛2 𝒏𝟐 light is reflected. ❑ tan 𝜃𝐵 = ⇒ 𝜽𝑩 = 𝒕𝒂𝒏−𝟏 Dr. Anupam Roy 35 𝑛1 𝒏𝟏  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 3) Polarization by reflection: Brewster’s law Example ❑ When the incident beam is in air (𝑛1 = 1) and the transmitting medium is glass (𝑛2 = 1.5) the Brewster angle is nearly 57°. ❑ For air (𝑛1 = 1) and water (𝑛2 = 1.33) interface (like the surface of a pond or a lake), 𝜃𝐵 is 53°. This means that when the sun is 37° above the horizontal, the light reflected by a calm pond or lake should be completely linearly polarized. ❑ Problem: A plate of flint glass (refractive index 1.67) is immersed in water (refractive index 1.33). Calculate the Brewster angle for internal as well as external reflection at an interface. Dr. Anupam Roy 36  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 4) Polarization by double refraction (Birefringence) ❑ Similar to reflection, refraction can also polarize the beam. ❑ Only optically anisotropic crystals (calcite, cellophane) polarizes the beam. (Optically anisotropic means that optical properties are not the same in all directions.) ❑ Light passes through optically anisotropic crystals shows “double refraction” or “birefringence”. ❑ Materials showing double refraction are called Birefringent material. Example: ❑ A dot mark on a paper appears as two dots when viewed through calcite crystal. ❑ Calcite crystal splits the incident beam into two due to double refraction. Dr. Anupam Roy 37  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 4) Polarization by double refraction (Birefringence) ❑ When an unpolarized light beam is incident on a calcite (CaCO3) crystal, it splits into two beams, each of which is linearly polarized. ❑ The ray that follows Snell’s law is called ordinary ray or o-ray. ❑ The ray that does not follow Snell’s law is called extraordinary ray or e-ray. ❑ If the crystal is rotated about the incident ray as an axis, then the e-ray rotates about the o-ray. Dr. Anupam Roy 38  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Remember: Snell’s law ❑ Snell’s law gives a relationship between the angles of incidence (𝜃1 ) and refraction (𝜃2 ) when a ray of light travels from a rarer medium of refractive index (𝑛1 ) to a denser medium of refractive index (𝑛2 ) ❑ 𝑛1 𝑠𝑖𝑛𝜃1 = 𝑛2 𝑠𝑖𝑛𝜃2 Dr. Anupam Roy 39  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 4) Polarization by double refraction (Birefringence) ❑ The velocity of e-ray is different from o-ray. ❑ If 𝑣𝑜 < 𝑣𝑒 , then the crystal is negative (for example, calcite). ❑ If 𝑣𝑜 > 𝑣𝑒 , then the crystal is positive (for example, quartz). ❑ There is a particular direction, called the Optic Axis, along which there is no splitting and 𝑣𝑜 = 𝑣𝑒 (only one ray is obtained that follows the Snell’s law). ❑ Optic axis: A special directions in a birefringent crystal along which two refractive indices are equal and both o- and e-rays in the direction of optic axis propagate with same velocity. Dr. Anupam Roy 40  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 4) Polarization by double refraction (Birefringence) ❑ Optic axis (Example): Calcite crystal normally used in the shape of rhombohedra bounded by six parallelograms with angles 102° and 78°. At corners A and H, angle of each of the three faces is obtuse and these points are known as blunt corners. ❑ A line passing through one of the blunt corners and is equally inclined to all the three edges meeting at that point gives the direction of optic axis. Any line parallel to this line is also known as optic axis. ❑ Birefringent crystals with only one optic axis are called uniaxial crystals (e.g., Calcite, Quartz). Crystals having two optic axes are called biaxial crystals (e.g., Mica). ❑ When the refractive index for o-ray (𝑛𝑜 ) is more than that for the e-ray (𝑛𝑒 ) – negative uniaxial crystal. (𝑛𝑜 > 𝑛𝑒 ). For positive uniaxial crystal, (𝑛𝑜 < 𝑛𝑒 ). Dr. Anupam Roy 41  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Polarization - Production of polarized light (methods to produce linearly polarized light) 4) Polarization by double refraction (Birefringence) ❑ The birefringent crystal splits an unpolarized light into e-ray and o-ray. ❑ the electric field of e-ray vibrate in the plane containing the optic axis and electric field of o-ray vibrates perpendicular to the optic axis. ❑ Therefore, due to double refraction, the unpolarized light beam splits into two components which are plane polarized. Dr. Anupam Roy 42  PH113: Physics (B.Tech. I) Module 1 Questions? Dr. Anupam Roy 43 PH113: Physics (B.Tech. I) Module 1 Date: 11.09.2023 Lecture: 3 (Tutorial Class) Dr. Anupam Roy 44  PH113: Physics (B.Tech. I) Module 1 Malus’s Law Dr. Anupam Roy 45 PH113: Physics (B.Tech. I) Module 1 Dr. Anupam Roy 46 PH113: Physics (B.Tech. I) Module 1 Dr. Anupam Roy 47 PH113: Physics (B.Tech. I) Module 1 Dr. Anupam Roy 48 PH113: Physics (B.Tech. I) Module 1 Date: 12.09.2023 Lecture: 4 Dr. Anupam Roy 49  PH113: Physics (B.Tech. I) Module 1 Module – 1 : Physical Optics Polarization, Malus’ Law, Brewster’s Law, Double Refraction, Interference in thin films (Parallel films), Interference in wedge-shaped layers, Newton’s rings, Fraunhofer diffraction by single slit, Double slit. Elementary ideas of fibre optics and application of fibre optic cables. Text book: T1: A. Ghatak, Optics, 6th Edition, 2017, McGraw-Hill Education (India) Pvt. Ltd. References: (1) Optics – E Hecht (2) Fundamentals of Optics – Jenkins & White Class structure: 4 Lectures including 1 Tutorial per week. (8 hours ~ 2 weeks for this module!)​ Dr. Anupam Roy 50  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference ❑ Superposition of two coherent wave (definite phase relation) (ideally from the same source) to form periodic variation (fringes) in the intensity. ❑ Fringe with highest intensity is due to constructive interference. ❑ Fringe with lowest intensity is due to destructive interference. Wave amplitudes add in Wave amplitudes cancel constructive interference to give zero output due to ❑ Interference can be produced by (a) division of wavefront and (b) destructive interference division of amplitude. ❑ Consider two waves as ❑ Principle of superposition: ❑ 𝑦1 = 𝐴 cos 𝜔𝑡 ❑ 𝑦 = 𝑦1 + 𝑦2 = 𝐴 cos 𝜔𝑡 + 𝐴 cos (𝜔𝑡 + 𝜑) ❑ 𝑦2 = 𝐴 cos (𝜔𝑡 + 𝜑) Dr. Anupam Roy 51  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film Constructive and destructive interferences ❑ Consider two waves as ❑ 𝑦1 = 𝐴 cos 𝜔𝑡 and 𝑦2 = 𝐴 cos (𝜔𝑡 + 𝜑) ❑ Principle of superposition: 𝑦 = 𝑦1 + 𝑦2 = 𝐴 cos 𝜔𝑡 + 𝐴 cos (𝜔𝑡 + 𝜑) ❑ Remember: path difference of 𝝀 (the wavelength) corresponds to a phase difference of 𝟐𝝅. ❑ For constructive interference, the phase difference between 𝑦1 and 𝑦2 is 𝜑 = 2𝑛𝜋, 𝑛 = 0, 1, 2, … …, i.e., the optical path difference (OPD) is 𝑙 = 𝑛𝜆 ❑ For destructive interference, the phase difference is 𝜑 = (2𝑛 + 1)𝜋, 𝑛 = 0, 1, 2, … …, i.e., the optical path 𝜆 1 difference (OPD) is 𝑙 = 2𝑛 + 1 = (𝑛 + )𝜆 2 2 Dr. Anupam Roy 52  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by superposition of waves ❑ Constructive interference: 1 + 2 ❑ Path diff.: 0, , 2, …., n ❑ Destructive interference: 1 + 3 ❑ Path diff.: /2, 3/2, …., (2n+1)/2 ❑ Partial constructive interference: 1 + 4 ❑ Path diff.: /4 …. 2𝜋 ❑ 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = × 𝑃𝑎𝑡ℎ 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜆 Dr. Anupam Roy 53  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (normal incidence) ❑ The film has parallel surface, and incident rays are normal to the surface ❑ For constructive interference, we should have: 2𝑛𝑑 = 𝑚𝜆, 𝑚 = 0, 1, 2, … … ❑ For destructive interference, we should have: 1 2𝑛𝑑 = (𝑚 + )𝜆, 𝑚 = 0, 1, 2, … … 2 ❑ The above relations are incorrect, because due to reflection from a rarer medium to a denser medium, a phase 𝜆 difference of 𝜋 (or an optical path difference, OPD of ) is introduced. 2 ❑ Optical path difference, OPD = actual distance × refractive index of medium ❑ The correct relations are: 1 ❑ For constructive interference, we should have: 2𝑛𝑑 = (𝑚 + )𝜆, 𝑚 = 0, 1, 2, … … 2 ❑ For destructive interference, we should have: 2𝑛𝑑 = 𝑚𝜆, 𝑚 = 0, 1, 2, … … Dr. Anupam Roy 54  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (oblique incidence) Source ‘S’ P ❑ The film has parallel surface, and incident rays are at certain angle to the surface. ❑ Light from a monochromatic source ‘S’ falling on a thin film creates two set of parallel rays. ❑ Wave reflected from the upper surface of the film interferes with the wave reflected from the lower surface. ❑ Each set, when collected through a lens can make constructive or destructive interference pattern (at P) depending upon the phase relation between them. Dr. Anupam Roy 55  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (oblique incidence) ❑ Optical path difference (OPD) can only be introduced in the region BCFD. ❑ Remember: OPD = actual distance × refractive index of medium ❑ 𝑂𝑃𝐷 = 𝑛2 𝐵𝐷 + 𝐷𝐹 − 𝑛1 𝐵𝐶 ❑ ∠𝐵𝐹𝐶 = 𝜃𝑖 and ∠𝐹𝐵𝐾 = 𝜃𝑟 ❑ From Δ𝐵𝐹𝐶, 𝐵𝐶 = 𝐵𝐹 𝑠𝑖𝑛𝜃𝑖 ……….Eqn (1) 𝐵𝐷 + 𝐷𝐹 = 𝐵’𝐹 ❑ From Δ𝐵𝐾𝐹, 𝐾𝐹 = 𝐵𝐹 𝑠𝑖𝑛𝜃𝑟 ……….Eqn (2) 𝐾𝐹 𝑛2 ❑ Substitute Eqn. 2 in Eqn. 1: 𝐵𝐶 = 𝑠𝑖𝑛𝜃𝑖 = 𝐾𝐹 (using Snell’s law) ……..Eqn (3) 𝑠𝑖𝑛𝜃𝑟 𝑛1 Dr. Anupam Roy 56  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (oblique incidence) ❑ Optical path difference: 𝑂𝑃𝐷 (Δ) = 𝑛2 𝐵𝐷 + 𝐷𝐹 − 𝑛1 𝐵𝐶 ❑ From Δ𝐵𝐹𝐶, 𝐵𝐶 = 𝐵𝐹 𝑠𝑖𝑛𝜃𝑖 [Eqn (1)] and from Δ𝐵𝐾𝐹, 𝐾𝐹 = 𝐵𝐹 𝑠𝑖𝑛𝜃𝑟 [Eqn (2)] ❑ Substituting Eqn. 2 in Eqn. 1, we get: 𝐾𝐹 𝑛2 ❑ 𝐵𝐶 = 𝑠𝑖𝑛𝜃𝑖 = 𝐾𝐹 (using Snell’s law) ……..[Eqn (3)] 𝑠𝑖𝑛𝜃𝑟 𝑛1 𝐵𝐷 + 𝐷𝐹 = 𝐵’𝐹 ❑ OPD: Δ = 𝑛2 𝐵𝐷 + 𝐷𝐹 − 𝑛1 𝐵𝐶 = 𝑛2 𝐵′ 𝐹 − 𝑛1 𝐵𝐶 𝑛2 = 𝑛2 𝐵′ 𝐹 − 𝑛1 𝐾𝐹 = 𝑛2 𝐵′ 𝐹 − 𝐾𝐹 𝑛1 = 𝑛2 𝐵′ 𝐾 = 𝑛2 𝐵𝐵′ 𝑐𝑜𝑠𝜃𝑟 = 𝑛2 2𝑑 𝑐𝑜𝑠𝜃𝑟. (cosine law) Dr. Anupam Roy 57  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (oblique incidence) ❑ Optical path difference: 𝑂𝑃𝐷 = 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 (cosine law) ❑ Due to reflection from a rarer medium to a denser medium, a phase difference of 𝜋 (or an optical path difference of 𝜆/2) is introduced. 𝜆 ❑ Ray reflected from the top surface of film undergoes a phase change of  (or path change ). So, total 2 𝜆 path difference is 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 + 2 𝜆 1 ❑ Condition for destructive interference is: 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 + = (𝑚 + )𝜆 for 𝑚 = 0, 1, 2, …. 2 2 ❑ Or, 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 = 𝑚𝜆 (condition for minima) 𝜆 ❑ Condition for constructive interference is: 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 − = 𝑚𝜆 for 𝑚 = 0, 1, 2, …. 2 1 ❑ Or, 2𝑛2 𝑑 𝑐𝑜𝑠𝜃𝑟 = (𝑚 + )𝜆 (condition for maxima in interference pattern) 2 Dr. Anupam Roy 58  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a plane parallel film (oblique incidence) Example ❑ A thin film of thickness 4 x 10-5 cm is illuminated by white light normal to its surface. Its refractive index is 1.5. Of what color will the thin film appear in reflected light? Solution ❑ Color in reflection will appear due to constructive interference. 1 ❑ Condition for constructive interference is, 2𝑛𝑑 𝑐𝑜𝑠𝜃𝑟 = (𝑚 + )𝜆 2 2400 ❑ By putting the values of 𝑛 (= 1.5), 𝜃𝑟 (= 0°) and 𝑑 (= 4 x 10−5 cm), and solving, 𝜆 = 𝑛𝑚 (2𝑚+1) ❑ Taking 𝑚 = 0, 1, 2, 3, we get 𝜆 = 2400, 800, 480, 343.1 𝑛𝑚 ❑ Wavelengths 2400, 800 and 343.1 nm are not in visible region. 480 nm is in blue spectral region. So, the color appear in interference is Blue. Dr. Anupam Roy 59  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a wedge-shaped film (two non-parallel reflecting planes) ❑ So far, we have assumed the film to be uniform thickness. ❑ Now, we will discuss the interference pattern produced by a film of varying thickness. ❑ For example, wedge shaped film, where two surfaces (top and bottom) are not parallel. ❑ Two rays are normally incident on the upper surface, and they get reflected from the lower surface almost normally. ❑ The path difference is still given by the cosine law. However, 𝑋𝐵′ for a small angle 𝜑 and almost a normal incidence, 𝑐𝑜𝑠𝜑 ≈ 1. 𝑡𝑎𝑛𝜑 = 𝑋𝐴′ 𝑋𝐵′ = 𝐵𝐵′ − 𝐴𝐴′ = 𝑋𝐴′ 𝑡𝑎𝑛𝜑 Dr. Anupam Roy 60  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a wedge-shaped film (two non-parallel reflecting planes) ❑ Consider a bright fringe at A. The optical path difference (OPD), ∆= 𝑛(𝐿𝑀 + 𝑀𝐴) ≈ 2𝑛𝐴𝐴′ (∵ 𝐴𝐴′ ≃ 𝐿𝑀 ≃ 𝑀𝐴) ❑ For the next bright fringe (at B), OPD, ∆ ≈ 2𝑛𝐵𝐵′ ❑ From the condition of bright fringe, 1 1 ❑ 2𝑛𝐴𝐴′ = (𝑚 + )𝜆 (set 𝜃 → 0 in 2𝑛𝑑 𝑐𝑜𝑠𝜃 = (𝑚 + )𝜆) 2 2 1 ❑ So, condition for maxima is 2𝑛𝑑 = (𝑚 + )𝜆 (consider 𝐴𝐴′ = 𝑑) 2 𝑋𝐵′ ❑ Similarly, condition for minima is 2𝑛𝑑 = 𝑚𝜆 𝑡𝑎𝑛𝜑 = 𝑋𝐴′ 𝑋𝐵′ = 𝐵𝐵′ − 𝐴𝐴′ = 𝑋𝐴′ 𝑡𝑎𝑛𝜑 Dr. Anupam Roy 61  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a wedge-shaped film (two non-parallel reflecting planes) ❑ Consider a bright fringe at A. The optical path difference (OPD), ∆= 𝑛(𝐿𝑀 + 𝑀𝐴) ≈ 2𝑛𝐴𝐴′ (∵ 𝐴𝐴′ ≃ 𝐿𝑀 ≃ 𝑀𝐴) ❑ For the next bright fringe (at B), OPD, ∆ ≈ 2𝑛𝐵𝐵′ ❑ From the condition of bright fringe, 1 1 ❑ 2𝑛𝐴𝐴′ = (𝑚 + )𝜆 (set 𝜃 → 0 in 2𝑛𝑑 𝑐𝑜𝑠𝜃 = (𝑚 + )𝜆) 2 2 1 3 ❑ 2𝑛𝐵𝐵′ = 𝑚 + 1 + 𝜆= 𝑚+ 𝜆 2 2 ❑ Subtracting, 2𝑛(𝐵𝐵′ − 𝐴𝐴′ ) = 𝜆 𝑋𝐵′ ❑ Using the relation, 𝑋𝐵′ = 𝐵𝐵′ − 𝐴𝐴′ = 𝑋𝐴′ 𝑡𝑎𝑛𝜑, we get 𝑡𝑎𝑛𝜑 = 𝑋𝐴′ ❑ 2𝑛 𝑋𝐴′ 𝑡𝑎𝑛𝜑 = 𝜆 𝑋𝐵′ = 𝐵𝐵′ − 𝐴𝐴′ = 𝑋𝐴′ 𝑡𝑎𝑛𝜑 Dr. Anupam Roy 62  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a wedge-shaped film (two non-parallel reflecting planes) ❑ 2𝑛 𝑋𝐴′ 𝑡𝑎𝑛𝜑 = 𝜆 ❑ Now, 𝑋𝐴′ = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑟𝑖𝑛𝑔𝑒 (distance between one bright/dark fringe to the next one) ❑ Let’s say, 𝑋𝐴′ ≡ 𝛽 ❑ Hence, 2𝑛 𝑋𝐴′ 𝑡𝑎𝑛𝜑 = 2𝑛𝛽 𝑡𝑎𝑛𝜑 = 𝜆 𝝀 ❑ In the limit 𝜑 → 0, 𝑡𝑎𝑛𝜑 ≃ 𝜑 and we have 𝜷 = 𝟐𝒏𝝋 ❑ This is the relation for the fringe width 𝑋𝐵′ ❑ 𝛽 represents the fringe width and we have assumed 𝜑 to be 𝑡𝑎𝑛𝜑 = very small. Such fringes are commonly referred to as fringes of 𝑋𝐴′ equal thickness. 𝑋𝐵′ = 𝐵𝐵′ − 𝐴𝐴′ = 𝑋𝐴′ 𝑡𝑎𝑛𝜑 Dr. Anupam Roy 63  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference by a wedge-shaped film (two non-parallel reflecting planes) 𝝀 ❑ In the limit 𝜑 → 0, 𝑡𝑎𝑛𝜑 ≃ 𝜑 and we have 𝜷 = 𝟐𝒏𝝋 ❑ This is the relation for the fringe width (𝛽) – width of the bright fringes or that of the dark fringes. ❑ Fringe width depends on the wavelength (𝜆) of incident light, the refractive index of the film (𝑛) and the angle of wedge (𝜑). ❑ To observe a bright or dark fringe of a particular order for a wedge-shaped film with small wedge angle (𝜑) and normal incidence of light, thickness (𝑑) must be constant. ❑ For a wedge-shaped film, 𝑑 remains constant only along lines parallel to the thin edge of the wedge. The bright and dark fringes are obtained as straight lines parallel to the thin edge of the wedge. 𝜆 ❑ At the thin edge, 𝑑 = 0 and therefore path difference = , (the condition of minimum intensity) and the 2 edge of the wedge-shaped film will be dark. Dr. Anupam Roy 64  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Thin Film Interference ❑Interference from parallel reflecting plates (normal incidence) ❑Interference from parallel reflecting plates (oblique incidence) ❑Interference from non-parallel reflecting plates (wedge-shaped planes) Dr. Anupam Roy 65  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings ❑There is a glass plate, on which a plano-convex lens is placed with the plane surface facing upward. ❑Consider a monochromatic light incident on the lens. ❑Then we obtain circular fringes, called “fringes of equal thickness”. ❑Light reflected from the curved surface of the lens and from the glass plate interfere to produce these fringes. Dr. Anupam Roy 66  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings ❑When a curved surface (Convex surface of a plano-convex lens) LOL’ is placed on a flat surface POQ, a wedge (with upper surface curved) forms QOL’. ❑At the point of contact, the thickness of air film is Zero ❑Light wave from source A falling on this surface results in reflected ray 1 and refracted-reflected ray 2. ❑Interference of two waves results in concentric ring of dark and bright fringes called “Newton’s Ring”. ❑At the point of contact, the fringe is completely dark, as it was in case of wedge-shaped film. Dr. Anupam Roy 67  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings ❑When a curved surface (Convex surface of a plano-convex lens) LOL’ is placed on a flat surface POQ, a wedge (with upper surface curved) forms QOL’. ❑Recall small (wedge) angle approximation. At the point of contact, the fringe is completely dark, as in case of wedge-shaped film. ❑For near normal incidence (𝜑 → 0), the optical path difference is 2𝑛𝑡 (𝑛 = refractive index of the lens, glass plate; see Fig). 1 ❑Condition for maxima (bright): 2𝑛𝑡 = (𝑚 + )𝜆 2 ❑Similarly, condition for minima (dark): 2𝑛𝑡 = 𝑚𝜆 Dr. Anupam Roy 68  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings 𝑅 2𝑅 2𝑅-𝑡 Upper 𝑅 surface of 𝑅-𝑡 the lens 𝑡 𝑡 𝑟𝑚 𝑚th bright fringe ❑ Upper surface of the lens (labeled). Radius of curvature of the curved surface of the lens is 𝑅. ❑ Line in red represents 𝑚th bright fringe of radius 𝑟𝑚 and 𝑡 is the thickness of air film where 𝑚th film is formed. Dr. Anupam Roy 69  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings 𝑅 2𝑅 2𝑅-𝑡 Upper 𝑅 surface of 𝑅-𝑡 the lens 𝑡 𝑡 𝑟𝑚 𝑚th bright fringe ❑ Using the Pythagoras theorem: 𝑅2 = 𝑟𝑚 2 + (𝑅 − 𝑡)2. ❑ Hence, 𝑟𝑚 2 = 𝑅2 − 𝑅 − 𝑡 2 = 𝑅2 − 𝑅2 + 𝑡 2 − 2𝑅𝑡 = 2𝑅𝑡 − 𝑡 2 = 𝑡 2𝑅 − 𝑡 ≃ 2𝑅𝑡. (since 2𝑅 >> 𝑡) Dr. Anupam Roy 70  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings 𝑅 2𝑅 2𝑅-𝑡 Upper 𝑅 surface of 𝑅-𝑡 the lens 𝑡 𝑡 𝑟𝑚 𝑚th bright fringe ❑ Using the Pythagoras theorem: 𝑅2 = 𝑟𝑚 2 + (𝑅 − 𝑡)2. ⇒ 𝑟𝑚 2 = 2𝑅𝑡. 𝑚𝜆 ❑ For the dark fringe: 2𝑛𝑡 = 𝑚𝜆 ⇒ 2𝑡 = 𝑛 𝑚𝜆𝑅 ❑ Hence, 𝑟𝑚 2 = 𝑅. 2𝑡 = ⇒ 𝑟𝑚 = 𝑚𝜆𝑅 𝑛 𝑛 Dr. Anupam Roy 71  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings 𝑅 2𝑅 2𝑅-𝑡 Upper 𝑅 surface of 𝑅-𝑡 the lens 𝑡 𝑡 𝑟𝑚 𝑚th bright fringe 𝑚𝜆𝑅 ❑ For the dark fringe: 𝑟𝑚 = 𝑛 1 𝑚+2 𝜆𝑅 ❑ Similarly, for the bright fringe: 𝑟𝑚 = 𝑛 ❑ Note: for the air film, 𝑛 = 1. Dr. Anupam Roy 72  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings ❑ Radii of the rings vary as the square-root of odd natural numbers (𝑛 = 0,1,2,3 …..). ❑ The rings will get closer and closer as their order increases. ❑ Radii of Newton’s rings depend on the wavelength of light used. ❑ Wavelength of light can be determined by measuring the radius (diameter) of ring and radius of curvature (𝑅) of lens. Dr. Anupam Roy 73  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Newton’s Rings ❑Homework ❑In terms of the diameter of the film: 𝐷𝑚 = 2𝑟𝑚 ❑Calculate the diameter of the 𝑚th bright ring, 𝐷2 𝑚 ❑Calculate the diameter of the (𝑚 + 𝑝)th bright ring, 𝐷2 𝑚+𝑝 𝐷2 𝑚+𝑝 − 𝐷 2 𝑚 ❑Subtracting these two one can express the wavelength in terms of diameters: 𝜆 = 4𝑝𝑅 Dr. Anupam Roy 74  PH113: Physics (B.Tech. I) Module 1 Tutorial Dr. Anupam Roy 75  PH113: Physics (B.Tech. I) Module 1 Next Class Questions? Dr. Anupam Roy 76 PH113: Physics (B.Tech. I) Module 1 Date: 15.09.2023 Lecture: 5 Dr. Anupam Roy 77  PH113: Physics (B.Tech. I) Module 1 Module – 1 : Physical Optics Polarization, Malus’ Law, Brewster’s Law, Double Refraction, Interference in thin films (Parallel films), Interference in wedge-shaped layers, Newton’s rings, Fraunhofer diffraction by single slit, Double slit. Elementary ideas of fibre optics and application of fibre optic cables. Text book: T1: A. Ghatak, Optics, 6th Edition, 2017, McGraw-Hill Education (India) Pvt. Ltd. References: (1) Optics – E Hecht (2) Fundamentals of Optics – Jenkins & White Class structure: 4 Lectures including 1 Tutorial per week. (8 hours ~ 2 weeks for this module!)​ Dr. Anupam Roy 78  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction ❑Consider a plane wave incident on a narrow slit of width b. ❑What should we expect according to the geometrical optics? ❑That only the region AB of the screen (SS’) to be bright and rest of the screen will be dark. ❑Now, if the slit-width is comparable to the wavelength, then the intensity in the AB region is not uniform, and there is also some intensity inside the geometrical shadow. ❑Smaller the slit-width, larger is the amount of energy reaching the geometrical shadow. ❑This spreading out of a wave when it passes through a narrow opening is usually referred to as diffraction and the intensity distribution on the screen is known as the diffraction pattern. ❑This spreading out decreases with decrease in wavelength. Dr. Anupam Roy 79  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction ❑This is the phenomenon of bending of light as it passes the edge of an object ❑Most prominent when aperture ≲ 𝜆. Diffraction always occurs but it’s effects are noticeable only when the wavelength is comparable to the size of the diffracting object. ❑The phenomena is caused by the interference of a large number of waves. Dr. Anupam Roy 80  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Interference vs Diffraction ❑Interference is due to superposition of two separate wavefronts originating from the same coherent source. ❑Diffraction is due to superposition of secondary wavelets originating from a wavefront. ❑The width of interference fringes may or may not be same. ❑Bands observed in diffraction are never of same width. ❑In interference, all bright fringes bears same intensity. ❑In diffraction, intensity of bright bands usually decreases with increase in order. ❑Intensity distribution of an interference fringe is uniform. ❑Diffraction bands does not have uniform intensity distribution. Dr. Anupam Roy 81  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction ❑Types of diffraction (based on the distance of source and observation screen from the diffracting aperture or obstacle): ❑Fresnel diffraction (either the source and or the screen, or both, are at finite distance from the aperture) ❑Fraunhofer diffraction (source and screen are at infinite distance from the aperture) Dr. Anupam Roy 82  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction ❑In Fresnel diffraction ▪ Source of light or the screen or both are, in general, at a finite distance from the diffracting obstacles. ▪ Superposing wavefronts are not plane, but either cylindrical or spherical. ❑In Fraunhofer diffraction ▪ Source and screen are at infinite distance from the aperture. This condition is achieved by putting suitable lenses between the source, the obstacles and the Screen. ▪ It is based on the concepts of secondary wavelets given by Huygens and their superposition. ▪ Superposing wavefronts are plane wavefronts. Dr. Anupam Roy 83  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑Recall Huygen’s Principle: Each point on a wave front can be thought of as a new source of wavelets, and the development of the new wave front at a later time is determined by the propagation of these wavelets. ❑Wavefront is a surface on which all waves have the same phase. Dr. Anupam Roy 84  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) P P0 S L1 L2 Slit Observation screen ❑A point source (S) of monochromatic light is placed in the focal plane of a converging lens L1, so that a plane wave is incident on a long narrow slit. ❑A converging lens L2 is placed on the other side of the slit. ❑The observation screen is placed at the second focal point of this lens ( L2 ). Dr. Anupam Roy 85  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑Diffraction pattern consists of a horizontal streak of light along a line perpendicular to the length of the slit (see Figure a). ❑Horizontal pattern is a series of bright spots. The intensity of the central spot is maximum and its peak is located at P0, which lies at the intersection of the axis of L1 and L2 with the observation screen is the brightest (see Figure b). ❑On the either side of this spot, few more bright spots symmetrically situated with respect to P0 are observed. Dr. Anupam Roy 86  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑ Intensities of other spots, on either side of the central spot, decrease rapidly as we move away from P0. ❑ The central maximum is called principal maxima and other maxima are called secondary maxima. The width of the central spot is twice the width of other spots. ❑ The central peak is symmetrical. But on either side of the central maximum, secondary maxima are asymmetrical. In fact, the positions of the secondary maxima are slightly shifted towards the observation point P0. Dr. Anupam Roy 87  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑ When a plane wavefront falls on the diffracting slit, each point of the aperture such as A1, A2, A3,...... becomes a source of secondary wavelets, which propagate in the direction of the point 𝑃𝜃. 𝑑 ❑ Optical paths are same for rays passing through a lens and getting focused to a single point. ❑ Let A1A2 = A2A3 =……..= ∆ ❑ Path difference between two consecutive rays = ∆ 𝑠𝑖𝑛𝜃. 2𝜋 ❑ So, the phase difference must be 𝜑 = Δ 𝑠𝑖𝑛𝜃. Dr. Anupam Roy 88 𝜆  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑ Suppose the 1st ray (starting at A1) be 𝑎 cos𝜔𝑡. ❑ Then the 2nd ray (starting at A2) is 𝑎 cos(𝜔𝑡 − 𝜑). ❑ Similarly, the other rays are 𝑎 cos(𝜔𝑡 − 2𝜑), 𝑎 cos(𝜔𝑡 − 3𝜑), …….etc. 𝑑 ❑ All the waves superpose on a point P on the screen. So the net field at P will be: ❑ 𝐸 = 𝑎 [𝑐𝑜𝑠𝜔𝑡 + cos 𝜔𝑡 − 𝜑 + cos 𝜔𝑡 − 2𝜑 + … … ….. ] ❑ We are considering 𝑛 sources A1, A2, ……, An, with a gap ∆ of between consecutive sources. At the end, we will set 𝑛 → ∞ and ∆→ 0 such that 𝑛∆= 𝑏 and the sources are continuous. Dr. Anupam Roy 89  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) ❑ 𝐸 = 𝑎 [𝑐𝑜𝑠𝜔𝑡 + cos 𝜔𝑡 − 𝜑 + cos 𝜔𝑡 − 2𝜑 + … … ….. ] = 𝑎 𝑅𝑒 [𝑒 𝑖𝜔𝑡 + 𝑒 𝑖 𝜔𝑡−𝜑 + 𝑒𝑖 𝜔𝑡−2𝜑 +⋯] ❑ Use de Moivre’s theorem: 𝑒 𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 ❑ Hence, 𝐸 = 𝑎 𝑅𝑒 [𝑒 𝑖𝜔𝑡 {1 + 𝑒 −𝑖𝜑 + 𝑒 −2𝑖𝜑 + ⋯ … + 𝑒 − 𝑛−1 𝑖𝜑 }] 𝜑 𝜑 𝜑 −𝑖𝑛 𝑖𝑛 −𝑖𝑛 1− 𝑒 −𝑖𝑛𝜑 𝑒 2 𝑒 2 − 𝑒 2 = 𝑎 𝑅𝑒 𝑒 𝑖𝜔𝑡. −𝑖𝜑 = 𝑎 𝑅𝑒 𝑒 𝑖𝜔𝑡. 𝜑 𝑖𝜑 𝑖𝜑 1−𝑒 −𝑖 − 𝑒 2 𝑒2 − 𝑒 2 ❑ Use de Moivre’s theorem: 𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃 = (cos 𝜃 + 𝑖 sin 𝜃) − cos 𝜃 − 𝑖 sin 𝜃 = 2𝑖 sin 𝜃 ❑ Now, look at the terms within the first bracket: Dr. Anupam Roy 90  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝜑 𝜑 𝜑 𝑛𝜑 −𝑖𝑛 2 𝑖𝑛 2 −𝑖𝑛 2 𝑖 𝑛−1 𝜑 sin 𝑒 𝑒 − 𝑒 − 2 ❑ Hence, 𝐸 = 𝑎 𝑅𝑒 𝑒 𝑖𝜔𝑡. 𝜑 𝑖𝜑 𝑖𝜑 = 𝑎 𝑅𝑒 𝑒 𝑖𝜔𝑡. 𝑒. 𝜑 2 −𝑖 − sin 𝑒 2 𝑒2 − 𝑒 2 2 𝑛−1 𝜑 𝑛𝜑 𝑖 𝜔𝑡− sin 2 = 𝑎 𝑅𝑒 𝑒 2. 𝜑 sin 2 𝑛𝜑 sin 𝑛−1 𝜑 ⇒ 𝐸=𝑎 2 𝜑 cos 𝜔𝑡 − 2 sin 2 𝑛𝜑 𝑛−1 𝜑 sin 2 ⇒ 𝐸 = 𝐸0 cos 𝜔𝑡 − , where 𝐸0 = 𝑎 𝜑 2 sin 2 ❑ 𝐸0 is the amplitude of the resulting field. Dr. Anupam Roy 91  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑛𝜑 𝑛−1 𝜑 sin 2 ❑ 𝐸 = 𝐸0 cos 𝜔𝑡 − , where 𝐸0 = 𝑎 𝜑 2 sin( 2 ) 2𝜋 ❑ But, 𝜑 = Δ 𝑠𝑖𝑛𝜃 𝜆 𝑛𝜑 𝑛 2𝜋 𝜋 𝜋 ❑ Hence, =. Δ 𝑠𝑖𝑛𝜃 = 𝑛Δ sin 𝜃 = 𝑏 sin 𝜃 (using 𝑛∆= 𝑏 when 𝑛 → ∞ and ∆→ 0) 2 2 𝜆 𝜆 𝜆 𝜋 sin 𝜆 𝑏 sin 𝜃 ❑ Using the above we get, where 𝐸0 = 𝑎 𝜋 sin( 𝜆 Δ 𝑠𝑖𝑛𝜃) ❑ Since, Δ → 0 we can rewrite sin 𝑥 ≈ 𝑥 for 𝑥 → 0 𝜋 sin 𝜆 𝑏 sin 𝜃 ❑ Hence, 𝐸0 = 𝑎 𝜋 𝜆 Δ 𝑠𝑖𝑛𝜃 Dr. Anupam Roy 92  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑛𝜑 𝑛−1 𝜑 sin 2 ❑ 𝐸 = 𝐸0 cos 𝜔𝑡 − , where the amplitude of the resulting field, 𝐸0 = 𝑎 𝜑. 2 sin( 2 ) 𝑛𝜑 𝜋 ❑ Using, = 𝑏 sin 𝜃 (𝑛∆= 𝑏 when 𝑛 → ∞ and ∆→ 0) 2 𝜆 𝜋 𝜋 sin 𝜆 𝑏 sin 𝜃 sin 𝜆 𝑏 sin 𝜃 Since, Δ → 0, ❑ Using the above we get, where 𝐸0 = 𝑎 𝜋 ≈ 𝑎 𝜋 sin 𝑥 ≈ 𝑥 for 𝑥 → 0 sin( 𝜆 Δ 𝑠𝑖𝑛𝜃) 𝜆 Δ 𝑠𝑖𝑛𝜃 𝜋 𝜋 sin 𝜆 𝑏 sin 𝜃 𝑛 sin 𝜆 𝑏 sin 𝜃 ❑ Or, 𝐸0 = 𝑎 𝜋 × = 𝑛𝑎 𝜋 (Why? Because 𝑛∆= 𝑏) Δ 𝑠𝑖𝑛𝜃 𝑛 𝑏 𝑠𝑖𝑛𝜃 𝜆 𝜆 sin 𝛽 𝜋 ❑ This can be expressed as: 𝐸0 = 𝐴 where 𝑛𝑎 = 𝐴 and 𝛽 = 𝑏 𝑠𝑖𝑛 𝜃 (in the limit 𝑛 → ∞ 𝛽 𝜆 and 𝑎 → 0 the product 𝑛𝑎 tends to a finite limit) Dr. Anupam Roy 93  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑛−1 𝜑 𝑛𝜑 ❑ Now, 𝐸 = 𝐸0 cos 𝜔𝑡 − ≃ 𝐸0 cos[𝜔𝑡 − ] (∵ for 𝑛 → ∞, 𝑛 − 1 = 𝑛) 2 2 sin 𝛽 ❑ Hence, 𝐸 = 𝐴 cos[𝜔𝑡 − 𝛽] 𝛽 ❑ Intensity pattern on the screen: (Intensity is given as square of amplitude) 2 sin 𝛽 2 𝑠𝑖𝑛2 𝛽 ❑ 𝐼 = 𝐸0 = (𝐴 ) = 𝐼0 2 (where 𝐼0 = 𝐴2 ) 𝛽 𝛽 Dr. Anupam Roy 94  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 2 𝑠𝑖𝑛𝛽 ❑ 𝛽 = 0 will give the central maximum where 𝐼 = 𝐼0 = 𝐴 (∵ lim = 1) 𝛽→0 𝛽 ❑ This implies that all diffracted waves arrive in phase at the point P0 (at 𝜃 = 0) and interfere constructively. 𝐴2 is then the value of the maximum intensity at the centre of the diffraction pattern. 𝑑 Dr. Anupam Roy 95  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Position of secondary maxima and minima: 𝑠𝑖𝑛𝛽 ❑ 𝛽 = 0 will give the central maximum where 𝐼 = 𝐼0 = 𝐴2 (∵ lim = 1) 𝛽→0 𝛽 ❑ The intensity is maximum for 𝜃 = 0 and 𝐼𝜃=0 = 𝐼0 = 𝐴2 ❑ The intensity gradually falls on either side of the principal maximum and becomes zero when 𝛽 = ± 𝜋. ❑ The angular half width of principal maximum is from 𝛽 = 0 to 𝛽 = 𝜋. ❑ The second minimum on either side occurs at 𝛽 = ± 2𝜋. Dr. Anupam Roy 96  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Position of secondary maxima and minima: ❑ 𝛽 = 0 will give the central maximum where 𝐼 = 𝐼0 = 𝐴2 ❑ The intensity is maximum for 𝜃 = 0 and 𝐼𝜃=0 = 𝐼0 = 𝐴2 ❑ General condition for minima: 𝛽 = 𝑚𝜋 where 𝑚 = ±1, ±2, ±3, ….. 𝜋 Or, 𝑏 𝑠𝑖𝑛 𝜃 = 𝑚𝜆 where 𝑚 = ±1, ±2, ±3, ….. (since 𝛽 = 𝑏 sin 𝜃) 𝜆 ❑ This is the condition for minima. The location of minima, determined by 𝜃 depends on the wavelength of light (𝜆) and the slit width (𝑏). Dr. Anupam Roy 97  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Position of secondary maxima and minima: ❑ 𝛽 = 0 will give the central maximum (for 𝜃 = 0) and 𝐼𝜃=0 = 𝐼0 = 𝐴2 ❑ General condition for minima: 𝑏 𝑠𝑖𝑛 𝜃 = 𝑚𝜆 where 𝑚 = ±1, ±2, ±3, ….. ❑ Location of minima, determined by 𝜃 depends on wavelength of light (𝜆) and the slit width (𝑏). 𝜆 ❑ The first minimum occurs at 𝜃 = ± sin−1. The second minimum occurs at 𝜃 = 𝑏 2𝜆 ± sin−1 and so on ……. 𝑏 𝑏 ❑ Note: sin 𝜃 ≯ 1, so the maximum value of 𝑚 is the integer which is less than (and closest to). 𝜆 Dr. Anupam Roy 98  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Direction and position of secondary maxima: 𝑠𝑖𝑛2 𝛽 ❑ To determine the positions of maxima, differentiate the equation 𝐼 = 𝐼0 2 with respect to 𝛽 𝛽 and set it equal to zero. 𝑑𝐼 ❑ That means: =0 𝑑𝛽 𝑑𝐼 𝑑 𝑠𝑖𝑛2 𝛽 2 sin 𝛽 cos 𝛽 2𝑠𝑖𝑛2 𝛽 ❑ = 𝐼0 2 = 𝐼0 − = sin 𝛽 (𝛽 − tan 𝛽) 𝑑𝛽 𝑑𝛽 𝛽 𝛽2 𝛽3 𝑑𝐼 ❑ We will set = 0. That means, sin 𝛽 (𝛽 − tan 𝛽) = 0 𝑑𝛽 Dr. Anupam Roy 99  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Direction and position of secondary maxima: 𝑑𝐼 ❑ =0 ⇒ sin 𝛽 (𝛽 − tan 𝛽) = 0 𝑑𝛽 ❑ The condition sin 𝛽 = 0, or, 𝛽 = 𝑚𝜋 (𝑚 ≠ 0) correspond to minima. ❑ The transcendental equation tan 𝛽 = 𝛽 gives the condition for maxima. ❑ 𝛽 = 0 will give the central maximum. The other roots can be found by determining the points of intersections of the curves 𝑦 = 𝛽 and 𝑦 = tan 𝛽. The intersections occur at 𝛽 = 1.43𝜋, 𝛽 = 2.46𝜋, etc., and are known as the first maximum, the second maximum, etc. Dr. Anupam Roy 100  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Diffraction from a single slit (Point source) 𝑠𝑖𝑛2 𝛽 ❑ Maxima and Minima of the intensity: 𝐼 = 𝐼0 2 𝛽 ❑ Direction and position of secondary maxima: ❑ Note: The intensity maxima do not fall exactly midway between two minima. For example, the first secondary maximum occurs at 𝛽 = 1.43𝜋 (rather than 1.50𝜋). Similarly, the second maximum occurs at 𝛽 = 2.46𝜋 (rather than 2.50𝜋) and so on. ❑ This means that the intensity curves for secondary maxima are asymmetrical and the positions of maxima are slightly shifted towards the centre of the pattern. 2 𝑠𝑖𝑛𝛽 2 sin(1.43𝜋) ❑ Intensity of second maxima is 𝐼 = 𝐼0 = 𝐼0 = 𝐼0 (0.0496) 𝛽 (1.43𝜋) ❑ Intensity of the first secondary maxima (nearest to central peak) is ~ 4.96% of the central maximum. Dr. Anupam Roy 101 PH113: Physics (B.Tech. I) Module 1 Date: 19.09.2023 Lecture: 6 Dr. Anupam Roy 102  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Two-Slit interference (Young’s double slit experiment) ❑ Optical path difference between the two rays reaching at point P is 𝑑 sin 𝜃. 2𝜋 ❑ Phase difference, φ = 𝑑 𝑠𝑖𝑛𝜃. 𝜆 ❑ Total amplitude at P: Arrangement for producing Young’s interference pattern 𝜑 𝜑 𝐸𝑝 = 𝑎 cos 𝜔𝑡 + 𝑎 cos 𝜔𝑡 − 𝜑 = 2𝑎 cos 𝜔𝑡 − cos 2 2 𝜑 𝜋 ❑ Let, = 𝑑 𝑠𝑖𝑛𝜃 ≡ 𝛾 2 𝜆 ❑ Then, 𝐸𝑝 = 2𝑎 cos 𝜔𝑡 − 𝛾 cos 𝛾 = (2𝑎 cos 𝛾) cos 𝜔𝑡 − 𝛾 Dr. Anupam Roy 103  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Two-Slit interference (Young’s double slit experiment) ❑ Intensity distribution for two-slit interference: 𝐼 = |𝐸𝑝0 |2 = (2𝑎 cos 𝛾) 2 ⇒ 𝐼 = 4 𝐼0 cos 2 𝛾 (where 𝐼0 = 𝑎2 ) 𝑠𝑖𝑛2 𝛽 ❑ Recall the intensity expression for single-slit diffraction: 𝐼 = 𝐼0 2 𝛽 sin2 𝛽 ❑ Next topic: Intensity expression for two-slit diffraction: 𝐼 = 4 𝐼0 cos 2 𝛾 𝛽2 ❑ Note: For the two-slit diffraction, the intensity distribution is a combination of intensity patterns of single-slit diffraction and two-slit interference. Dr. Anupam Roy 104  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Two-Slit Fraunhofer Diffraction ✓ Fraunhofer diffraction pattern produced by a single slit of width 𝑏 and found that the intensity distribution consists of maxima and minima. ❑ Now, we will study the Fraunhofer diffraction pattern produced by two parallel slits (each of width 𝑏) separated by a distance 𝑑. ❑ In this case, the resultant intensity distribution is a product of the single-slit diffraction pattern and the interference pattern produced by two point sources separated by a distance 𝑑. Dr. Anupam Roy 105  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Two-Slit Fraunhofer Diffraction ❑ Assumption: two slits are identical in all respect. ❑ Source slit and the lens are symmetrically placed relative to two slits. ❑ We assume that the slits consist of a large number of equally spaced point sources and that each point on the slit is a source of Huygens’ secondary wavelets. ❑ Let the point sources be at A1, A2, A3, ……(in the 1st slit) and B1, B2, B3, …… (in the 2nd slit). ❑ We assume that the distance between two consecutive points in either of the slits is ∆. Dr. Anupam Roy 106  PH113: Physics (B.Tech. I) Module 1 (Physical Optics) Diffraction: Two-Slit Fraunhofer Diffraction ❑ Let’s consider that the diffracted rays make an angle 𝜃 with the normal to the plane of the slits. ❑ The path difference between two consecutive rays reaching the point P will be ∆ sin 𝜃. ❑ The field produced by the 1st slit at the point P will, therefore, be given by sin 𝛽 Where, 𝛽 = 𝜋 𝑏 𝑠𝑖𝑛𝜃 (𝜃 is the diffraction angle). 𝐸1 = 𝐴 cos(𝜔𝑡 − 𝛽) 𝜆 𝛽 Phase difference between the rays coming from ❑ Similarly, the 2nd slit will produce a field at P: pair of corresponding points on the slits (e.g., A1 sin 𝛽 2𝜋 𝐸2 = 𝐴 cos(𝜔𝑡 − 𝛽 − Φ) and B1, A2 and B2, etc.) is: Φ = 𝜆