Concepts of Engineering Physics 14 11 2017-1.pdf

Concepts of ENGINEERING PHYSICS NARENDRA MATHAKARI 10 All of physics is either impossible or trivial. It is impossible until you understand it, and then it becomes trivial. Ernest Rutherford...

Concepts of ENGINEERING PHYSICS NARENDRA MATHAKARI 10 All of physics is either impossible or trivial. It is impossible until you understand it, and then it becomes trivial. Ernest Rutherford (Nobel Laureate, 1908) 11 About the cover page The cover page shows the photographs of two huge facilities constructed by Physicists for detection of Higg’s Boson and the Gravitational waves. Both the facilities are billion dollars projects. As such, year 2012 and 2015 have been unique as regards to Physics. In 1912, Physicists discovered the theoretically proposed Higgs Boson in world’s largest accelerator, the Large Hadron Collider (LHC) sitting in a circular tunnel 27 km in circumference! LHC straddles the Swiss and French borders on the outskirts of Geneva. The LHC is designed to collide two counter rotating beams of protons or heavy ions, with a tremendous energy of 7TeV. The beams move around the LHC ring inside a continuous vacuum guided by superconducting magnets maintained by a huge cryogenic system. To pinpoint the smallest fragments of the universe, Physicists have to build the biggest machine in the world! The second photograph shows LIGO (Laser InterferometerGravitational Wave Observatory) another huge facility at Caltech where Gravitational waves, the weakest waves in the Universe have been successfully detected. The length of each of the arms in LIGO is 4 km and consequently, the sensitivity of the huge LIGO is 1 part in 1021! LIGO is basically an advanced Michelson’s interferometer, whose arms are considerably long..4 kms 12 Preface Physics: the word derives itself from the Greek language. Usually translated into English as “nature”, this ‘natural philosophy’, started dominating Science and Technology from its beginning itself. It seems that, all the disciplines in science and technology are linked with Physics. A few examples, which support this argument, are, Chemical Physics, Physical Chemistry, Geophysics, Biophysics, Astrophysics, Mathematical Physics and off course Engineering Physics. While the contributions of Newton, Gibbs, Carnot, Bernoulli, Young, Maxwell, Hertz, Gauss, Kirchhoff, Tesla, Joule, Ampere, Archimedes, Edison, and Faraday in the classical era remain unforgettable, the contributions of A few Nobel Laureates in the twentieth and the current century are also momentous. The first Nobel Prize bestowed on Roentgen in 1901 and the recent one on Andre Gim, and Nokolosov, for the discovery Graphene, are just a few examples. In short, the contributions of the Physics in technology leave no doubt for an enlightened. This books aims at providing the elements of Physics which are required to engineers. Dr. Narendra Mathakari, [email protected] [email protected] 13  Before beginning, you may learn a few definitions Science Systemized knowledge derived through experimentation, observation, and study. Also, the methodology to acquire the knowledge Technology Definition 1: The application of scientific advances to benefit the humanity Definition 2: The practical application of science to commerce or industry Engineering Definition 1: The design, construction and operation of structures and machines, using scientific principles. Definition 2: Science, discipline, art and profession of acquiring and applying technical, scientific and mathematical knowledge to design and implement materials, structures, machines, devices, systems, and processes that safely realized desired objective or inventions. Definition 3: The profession in which a knowledge of the mathematical and natural sciences gained by study, experience, and practice is applied with judgment to develop ways to use economically the materials and forces of nature for the benefit of mankind 14 Contents CHAPTER 1 Interference A basis of high precision instruments called Interferometers CHAPTER 2 Diffraction Diffraction plays a major role in high resolution optical instruments CHAPTER 3 Polarization Restricting the vibrations of light leads to applications CHAPTER 4 Sound Engineering-I (Architectural Acoustics) How to design enclosures with superior acoustics Sound Engineering-II (Ultrasound) The ‘soundless sound’ finds several applications in the technology CHAPTER 5 Lasers A coherent and amplified light; basis of photonics 15 CHAPTER 6 Semiconductor Physics Fermi level determines the behavior of semiconductors and devices CHAPTER 7 Basics of Quantum Mechanics (Wave particle duality) Energy and matter are quantized CHAPTER 8 A few applications of Quantum Mechanics (Wave-function& wave- equation) Probing into the Physics of atoms, molecules and photons CHAPTER 9 Superconductivity Zero electrical resistance, but not yet at room temperature CHAPTER 10 Elements of Nanotechnology Tiny aggregates of atoms show new horizons to the technology 16 Albert Einstein (1879-955):Underestimated at school level, and then working as a clerk in the Swiss patent office, he at the age of 26, published his first three seminal papers, interpreting Brownian motion, relativity and the concept of quanta initiated by Max Planck. His contributions are too many to accommodate here, a few of which are E = mc2, an acceptable interpretation of gravity as warping of space-time, concept of stimulated emission etc. He won Nobel prize in Physics for the quantum explanation of the Photoelectric effect; however, the mastermind deserved more than one. He correctly interpreted Gravity in terms of warping of the space-time. One of his endeavors, the unification of the four fundamental forces in the Universe was only partially fulfilled.In the Engineering Physics course, a thorough discussion of Quantum Physics will follow, encompassing the applications such as electronic configuration of the atoms, LASER, TRANSISTOR, Scanning Tunneling Microscope, Josephson junctions, SQUIDS etc. But, it is interesting to note that though he was the originator of Quantum Physics, never accepted it until his death, Probably only once, the Nature deceived the Genius. In 1917, he declared, “for rest of the life I would reflect on what light is”. 17  Optics For the rest of my life I will reflect on what light is Albert Einstein (1917) The photograph on the left shows one of the most versatile interferometer, the Michelson interferometer. There are several interferometers which form the basis of interferometry. The photograph on the right shows optical fibers which are essential components of modern communication systems. These are two out of several applications of Photonics, a rapidly emerging branch of Physics. The exact nature is of light is yet to be understood. Thus, conventionally Optics, which describes light and its applications, divides itself in to three branches, called Ray- Optics (reflection, total internal reflection, refraction, double refraction etc.), Wave-Optics (interference, diffraction and polarization) and Quantum-Optics (photoelectric effect, optoelectronics, lasers etc). Optics finds a few notable applications in day-to-day life as well as technology. A few of these are such as Optical Engineering, i.e. design and fabrication of several optical devices and instruments such as lenses, mirrors, gratings, prisms, interferometers, diffractometers, polarimeters, cameras, video cameras, movie projectors etc., Further, the disciplines appearing on the frontiers of current Physics, such as Photonics also require a thorough understanding of Optics. 18  CHAPTER 1 Interference Interestingly, the oily films spread on the roads in rainy days, or the soup bubbles show beautiful colors. The pattern changes when viewed at different angles. This is due to thin film interference, where the thin film behaves like a ‘natural interferometer’. An extension of this concept gives birth to several interferometers, used for high precision measurements. What is an interferometer and what are its uses? The answer to this question is in this chapter 19  Index 1.1 INTRODUCTION How to produce a steady state and sharp interference pattern 1.2 THIN FILM INTERFERENCE How thin film behaves as a natural interferometer 1.3 NEWTON’S RINGS (NEWTON’S INTERFEROMETER) Circular fringes due to circular symmetry 1.4 MICHELSON’S INTERFEROMETER (Optional) Circular fringes, but where is the circular symmetry? 1.5 INTERFERENCE COATINGS How to minimize the unwanted reflection or transmission of light 20  Thomas Young (1773-1829): ThePhysicist with versatile intelligence was the first to demonstrate interference. He learned to read at the age of 2 and said to have read the complete Bible, twice, at the age of 6. He was a Physicist of high caliber as well as a Physician. He made significant contributions in Physics as well as Physiology, a few of which include understanding of the Physiology of Human eye, Physics of color vision, concept of modulus of elasticity known due to his name etc. He also gave an experimental foundation to the wave theory of light, which was based on his double slit experiment. What follows is a rapid revision of his experiment, necessary before going ahead. 1.1 INTRODUCTION How to produce steady state and sharp interference pattern The first demonstration of the interference occurred in 1801, due to Thomas Young, who, with the help of his famous double slit experiment, produced a well-defined interference pattern. He also proposed a theory of interference. His explanation was based on an assumption that light is a wave. Indeed all the phenomena related to interference conclusively prove that light is a wave. In day to day life interference occurs in many situations. An oil film spread on the road in rainy days appears colored. This is due to interference. Interference is also responsible for the colored appearance of soap bubbles when illuminated by white light. Interference also has some interesting practical applications. It is the basis of high precision interferometers which are used for various measurements. Antireflection and anti-transmission coatings and interference filters are also based on interference. Interference is also used to inspect the preciseness of the optical devices such as glass plates, lenses etc. Interference is the modification in the intensity when two or more waves are superimposed. It occurs only at a point where the waves overlap. As the waves surpass the point, they travel independently and unaffected. The principle of superposition is given by 𝑦 = 𝑦1 ± 𝑦2 …(1.1) Wherey1and y2are the instantaneous displacements of first and second wave at the point of superposition.y is the resultant displacement. The ± sign indicates that constructive as well as destructive interference are possible. Let 𝑦1 = 𝑎1 𝑠𝑖𝑛𝑤𝑡 and 𝑦2 = 𝑎2 𝑠𝑖𝑛(𝑤𝑡 + 𝜙) Where a1and a2are the amplitudes, 𝑤 is the frequency and 𝜙 is the phase difference between the first and second wave. 21 By using the principle of superposition, it can be shown that 𝐼 = 𝑅2 = 𝑎12 + 𝑎22 + 𝑎1 𝑎2 𝑐𝑜𝑠𝜙 …(1.2) Where R is the resultant amplitude and I is the resultant intensity. Eq. (1.2) can be analyzed with two special cases If the phase difference𝜙 = 0, ± 2𝜋, ± 4𝜋, ± 6𝜋, … … ± 2𝑛𝜋, then 𝑐𝑜𝑠𝜙 = 1. Then we have 𝐼 = 𝑅2 = 𝑎12 + 𝑎22 + 2𝑎1 𝑎2 cos(2𝑛𝜋) …(1.3) ⇒ 𝐼 = (𝑎1 + 𝑎2 )2 > 𝑎12 + 𝑎22 ⇒ 𝐼 > 𝐼1 + 𝐼2 𝑖𝑓 𝑎1 = 𝑎2 = 𝑎 𝑡ℎ𝑒𝑛 𝐼 = 4𝑎2 This indicates that if the phase difference between two waves is even multiple of𝜋, then the resultant intensity is greater than the sum of the individual intensities. Thus intensity is enhanced. Such interference is called as constructive interference. The fringe with enhanced intensity is called maxima. 𝜆 We know that the phase 𝜋 corresponds to path 2. Thus the condition of constructive interference in terms of 𝜆 is 𝜆 Path difference for constructive interference = 2𝑛 2 = 𝑛𝜆 If the phase difference 𝜙 = 𝜋, 3𝜋, 5𝜋 ….. (2𝑛 ± 1)𝜋 then𝑐𝑜𝑠𝜙 = −1. Then we have 𝐼 = 𝑅2 = 𝑎12 + 𝑎22 + 2𝑎1 𝑎2 cos(2𝑛 ± 1)𝜋 ⇒ 𝐼 = 𝑎12 + 𝑎22 − 2𝑎1 𝑎2 ⇒ 𝐼 = (𝑎1 − 𝑎2 )2 < 𝑎12 + 𝑎22 ⇒ 𝐼 < 𝐼1 + 𝐼2 If 𝑎1 = 𝑎2 = 𝑎, then 𝐼 = 0 Thus if the phase difference between the two waves is (2𝑛 ± 1)𝜋 , then the resultant intensity is less than the sum of individual intensities. If the amplitudes of the two waves are equal then the resultant intensity is zero. Thus the intensity is decreased. Such interference is called as destructive interference. The fringe with decreased intensity is called as minimum. The condition of the destructive interference in terms of 𝜆 is 22  𝜆 Path difference for destructive interference = (2𝑛 ± 1) 2 Figure (1.1) shows the interference pattern with maxima and minima Figure (1.1): A typical interference pattern Apart from the conditions of maxima and minima, other path differences are also possible. Thus depending upon the path difference, the resultant intensity may take any value from 0 to 4a 2. It is to be noted that, in interference, the energy is neither created nor destroyed. At maximum there is an enhancement of intensity. But this is not the created energy. In minima there is decreased intensity. But this is not the destroyed energy. Indeed, in interference, there occurs a redistribution of the energy. The intensity which was supposed to be at minimum is transferred to maximum. Thus interference can also be defined as the redistribution of energy due to superposition of the waves. Fig 1.1 shows a well-defined interference pattern consisting of maxima and minima. For obtaining a well-defined, sharp and steady state interference pattern, some conditions have to be satisfied. These are i. The waves must be coherent. This means that the phase difference between the interfering waves at any point should remain constant with respect to time. If there is no coherence, the phase difference and hence the intensity will change with respect to time. The pattern will not be a steady state pattern. Coherence is the prime condition to be satisfied for getting a steady state interference pattern ii. For having a sharp, distinct and well defined interference pattern, the waves must be monochromatic and they should have same amplitude. If these conditions are not satisfied then the intensity at maxima will not be maximum and the intensity at the minima will not be zero. Thus a distinct and sharp interference pattern cannot be obtained 23  iii. There must be a path difference. iv. There should be a systematic and gradual variation of the path difference. For having zeroth, first, second, third maxima etc., the path difference should gradually increase from 0, 𝜆, 2𝜆, 3𝜆…. Similarly, for having the first, second, third minima at 𝜆 3𝜆 appropriate places, the path difference at these points should take values from 2, 2 , 5𝜆 etc. Thus for having alternate maxima and minima, there needs to be a gradual and 2 systematic variation of the path difference. v. The path difference should vary only with the position and not with time. Two separate and independent ordinary sources of light are always incoherent. If a screen is illuminated by such sources, it does not show a steady state interference pattern, as the phase difference between the waves at any point keeps on changing rapidly and randomly. Thus intensity changes several times in a second. Therefore the screen appears to be uniformly illuminated. The prime condition for interference is coherence. Two independent sources of light can be monochromatic but not coherent (laser is an exception). Coherence is possible when the two sources are derived from the same source. This can be done by using two techniques, namely division of wave-front and division of amplitude. Fig.1.2.shows the technique of division of wavefront. A point monochromatic source of light emits primary (spherical) wavefront. Two point like slits become two sources of light, which emit two secondary (spherical) wavefronts. Point sources (spherical wavefronts) are employed in this technique. Both these wavefronts are coherent as they are derived from the same source. The technique of division of the wavefront is used in Young’s double slit experiment Figure 1.2Division of wavefront Fig.(1.3) depicts the technique of division of the amplitude (intensity). A semi-silvered glass plate is inclined at 45o. A ray of light having amplitude A is incident on this glass plate. As the glass plate is semi-silvered, part of this ray is transmitted and part is reflected. The intensities of 𝐼 both these rays are approximately equal to 2. This technique requires extended source (parallel wavefront). The division of amplitude technique is used in thin film interference, Newton’s 24 ringsexperiment and Michelson’s interferometer. Apart from these two techniques, refraction and reflection are also used to obtain coherent sources. Fresnel’s biprism and Lloyds mirror are based on these principles. In this chapter we will learn the interference based on thin films and its applications Fig 1.3 Division of amplitude If all the conditions required for generating steady state and sharp interference pattern are satisfied then a well-defined and steady state pattern can be produced. The arrangement which takes care of all these conditions and which generates a steady state and sharp interference pattern is called as interferometer. There are varieties of interferometers used for various applications. Indeed the Young’s double slit apparatus can be called as ‘Young’s interferometer’. In the subsequent sections, we will see that thin film behaves as a natural interferometer. Newton’s rings apparatus may be treated as Newton’s interferometer. The other examples are Michelson’s interferometer, Fabry Perot interferometer etc. The Newton’s interferometer and Michelson’s interferometer are based on the thin film interference. 1.2 THIN FILM INTERFERENCE(Compulsory but derivation is optional) How thin film behaves as a natural interferometer Thin film is a transparent film having thickness slightly above the average wavelength of light. The average wavelength of visible light is roughly 5500 A o (0.55 µm). Thus in optics, a film having a thickness in mm is also considered as a thick film. There are several examples of thin films in the real world. These are oil films on the roads in rainy days, soap bubbles etc. Thin films are encountered in technology also. The antireflection coatings on camera lenses and solar cells, anti-transmission coatings on invisible glasses, interference filters etc are based on thin film interference. When exposed to light, the thin film produces an interference pattern. The thin film interference was first observed by Newton and Robert Hook. It was Thomas who gave the correct explanation to the phenomenon. 25  Fig 1.4 shows the ray diagram of the thin film interference. A thin film having uniform thickness t, refractive index  is exposed to a monochromatic ray of light from an extended source having wavelength λ. The media above and below the film have refractive indices lower than that of film. (Freely suspended soup bubble in air is an example of this case). This is one out of a few more cases which will be discussed later. Figure 1.4 Thin film interference; the diagram is idealized and represents one out of a few more cases A monochromatic ray from an extended source is incident on the film at the point A, at an angle of Iand is reflected ati itself. During refraction, it enters into a denser medium at an angle of r. On the second interface at point B, the ray is incident at r and is reflected at r itself. Such reflections and refractions continue within and outside the film as shown in the Fig 1.. During every reflection and the refraction, the light suffers a loss in the intensity and thus the amplitude of various rays in the reflected and the transmitted system decreases gradually. Thus, here the conditions required for generating sharp fringes are not exactly satisfied. However, ray I and II (and consecutive rays) in the reflected as well as transmitted system are coherent as these are derived from the single ray. (This is amplitude division technique). Ray I and II, though appear parallel, can still interfere if watched through the lens in the eye. Or otherwise, if observed through the microscope, the lens will make them interfere. Now the interference of ray I and II will decide the intensity of the point A. Note that the rays next to the ray II are weak in intensity and therefore can be ignored. Point A will appear bright if I and II interfere constructively and it will appear dark if destructive interference takes place. As we know, the nature of the interference is essentially governed by the path difference between the interfering rays. 26  Ray II, before it arrives at C, travels a distance of AB + BC, while with respect to the line CD, ray I travels a distance of only AD. Additionally, as the distance traveled by the ray II is through the denser medium, it becomes slower and spends more time in the medium depending upon. Thus the geometrical path (𝐴𝐵 + 𝐵𝐶 ) should be converted in to an optical path 𝜇 (𝐴𝐵 + 𝐵𝐶 ). In the other words, ray II suffers a decrease in the wavelength as it travels through the denser medium. The decrease is governed by . Thus the optical path of ray II with respect to the line CD is larger than ray I. (refer appendix I to chapter 1). Thus we get This Derivation is Optional PDR,, I , II   ( AB  BC )  AD...(1.4) For the generalization of the Eq(1.4), we need to express PD in terms of t,  and i (or r). The geometry of the Fig 1.4 assists in doing the task Ideally, t and are same at all the points; thus the symmetry suggests, AB =BC  AB + BC = 2AB From the geometry of the triangle ABE, we have t AB  …(1.5) cos r From the geometry of the triangle ADC, we have AD  AC sin i AD  ( AE  EC ) sin i AD  2 AE sin i From triangle AEB, AE  tan r  t Substituting AE, we get 𝐴𝐷 = 2𝑡𝑎𝑛𝑟 × 𝑡 × 𝑠𝑖𝑛𝑖 Using Snell’s law AD  2 tan r  t   sin r ….(1.6) 27 Substituting AD and AB from eqn 1.5 and 1.6 into 1.4, we have 2t PDR , I , II   ( )  2t  tan r  t   sin r cos r 2𝜇𝑡 2𝜇𝑡 𝑃𝐷𝑅,𝐼,𝐼𝐼 = ( )−( ) 𝑠𝑖𝑛2 𝑟 𝑐𝑜𝑠𝑟 𝑐𝑜𝑠𝑟 2𝜇𝑡 𝑃𝐷𝑅,𝐼,𝐼𝐼 = (1 − 𝑠𝑖𝑛2 𝑟) 𝑐𝑜𝑠𝑟 2𝜇𝑡 𝑃𝐷𝑅,𝐼,𝐼𝐼 = (𝑐𝑜𝑠 2 𝑟) 𝑐𝑜𝑠𝑟 Read from here (Compulsory) PDR,I ,II  2t cos r …(1.7) As the ray I is reflected from the denser medium, according to Stokes law, it’s phase is reversed 𝜆 by = 180𝑜 (𝑡ℎ𝑎𝑡 𝑖𝑠 𝑏𝑦 2). However, this is not the case with ray II, as it is transmitted at A, reflected at the point B , but due to the rarer medium, and then transmitted at C. (According to Stoke’s law, there is no phase reversal during transmission, or reflection, but due to rarer medium. (For a detail account of Stokes law, refer Appendix II to chapter 1). As ray I is  reversed in phase and ray II is not, in this case, there is an additional path difference of  2 between ray I and II. Thus,  PDR ,I ,II  2t cos r  ….(1.8) 2 Eq. (1.8) accounts for the path difference between consecutive reflected rays. Note that the 𝜆 presence or absence of the factor± 2 in such equation essentially depends upon the situation i.e. the relative denseness of film w.r.t. to the media above and below it. We now use the Eqns(1.7) and (1.8) to identify some interesting characteristics of thin film interference. i. The interference patterns of the reflected side and transmitted side of the thin film are always complimentary As t and are assumed to be same throughout, the geometry of the triangles ABC and BCF and triangles ACD and BFG is same. Thus the geometrical path difference between reflected rays I and II and transmitted rays I’ and II’ is same. However, the interference patterns i.e. occurrence 28 of maxima (or minima) from the reflected side and the transmitted side must complement each other, otherwise the optical energy will not be conserved. Thus for transmitted rays we have PDI ' , II i  2t cos r ….(1.9) Note that if Stokes law is carefully applied for the reflected rays and transmitted rays, then if   is present in the equation for the path difference for reflected rays, and then it will be 2 obviously absent for transmitted rays and vice versa. If a point appears bright at a given angle in the reflected system, then it appears dark at the same angle in the transmitted system and vice versa. Note that in the transmitted system, ray I’ does not undergo any phase reversal, as it is just transmitted at point A and B. Ray II’ is reflected at B and C and transmitted at F. As the reflection at B and C is due to rarer medium, there is no phase reversal. Thus both I’ and II’ do  not undergo phase reversal and therefore the term  is absent in the transmitted system. The 2 nature of interference (constructive or destructive) between I and II or I’ and II’ decides the intensity at the point A (bright or dark). Similar discussion is possible for the rays incident at the points, other than A. In the present case (when the medium above and below the film is rarer than the film) for reflected system For constructive interference 𝜆 𝜆 2𝜇𝑡𝑐𝑜𝑠𝑟 ± = 2𝑛 2 2 For destructive interference 𝜆 𝜆 2𝜇𝑡𝑐𝑜𝑠𝑟 ± = (2𝑛 ± 1) 2 2 Note that, depending upon the situation, that is relative denseness of the film with respect to the 𝜆 medium on the top or bottom, the factor ± may or may not be present on LHS 2 ii. An extremely thin or extremely thick film cannot produce interference pattern For, ex. if the film is too thin (𝑡 ≪ 𝜆)then t 0, thus the P. D., irrespective of any point on the  film has same path difference, say  (or  zero). Thus all points appear dark in the reflected 2 system and bright in the transmitted system (or vice versa). However, this is not the interference pattern, as it requires the presence of alternate maxima and minima. In case of the films having their thickness too large as compared to wavelength of light also the interference pattern cannot be observed. Owing to very large value of t, the P.D. at every point becomes be very large as compared to wavelength of any color in the light. Thus, for every color, there exist a few integers 29   for which the P.D. is an odd multiple of and there also exist a few integers for which the P.D. 2  will be an even multiple of. Thus, at every point the constructive as well as destructive 2 interference of any color is possible. Thus interference pattern will not be observed. iii. Fizau’s and Haindinger’s fringes: If we expect the thin film to produce an interference pattern, then it should be able to produce alternate maxima and minima. Thus the path difference should gradually pass through the even  and odd multiples of. As, it can be observed, it is not possible in the case depicted in the Fig 2 1.4. Here every parameter including t,  and r remains the same at all the points. The systematic and gradual variation of path difference requires that either t should vary by keeping r same or vice versa. Both these cases are depicted in the Fig 1.5a and b. Figure 1.5: Fizau’s and Haindinger’s fringes As it can be observed, in the first case,t increases gradually and, as the wavefront is parallel, r remains the same. As the variation of the t occursin horizontal (𝑋)direction, the P.D. and the change in the intensity of the fringe occurs in horizontal (𝑋)direction. However, 30 tremainssame along a direction parallel to edge of the film and perpendicular to the plane of the paper. Thus P.D and consequently the intensity of the fringe remain same along the edge of the film(𝑍 ). These fringes, which are parallel to the edge of the film, equidistant, and in the horizontal plane, are referred as Fizau’s fringes. In Fig 1.5.b, as the source is a point source, it emits a spherical wavefront and thus the rays are incident on the film along various cones. Thus, for each cone,r remains constant over a circle. Thus PD remains constant over a circle. Owing to this circular symmetry, if observed from the top, the fringes will appear concentric and circular. This are referred as Haidenger’s fringes. Note that in Fizau’s fringes t varies andrremains the same,while in Haidenger’s fringes r varies and t remains the same. We can think about a situation, where t as well r is varied, but then interference pattern will be complicated iv. Why oil films on wet roads and soap bubbles appear colored: Eqn. 1.8 also explains why the oil films spread on the road in rainy days or the soap bubbles appear colored. Such films are exposed to the ambient light, which is polychromatic. Further the parameters like t, r and  vary randomly. Thus the P.D. varies from region to region in a random manner. Such P.D. satisfies the conditions of constructive and/or destructive interference for different colors in the different regions in a random manner. Thus some colors are enhanced in some regions and some are suppressed in the other regions. We thus see a random, but beautiful distribution of colors. Colors in the soap film are due to thin film Wings of the Morpho butterfly. The colors are interference. Why there are no colors near due to thin fil interference. The colors change the upper edge? when viewed at different angles. Why? Interestingly, the appearance changes with the angle of viewing (variation of r). It is also needless to mention that the pattern of colors observed in the reflected system will exactly complement the pattern in the transmitted system. 31 ___________________________________________________________________________ Example (1.1): A thin is film spread on a road which is optically denser than the film. Thickness of the film is 0.55 m, while its refractive index is 1.34. Why does the film appear greenish, when viewed in the reflected mode at 68o? The wavelength of the green light is 5500 Ao Solution: As the film is denser than air and as road is denser than the film, the ‘Stoke’s phase reversal  occurs for both the rays, and the factor  disappears form the eqn for P.D.. Thus, the equation 2 for the P.D. for reflected rays is P.DR, I, II = 2t cos r The film will appear greenish, if constructive interference occurs for the green color Thus, 2t cos r  n As the film is thin, considering least possible value of n (=1) 2t cos r   Angle of viewing is same as angle of incidence, thus 2t cos i   2 1.34  5500 cos i  5500 i  68o Thus constructive interference occurs in this case when the angle of viewing is 68 o. The film thus appears greenish, when viewed at 68o. The film, at the same angle but from the opposite side will appear less greenish i.e. more purplish. ___________________________________________________________________________ Example 1.2: A thin film of CCl4having refractive index 1.46 and thickness 0.1068 µm is spread on water having refractive index 1.33. If viewed at 45o which color will be seen enhanced? Solution: Angle of viewing = angle of incidence 32  𝑠𝑖𝑛𝑖 𝜇= 𝑠𝑖𝑛𝑟 𝑠𝑖𝑛45 ⇒ 1.46 = 𝑠𝑖𝑛𝑟 𝑠𝑖𝑛45 ⇒ 𝑠𝑖𝑛𝑟 = 1.46 𝑠𝑖𝑛45 ⇒ 𝑠𝑖𝑛𝑟 = 1.46 𝑠𝑖𝑛45 ⇒ 𝑠𝑖𝑛𝑟 = 1.46 ⇒ 𝑠𝑖𝑛𝑟 = 0.48 ⇒ 𝑟 = 28.97𝑜 We have 𝜆 𝜆 2𝜇𝑡𝑐𝑜𝑠𝑟 ± 2 = 2𝑛 2 (Constructive interference) (Note that as the substrate (water) is rarer than the film, and as the medium above the film is air, 𝜆 the Stoke’s factor ± 2 is present on the LHS) 𝜆 ⇒ 2𝜇𝑡𝑐𝑜𝑠𝑟 = (2𝑛 ± 1) 2 As the film is thin, we will choose the smallest value of n for which the RHS will be nonzero. The n = 0. Subsequently, we will prove that n = 1 is not possible 𝜆 ⇒ 2𝜇𝑡𝑐𝑜𝑠𝑟 = 2 𝜆 ⇒ 2 × 1.46 × 1068 × 𝑐𝑜𝑠28.97 = 2 𝜆 = 5456 𝐴𝑜 ⇒ Green color will be enhanced Let us choose n = 1 33  𝜆 2𝜇𝑡𝑐𝑜𝑠𝑟 = (2𝑛 ± 1) 2 3 ⇒ 2 × 1.46 × 1068 × 𝑐𝑜𝑠28.97 = 𝜆 2 ⇒ 𝜆 = 1818.89 𝐴𝑜 This wavelength is beyond the visible spectrum, therefore it is necessary choose n=0. Here 𝑛≥1 has no significance. __________________________________________________________________________ 1.2 WEDGE SHAPED FILMS(Compulsory, but derivation is optional) Fringe width depends upon the wedge angle Following Ray Diagram is compulsory Refer the ray diagram as shown in Figure 1.6 Figure 1.6 Ray diagram for the Wedge shaped film 34 Fringe width The P.D. between the reflected ray I and II can be shown to be equal to  …(1.10) P.D.W ,R ,I ,II  2t cos(r   )  2 (This formula is derived in Appendix III to chapter 1) Optional For normal incidence 𝑟 → 0. Thus eqn (1.10) simplifies to  …(1.11) P.D.W , R , I , II  2t cos   2 This path difference will produce nth bright fringe, if tnsatisfies the following Eqn.  …(1.12) P.D.W , R , I , II  2tn cos    n 2 Similarly for the (n+1)thbright fringe  …(1.13) P.D.W , R , I , II  2t( n 1) cos    (n  1) 2 Figure 1.7 Fringe width 35 Eqn (1.13) shows that at 𝑡 = 0, 1st dark fringe will be produced. Thus, in between nth and (n+1)thbrightfringe, (n+1)thdark fringe will appear. Consider Fig. 1.7. We observe that t n  xn tan  and t n1  xn1 tan  Substituting both these identities in eqn 1.12 and 1.13, we get  …(1.14) P.D.W , R , I , II  2xn tan cos    n 2  …(1.15) P.D.W , R , I , II  2x( n 1) tan cos    (n  1) 2 Subtracting eqn (1.14) from (1.1.15) and rearranging, we get, Compulsory    …(1.16) xn 1  xn    (air ) 2 tan cos  2 sin  2 Above formula represents the width of a dark fringe. Similar procedure for the bright fringe will yield the same result. Note that this formula represents only approximate fringe width, as it has been obtained with a few assumptions. _____________________________________________________________________________ Example 1.3: As shown in the Fig. 1.6, a wedge is formed by separating two glass plates by an extremely thin wire kept at a distance of 10.0 cm from the edge. When illuminated by sodium light of wavelength 5890 Ao the width of the Fizau’s fringes is measured to be 2.945 mm. Calculate the diameter of the wire Solution:  fw  2 tan  5890  10 10 2.945  10 3  2  1  tan  tan   1  10  4 Figure 1.8 Wedge film for Ex 1.3   5.7296  10 3 deg From the figure 36  d  X tan  d  10.0  1  10  4 cm d  1  10 3 cm d  10m ________________________________________________________________________ Thus, by measuring the fringe width, we can measure the extremely small dimensions such as diameters of thin wires. This method can also be extended to any thin spacer that can be used to form the wedge. From the Eqn. 1.16), one may note that thinner the dimensions of the object, more it is easy to measure it with the help of this technique. This is because the fringe width increases with the decrease in the dimensions of the object to be measured. On the contrary, the measurement by conventional techniques such as screw gauges and Vernier becomes more difficult as the object becomes thin. We thus conclude that optical measurements are more convenient and precise than the conventional measurements. Fringes due to wedge shaped film (Regular and Irregular): The Optical flatness of a glass slab can be checked. Sir Isaac Newton (1642-1727) While the fall of an apple remains an ordinary phenomenon for a layman, it was not so for him, as the phenomenon inspired him to discover the universal law of Gravitation. He also concluded that, the force which pulls apple down also makes Earth rotate around Sun and Moon around the Earth. While his several contributions such as mechanics, optics and gravitation are noteworthy, it is suffice to mention that he is also claimed to be an originator of calculus. He acquired the Lucasian Professorship of Mathematics in Cambridge University, just at the age of 26, and was elected as a Fellow of Royal Society at the age 30. In his famous book named PRINCIPIA, he published the three laws of motion and the Universal law of Gravitation. In another publication named OPTICKS, he presented Physics of the spectrum, interference, color vision and rainbow. Herewith, we discuss the circular and concentric interference fringes known by his name, as he was the first to observe. 37 Compulsory, but derivations are optional 1.3 NEWTON’S RINGS (NEWTON’S INTERFEROMETER): Circular fringes due to circular symmetry The thin film may be treated as a ‘natural interferometer’ as it produces an interference pattern. Our previous discussion of the thin film interference is oversimplified, however, in nature, the cases of thin films can be quite complicated. However, at this stage it is possible to verify whether the concept of the thin film interference can be extended to design an interferometer. One such interferometer was ‘designed’ by Isaac Newton. He didn’t call his system as an interferometer, and as he proposed the corpuscular theory of light, neither did he understand the Physics behind the production of Newton’s rings. The phenomenon was correctly explained by Thomas Young using his wave theory of light. Consider Fig. 1.6. The figure is exaggerated because, the planoconvex lenses involved in the practice have extremely small curvature, and the incident ray falls almost along the normal. Figure 1.9The ray diagram for Newton’s rings It can be understood that, in this case the contours of constant thicknesses i.e. constant path difference form circles. Thus the Fizau’s fringes in this case are circular and concentric. The film involved here is a special case of wedge shaped films. Thus, we can proceed ahead with eqn (1.10)  P.D. I , II  2t cos(r   )  …(1.17) 2 Optional Some assumptions such as 0, r0 will simplify the discussion. Thus we have, for nth dark ring. 38    P.D. I , II  2t n   (2n  1) 2 2 We get, P.D.I , II  2t n  n (nth dark ring) …(1.18) The geometry of the Figure 1.9 helps to express tnin terms of parameters of interest such as R, the radius of curvature of the planoconvex lens and Dn, the diameter of the nth dark ring. Figure 1.7 clearly indicates that R 2  R  t n   rn 2 2 On solving the above eqn and by approximating tn2 0, we get Figure 1.9Howtnrelates with R and Dn Dn2 tn  8R Substituting above tn in the eqn (1.13), we get Dn2 2  n …(1.19) 8R 39 Compulsory Thus, 4 Rn Dn2  …(1.20)  From above eqn , we get 4 Rn 2 R Dn   2n    Dn  2n …(1.21) A similar derivation for the mth bright ring yields 2 R Dm  2m  1   Dm  2m  1 …(1.22) From eqns 1.21 and 1.22, we may be tempted to conclude that, the diameters of dark rings are proportional to the square root of even natural numbers, and the diameters of bright rings are proportional to square root of odd natural numbers. However, this need not be the case always, as the Stoke’s factor in the eqn 1.13 will disappear in a few situations, say if the  increases gradually from planoconvex lens to the wedge to the optical flat or it may decrease also. On disappearance of the Stoke’s factor, the equations 1.21 and 1.22 will complement each other. Thus, it is better to state that diameters of Newton’s rings are proportional to the square root of the natural numbers. The wedge angle in this case increases gradually, and thus the width of Newton’s rings decreases with the sequence number of the ring. Further, it is difficult to keep the center of the fringes dark or bright permanently. Thus it is difficult to decide sequence number of the ring. Thus eqn1.16 will not yield the same accuracy, and the accuracy will depend upon which ring is measured. Thus to have better accuracy, qn 1.16 can be written for the mthdark ring and then the   difference, Dm  Dn can be obtained. Thus, 2 2 40  4 Rm …(1.23) Dm2   Taking the difference of the eqn 1.23 and 1.20 and rearranging,  Dm2  Dn2  …(1.24) R 4m  n  Eqn 1.21 suggests a few applications of Newton’s interferometer. If the diameters D mandDnof the mthandnthdark rings are measured, and if  is known, then the radius of curvature R of a planoconvex lens can be calculated provided  is known. R can also be measured by spherometers, thus  can also be calculated. Further, if R, and  are known then  can be evaluated. Eqn. (1.20) suggests that if the medium having refractive index  is replaced by the air, then the diameter of Newton’s ring will increase. By choosing new symbols, we can thus write 4 Rn and Dn  4Rn ' Dn2  2  Thus we have, Dn2  ' …(1.25) Dn2 Newton’s rings Newton’s rings due to a lens during the manufacturing Process. What is signified by the irregularity? 41 _For better accuracy, Dm2  Dn2 …(1.26)  2' 2' D m  Dn Thus the Newton’s interferometer also provides a method to measure the refractive index of liquids. It is needless to mention that accurate measurements require a monochromatic source. ____________________________________________________________________________ Example1.4The diameters of the fifth and the third dark rings as measured by Newton’s interferometer are 0.343 cm and 0.266 cm respectively. If the setup is operated by a sodium light of wavelength 5890 Ao , calculate the radius of the curvature of the Planoconvex lens. Assume that the setup is kept in the air. We have  Dm2  Dn2  R 4m  n  R  1  0.3432  0.2662  4  (5  3)  5890  108 R  99.52cm ______________________________________________________________________________ Example (1.5): The diameters of the 5th and 3rd ring measured, when the Newton’s ring set up is kept in a 30% sugar solution are 0.292 cm and 0.226 cm respectively. When the sugar solution is replaced by air, the diameters increase to 0.343 cm and 0.266 cm. respectively. What is the refractive index of the 30 % sugar solution? We have Dm2  Dn2  ' ' Dm2  Dn2  0.343 2  0.266 2  0.292 2  0.226 2    1.37 ______________________________________________________________________________ 42 Compulsory Characteristics of Newton’s rings i. Newton’s rings on reflected side are complementary to the Newton’s rings on transmitted side ii. If the glass plate in the Newton’s ring set up is replaced by the Mirror, then Newton’s rings fade out and a uniform illumination is observed. iii. If the Newton’s ring set up is illuminated by white light then a few colored rings near the center are observed. iv. When there is air gap at the center, the ring at the center may appear bright.. v. If the lens is gradually lifted up, then the Newton’s rings are shifted outwards vi. If the monochromatic source in the setup is replaced by a source of higher wavelength, then the diameters of Newton’s rings are increased. vii. If the planoconvex lens in the setup is replaced by the planoconvex lens of higher radius of curvature then the diameters of the rings will increase Albert A. Michelson (1852-1931) His specialty was high precision optical instruments. For several years, his measurement of the speed of the light was considered to be authentic. He redefined the meter in terms of the wavelength of light. Using his interferometer, he measured the diameters of the stars, which appear just like point objects on the Earth. One of his greatest achievements is confirming the absence of ‘ether’, which was then thought to a medium essential for propagation of light. If ‘ether’ existed, then his interferometer would show a fractional fringe shift of 0.36, which appears to be ignorable. However, he claimed that his interferometer was accurate enough to record such a small fringe shift if it had really occurred. Repeated experiments at different places and in the different seasons showed no fringe shift, which left no doubt about the absence of hypothetic ‘ether’. Thus light can travel in vacuum also. This was one of the concepts that became the basis for theory of relativity, credited to the contemporary Physicist, Albert Einstein. Albert Michelson was awarded the prestigious Nobel prize in Physics in 1907. (Entirely optional) 1.4 MICHELSON’S INTERFEROMETER Circular fringes, but where is the circular symmetry? We now discuss an interferometer having extreme precision and versatility, which was designed by Albert Michelson. Consider a set up in the Figure 1.10. It is observed that ray incident on the beam splitter divides itself into two rays of approximately same amplitude. These 43 rays reunite at the point of their origin itself, when reflected by the Mirrors M1 and M2. If the Michelson’s interferometer is aligned perfectly, then it produces circular fringes. Michelson’s interferometer has a few important applications. One such application is the measurement of wavelength of monochromatic light. Note that, both the rays 1 and 2 travel a double distance, from the point of their origin to the point of interference. As noted earlier the point of origin and interference is same. Now if M1 is moved up by /2, then the path difference between the ray 1 and 2 will increase by . Thus the point which originally corresponded to nth fringe will now correspond to (n+1)thfringe. Thus (n+1)thfringe will replace the nth fringe. In the similar manner (n+2)thfringe will replace (n+1)thfringe. Thus a crosswire focused on any fringe will encounter an inward shift of 1 fringe. This leads to a conclusion that if M1 is moved through a distance of X, then the path difference between ray 1 and 2 will increase by 2X. If this corresponds to the shift of N fringes, then we can write Figure 1.19 The schematic of the Michelson’s interferometer 2 X  N … (1.26) 44 Above eqn, though appears simple, depicts two major applications of Michelson’s interferometer. If the distance moved by M1, i.e. X is measured with the help of a screw gauge attached to it and if the shift in the fringes, N is counted then  can be calculated. If  is known then, on counting the shift N in the fringes, the unknown distance X moved by M1 can also be measured to a great precision. In the similar manner the change in X can also be measured by using 2X  N. Indeed, the Michelson’s interferometer plays a major role in the cases where the unknown distances or the change in the distance are too small to be measurable by conventional techniques. If a transparent film of thickness t and refractive index I s inserted in the path of any arm of the Michelson’s interferometer, the owing to the denseness of the film, the path difference will increase by (2t – 2t) i.e. 2t(( – 1).This will also lead to shift of say N fringes. We know that one fringe shifts if the path difference increase by . Thus, 2t  1  N …(1.27) Above eqn emphasizes importance of the Michelson’s interferometer. If  is known and if the shift in the fringes after insertion of the film is counted, then the thickness t of the film can be calculated if its refractive index  is known, otherwise if t is known then  can be calculated. Eqn 1.27 is applicable to liquids, gases and air also. In such case t will represent the thickness of the cell in which these species are filled. Michelson’s interferometer can also be used to measure the resolution of the spectral lines from a bi-chromatic source. In this case, two sets of circular fringes will be observed, one corresponding to 1 and another to 2. If, the bi-chromaticity of the source is extremely small, then 1 and 2 will have extremely closer values. Thus the fringes corresponding to these wavelengths will overlap. Off course, one to one overlap is never possible, as the wavelengths are not equal.. If the maxima in one fringe pattern overlap on the minima of another one, then the composite interference pattern will almost fade out. If the maxima of both the fringe patterns overlap, then the composite fringe pattern will show distinct fringes. Now consider a case where the pattern is distinct. Let the corresponding position of M1 be X1. Now if M1 is moved up, then the fringes in both the interference patterns will start getting shifted, but not at the same rate. According to eqn 1.24, the fringes corresponding to lower wavelength, say 1 will shift faster than 2. Now if the movement of M1 is continued, then obviously the next stage will be corresponding to the disappearance of the fringes. If M1 still continues to move, then the fringe pattern will once again become distinct. Let the position of M1 corresponding to this distinct condition be X2. Considering the fact that M1 has moved such that the fringe pattern has shifted from a given distinct condition to the immediately the next one, we can write 2 X 2  X 1   2 X  n2 and …(1.28) 2 X 2  X 1   2 X  n  1)1 … (1.29) 45 Eqns 1.28 and 1.29 are simultaneous, thus n 2  (n  1)1 n 2  n1  1 n(2  1 )  1 1 n (2  1 ) Substituting n in eqn 1.29 12 2av …(1.30) 2X   2  1  2  1  Thus, if avis known by any other technique, then (2 - 1) can be calculated and the bi-chromatic source can be perfectly resolved. In fact, this technique also helps us to know, whether the source is monochromatic or bi-chromatic. Albert Michelson carried out many enterprising experiments with his unique interferometer. One of these was to standardize a meter. For this purpose, he used a standard unit of length called as etalon. The shortest etalon had a length 0.390625 mm. M1 was moved from one end of the etalon to another one, and the shift in the fringes was counted. Then an etalon having its length double than that of the first etalon was experimented and the shift in the fringes was measured. In second case exactly double shift in the fringes was expected. Thus the average of experimentally measured fringes shifted in the case of first and the second etalon was calculated. Then, third etalon having its length double to that of the second etalon was experimented and the shift in the fringes was counted. An average of the shift in fringes of three successive etalons increased the accuracy by threefold. In the same way, etalons of increasing lengths were experimented. The length of last etalon was 10 times that of the first etalon. The average of the counted and the expected fringes increased the accuracy by tenfold. These measurements were then generalized to 1 meter. The results were 1 meter = 1,553,163.5R (Where R is the wavelength of the Cadmium red line = 6438.4722 A0) 1 meter = 2,083,372.1B (Where B = the wavelength of the Cadmium blue line = 4799.9107 Ao) And, 46  1 meter = 1,966,279.7G (Where G = the wavelength of the cadmium green line = 5085.8240 Ao) Michelson also used his interferometer along with Edward Morely to verify the then hypothesized ether. The idea was that, as the arms of the interferometer had equal lengths and as they were perpendicular to each other, the distance traveled by the ray 1 and 2 in the ether passing in a direction opposite to the rotation of the earth would differ. This would result in a shift in the fringe pattern. Mathematics indicated that a fractional shift of 0.36 would occur. But this was not observed. The experiment was repeated at various places and in the various seasons, but in any case the expected fringe shift was not observed. Michelson claimed that his interferometer was accurate enough to record the fringe shift of 0.36 if it really occurred. The absence of any fringe shift convinced the Physicists about the absence of the ether. Thus light could travel in vacuums also. This conclusion convinced Albert Einstein to propose special theory of relativity. Michelson also measured the diameters of the stars to a great precision using his interferometer. In Caltech, an advanced version of Michelson’s interferometer is being installed. This interferometer proposes to detect the gravitational waves if they exist. The undulations created by the gravitational waves in the path of ray 1 and 2, would cause shift in the fringes. If this is observed then the existence of gravitational waves will receive an experimental confirmation. It may be noted that the length of each arm in this interferometer will be 4.0 km. The name of this lab is LIGO (Laser Interferometer Gravitational-Wave Observatory, http://www.ligo.caltech.edu/ ) As mentioned in the beginning of this chapter, several interferometers exist, each having its own applications. One of them is Fabry-Perot Interferometer. However, their discussion is beyond the scope of the first year engineering course. ____________________________________________________________________________ Example 1.21 Assume that the movable mirror in the Michelson’s interferometer is moved through the L.C. of the micrometer screw gauge attached to it, which is 10 µm, i.e. 0.001 cm. If the interferometer is operated using sodium light having wavelength 5890 A o, then how many fringes will be shifted? Solution: We have 2 X  N 2  0.001  N  5890  10 8 N  0.02945 This is impractical answer. Thus, we should assume realistic data 47 _____________________________________________________________________________ Example 1.22 Through what distance the movable mirror in the Michelson’s interferometer should be moved so that at least one fringe is shifted? The source used is sodium having wavelength 5890 Ao. Solution: 2 X  N 2 X  1  5890  10 8 cm We have X  29.45m X  0.02945mm Considering that the L.C. of the micrometer screw gauge is 0.001 cm, we need to approximate the above answer to X = 0.0295 mm. ______________________________________________________________________________ Compulsory 1.4 INTERFERENCE COATINGS How to minimize the unwanted reflection or transmission of light A camera can produce better images if it receives maximum light from the object to be photographed. Solar cells can also give better electrical output if maximum light passes in. But this is prohibited by the reflection of light. For ex. in camera, the reflection of the light from the lens is nearly 4% and thus it reduces the transmitted intensity to 96%. Can this problem be solved using the interference of thin films? Further, as we may have experienced, certain glasses, especially those used in the windows or the doors of the commercial shops, or windows of automobilesor sometimes in buildings are excessively reflective, but only from one side. The excessive reflectivity doesn’t allow the observer to look on the other side. While, as the reflectivity is only from only one side, the observer at the non-reflective side can look through the glass. The rhinestones can also be provided with attractive looks by using such techniques. The reflectivity of the mirrors can also be enhanced to it’sideal value (100 %). Interference filters which can pass an extremely small narrow band of the wavelengths can also be designed using such concepts. In short, what we would like to discuss here are Anti-Reflection Coatings (ARC) and the High Reflection Coatings/Anti-Transmission coatings (HRC/ATC). Consider the Figure 1.10 We start with the usual eqn. of thin film interference, 48  Figure 1.10 Interference coatings (a) ARC and (b) HRC/ATC  …(1.31) PDR ,I ,II  2t cos r  2 In general, the incident ray falls very close to the normal, thus r0. Note that in the case depicted in Figure 1.9, the Stoke’s factor is absent in the above eqn. A destructive interference between reflected rays suppresses the reflection. Consequently the transmission enhances. Thus  PD I , II  2t cos r  2n  1 …(1.32) 2 If the film is thin enough, we can safely assume 𝑛 ≈ 0. Thus  t ARC  …(1.33) 4 Thus, a film having thickness t, which satisfies above eqn behaves as ARC. ARCs are made of MgF2 (µ= 1.38), SiO2 ((µ= 1.49). The discussion of HRC/ATC and ARC is not exactly opposite. We just need to replace the film in ARC by a material denser than the substrate. The substrate is generally the glass. In such case, the Stokes’s factor /2 is present in the eqn 1.31 (as suggested by Stoke’s law). Further, a constructive interference between the reflected rays will make the film more reflective and less transmittive. Thus, we have  PD ' '  2t cos r   n I , II 2  2t cos r  2n  1 2 49 For thin films, it is safe to assume n ≈ 0. Generally, the incident ray passes close to normal. Thus 𝑟 → 0and we get  t HRC / ATC  …(1.34) 4 Ift satisfies above eqn, the coating behaves as HRC/ATC. HRC/ATC does not work in opposite direction, as in such case when light passes through the glass, its wavelength, λ will decrease 𝜆 substantially(𝜆′ = ) and Eqn.(1.34) will not work for the decreased wavelength. Generally the 𝜇 materials used for HRC/ATC are TiO2 (µ = 2.4), ZnS (µ= 2.32). By using multi-layer coatings, it is possible to block the transmission of all the wavelengths except one. Thus an extremely narrow band of wavelengths (with λ ≈ 11 Ao)can be transmitted. Such coatings are called as interference filters. Note that the eqn (1.33) for ARC and eqn (1.34) for HRC/HTC appear same, but they are not. We also observe that in both the cases t depends upon λ Camera lens with and without antireflection Rhinestones coated with high reflection coatings coatings. Look at the bright colors Window of a car coated with high reflection Glasses coated with high reflection coatings coatings are also used in buildings RAPID REVIEW Interference is not just a natural phenomenon responsible for colors in the oily films and the soap 50 bubbles. It forms the basis of instruments with extreme precision called as interferometers. Both the interferometers that we have gone through i.e. Newton’s and Michelson’s interferometer are based on thin film interference. Thin film itself is an interferometer. There are two kinds of thin films as mentioned below.  Thin film of constant thickness, PD  2t cos r  2  Wedge shaped film PD  2t cosr     2 Thin films can produce two kinds of fringes; one is Fizaue’s fringes, where t is varied and r is kept constant and another is Haidinger’s fringes, where t is kept constant and r is varied. Newton’s interferometer involves a wedge film with circular symmetry. The measurement of diameter of Newton’s rings leads to three applications given below. R  Dm2  Dn2     Dm2  Dn2 ' '  4(m  n)  and Dm2  Dn2  Michelson’s interferometer surpasses the Newton’s interferometer in many respects. It is more versatile, more accurate and more sensitive. Its few applications are based on the following eqns 2av 2 X  N , 2t  1  N and 2 X     Interference also forms the basis of ARC and HRC/ATC, the formula is t ARC / ATC . For 4 ARC, the coating is rarer than the substrate, while for HRC/ATC is the case is opposite. APPENDIX I TO CHAPTER 1 The Optical Path Consider Figure 1.11,which is self-explanatory We have d v t vt  d c t  d  ct  d 51  Thus Optical path = ×Geometrical path Thus optical path represents the distance travelled by the light with speed c, in a time t, which the light spends while travelling through a medium of thickness d with lowered speed 𝑣. In short, a path difference is created due to slow speed of the light in the denser medium. As an alternative explanation, we have Figure 1.11The concept of Optical path  '   '       n '    n   n   n ' n    n  ' Optical path = ×Geometrical path Note that owing to the decreased speed in the denser medium, light carries a smaller wavelength ’, while passing through the denser medium 52 APPENDIX II TO CHAPTER1 The Stoke’s Law 1. If a wave suffers a reflection from denser to rarer medium its phase is reversed 2. No phase reversal occurs during the transmission. 3. No phase reversal occurs if the wave is reflected from rarer to the denser medium Figure 1.10 Physics behind the Stoke’s law 53  APPENDIX III TO CHAPTER 1 Derivation of Path Difference in case of Wedge-shaped film Consider the following figure. We have Figure 1.12 Ray diagram for the Wedge shaped film PD I , II   ( AE  EB  BC )  AD PD I , II   ( AE  EB  BO ' )  AD   PD I , II   AE  EO '  AD PD I , II   AE  cos r     AC sin i PD I , II   AC sin r  2t cos r     AC sin r PD I , II  2t cos r    With Stoke's law  PD I , II  2t cosr     2 54  QUESTIONS It is only the first step that takes the effort. –Marquise du Deffand General Fundamentals of Optics 1. Consider any optical instrument, say the Human eye, and identify any three laws of Optics involved in its functioning. 2. Indicate how the design of such instrument requires precision optical engineering. 3. Mention any one application of the optics in the daily life. 4. Light is an electromagnetic wave. What is an electromagnetic wave? 5. According to particle character of the light, it consists of photons. What do you mean by a photon? 6. What do you mean by coherence? 7. Name as many monochromatic sources as known to you. 8. Name the best coherent source known to you. 9. Name as many polychromatic sources as known to you. Basics of Interference 10. Interference explores the wave character of light. How? 11. Interference does not reveal whether light is longitudinal or a transverse wave. How? 12. What do you mean by a steady state interference pattern? 13. What conditions need to be satisfied for generating a steady state interference pattern? 14. What do you mean by a sharp interference pattern? 15. What conditions need to be satisfied for generating a sharp interference pattern? 16. The generation of a well-defined interference pattern requires a systematic and gradual variation of the path difference. What will happen if the path difference varies randomly? 1.1 Thin Film Interference 17. A thin film needs to be thin, why? 18. A thin film satisfies all conditions for generating a well-defined interference pattern except one. What is it? Why? 19. A film excessively thinner or thicker than the wavelength of the light cannot produce interference pattern. Why? 20. The fringes in case of a thin film of uniform thickness exposed to a point source are circular. Why? 21. The fringes in case of a thin film of gradually increasing thickness exposed to a broad source are straight. Why? 22. A thin film can produce the interference pattern, when either its thickness is varied by keeping the angle of incidence same, or the angle of incidence is varied by keeping thickness same. Is it possible to produce a well-defined interference pattern with the 55  combination of both these situations at once? Why? Why not? 23. Is it really necessary to expose the thin film to a monochromatic source in order to produce an interference pattern? Explain. 24. Evaluate whether the Stokesfactor will be present or absent in the following situations. Also explain, why a. A thin film existing in between denser media b. A thin film existing in between rarer media c. Medium above the film is denser and the medium below the film is rarer d. Medium below the film is rarer and the medium above the film is denser 25. If a film appears dark from the reflected side, then at the same point it appears bright when viewed at the same angle but on the transmitted side. Is it ever possible to create a situation, where this does not happen? Why? Why not? 26. Can a thin film of randomly varying thickness and randomly varying refractive index produce an interference pattern? Why? Why not?. 27. A film excessively thinner or thicker than the wavelength of the light cannot produce interference pattern. Why? 28. The fringes in case of a thin film of uniform thickness exposed to a point source are circular. Why? 29. The fringes in case of a thin film of gradually increasing thickness exposed to a broad source are straight. Why? 1.2 Wedge shaped films 30. Why it is necessary for the wedge shaped film to have a very small wedge angle? 31. Can a wedge film produce a well-defined interference pattern when exposed to a white source? Why? Why not? 32. The formula for the fringe width of the wedge shaped film in this chapter is an approximate formula. Why? 33. Can a wedge film produce a well-defined interference pattern, if exposed to a broad source? Why? Why not? 34. A wedge shaped film can be conveniently used to inspect the diameters of the thin wires. How? 1.3 Newton’s Rings 35. Newton’s rings are concentric and circular. Why? 36. Newton’s rings become thinner, as one moves away from the center. Why? 37. Newton’s rings disappear if the Plano-convex lens is kept on the mirror instead of the glass plate. Why? 38. The Plano-convex lens used for producing the Newton’s rings should have small curvature. Why? 39. How curvature and radius of curvature are related to each other? Are they directly proportional or indirectly proportional to each other? 40. Can Newton’s rings be observed using bio-convex lens? Why? Why not? 41. Newton’s rings show an outward shift if the lens is lifted up. Why? 56  42. Sometimes, Newton’s rings show a bright center instead of dark and sometimes vice versa. Why? 43. If Newton’s rings are observed with white source, then they appear colored. Why? 44. The colored Newton’s rings exist only near the center. Why? 45. Newton’s rings shrink in diameters when air is replaced by a denser medium. Why? 46. For better accuracy, the diameters of at least two rings are measured instead of one. Why? 47. The center of Newton’s rings is not pinpointed, but is bulged. How? 48. The perfection in the devices like lenses and glass plates can be tested with Newton’s interferometer. How? 49. Newton’s rings cannot resolve a bi-chromatic light. Why? 50. Newton’s rings disappear if the plano-convex lens is kept on the mirror instead of the glass plate. Why? Optional 1.4 Michelson’s Interferometer 51. In what ways Michelson’s interferometer is better than the Newton’s interferometer? 52. The lengths of the arms in the Michelson’s interferometer should not be equal. Why? 53. Is it really necessary for the Michelson’s interferometer to have compensatory plate? Why? Why not? 54. Does the accuracy of the Michelson’s interferometer depend upon the least count of the screw gauge? Why? Why not? 55. What is the smallest length or the change in length measureable by Michelson’s interferometer? 56. Michelson’s interferometer is considered as a precision instrument for measuring dimensions or the change in dimensions. Why? 57. If the ‘ether’ existed, then the fractional fringe shift encountered in Michelson’s Morley experiment would be 0.36. What do you mean by a fractional fringe shift? How can it be measured? 58. For the standardization of meter, Michelson used cadmium source. This source is polychromatic. How can a meter be standardized by using a polychromatic source? 59. What is the smallest resolution of a bi-chromatic source that can be resolved by using Michelson’s interferometer? 60. Michelson’s interferometer can resolve sodium source, but Newton’s interferometer cannot. Why? 61. The sensitivity of the Michelson’s interferometer increases with the length of the arms. How? 62. List any two interferometers other than Newton’s interferometer and Michelson’s interferometer. 63. Michelson’s interferometer can be used to e. Inspect the displacement in the walls of the Dam f. Young’s modulus of glass or metals 57  g. Coefficient of the thermal expansion of glass or metals Identify any one application of your choice and indicate how it is possible Compulsory 1.5 Interference Coatings 64. Antireflection coatings (ARC) enhance the transmission of light just from 96 % to ~ 100%. Is there really any advantage? 65. The ARC is made up of materials denser than the substrate, while HRC/ATC is made of materials rarer than substrate. Can this be made possible in opposite manner? How? 66. Lenses of most of the cameras appear purplish. Why? 67. The purplish tint is not observed if viewed at the different angels. Why? 68. The rhinestones can be made more attractive with HRC/ATC. How? PROBLEMS The significant problems of our time cannot be solved by the same level of thinking that created them. - Albert Einstein 1.1 Thin film interference 1. Calculate the range of the angles in which the thin film having a thickness 0.55 𝜇𝑚 being viewed at 68o will appear greenish. The wavelength of the green color varies from 4950– 5700 Ao. 2. At what angle should film in the problem 1 be viewed, so that it appears reddish? The wavelength of red light is 6320 Ao. 3. If the film is suspended on a dry road, it does not appear colored. Why? 4. What color, the film in the problem 1 will show, if viewed at the same angle, but from the transmitted side? 5. The range of visible light is 3800-4500 Ao (violet), 4500-4750 Ao (blue), 4760-4950 Ao (cyan), 4950-5700 Ao (green), 5700-5900 Ao (yellow), 5900-6200 Ao (orange), 6200-7500 Ao(red). Can this range be resolved by a thin film having thickness 0.55 m? Why? Why not? Assume that the film has refractive index of 1.35 and is denser than road on which it is spread. 6. Which two colors can just be resolved by thin film of thickness 1389 Ao and refractive index 1.35 at 45o? Assume that the thin film is denser than the substrate. 7. What should be the minimum thickness of the thin film which just resolves violot and the blue color at 45o. Given the refractive index of the film is 1.35 and it is spread on the substrate which is denser than the film. 58 1.2 Wedge shaped films 8. Calculate the smallest dimensions that can be measured by using the technique of wedge film. Assume that the source used is sodium, having wavelength 5890 A o and the object is kept at a distance of 10.0 cm from the wedge. The L.C. of the traveling microscope used to measure the fringe width is 100 µm (0.01 mm). 1.3 Newton’s rings (Newton’s interferometer) 9. If the L.C. of the traveling microscope used in the Newton’s interferometer is 100 µm, then what is the smallest radius of curvature of the Planoconvex lens that can be measured? The medium involved in the set up is air. 10. It is known that the refractive index of the sugar solution increases with its concentration. When the Newton’s rings set up is kept in a sugar solution with 40 % concentration having refractive index 1.40, the diameter of a certain ring is measured to be 0.290 cm. Now, when the 40 % sugar solution is replaced by 68% sugar solution, the diameter of that ring acquires a value of 0.285 cm. What is the refractive index of the 68% sugar solution? 11. Newton’s rings are being observed with the help of a planoconvex lens of radius of the curvature 100.0 cm and the sodium light having wavelength 5890 A o. The set up is kept in the air. Calculate the diameter of the 5th dark ring in such case. We know that calculators can provide the answers up to several decimal places. However, if the L.C. of the traveling microscope used in the interferometer is 100 µm, up to what decimal place the answer needs to be rounded off? 12. The diameter of the 5th dark ring measured using Newton’s interferometer in air is 0.292 cm. Considering that the L.C. of the traveling microscope is not more than 100 µm, What is the smallest refractive index measurable by using Newton’s interferometer? 13. Assume that the Planoconvex lens and the glass plate are kept in a system which can be evacuated. Initially the system contains air, the refractive index of which is accurately known to be µ = 1.00027717. The diameter of the 5th Newton’s ring is measured to be 0.592 cm. Assume that the interferometer is operated using sodium light having wavelength 5890 Ao. Now the air is slowly evacuated. To what extent the diameter of 5 th Newton’s ring will change? Will it increase or decrease? 14. The diameter of the 5th dark ring in presence of air is 0.592 cm. When the air is slowly evacuated, the diameter decreases to 0.591 cm. Calculate the refractive index of the air. Optional 1.4 Michelson’s Interferometer 15. Assume that you are operating the Michelson’s interferometer with the cadmium source emitting red, blue and green light having wavelengths 6438.4722 A0, 4799.9107 Aoand 5085.8240Ao respectively. If the movable mirror in each case is moved through the same distance then will the number of fringes shifted change in each case? 59 16. In above problem the wavelengths of cadmium are specified to the fifth decimal place. For ex. the wavelength of the blue line in cadmium is 4799.9107 Ao. Can such accuracy be achieved using Michelson’s interferometer? Verify 17. How many fringes will be shifted if the movable mirror in the Michelson’s interferometer is moved through a distance of 0.0295 mm. The source used is sodium-having wavelength 5890 Ao. Sodium is a bi-chromatic source, and its next wavelength is 5896 Ao. How many fringes will be shifted in case of second wavelength? Comment on the result 18. Do you feel that the distance moved by the mirror and hence, the fringes shifted are too many to distinguish between the sodium line? Try the smallest possible distance so that only one fringe is shifted. 19. What is the smallest length or the change in the length that can be measured using, Michelson’s interferometer, if operated by He-Ne laser having wavelength 6328 Ao. 20. Calculate the smallest thickness of the thin film that can be measured by using Michelson’s interferometer. The refractive index of the thin film is 1.38 and the source is He-Ne laser having wavelength 6328 Ao. 21. A glass cell having width 10.0 mm filled with air is inserted in the path of one of the interfering beams. The interferometer is operated with laser having wavelength 6328 A o. As the air is slowly evacuated the fringes start getting shifted. In which direction the fringes will shift? After the complete removal of the air 8.76 fringes are shifted. What is the refractive index of the air? 22. Can the thickness of the glass cell in the above problem be selected arbitrarily? Why? Why not? 23. Consider a cell having thickness 1.00 cm filled with water having refractive index 1.33. The interferometer is operated using laser light 6328 Ao. The water is removed slowly. How many fringes will be shifted when water is removed completely from the cell? 24. Michelson’s interferometer is used to resolve the sodium doublet, with an average wavelength of 589