Optics PDF - Physics Textbook 2025

 Optics Optics Prof. Abed Nasr 2025 i  ‫‪Optics‬‬ ‫بسم هللا الرحمن الرحيم‬ ‫إن السمم والبصمم والفؤاد كل‬ ‫أولئك كان عنه سئوال‬ ‫صدق هللا العظيم‬ ‫‪ii‬‬ ...

 Optics Optics Prof. Abed Nasr 2025 i  ‫‪Optics‬‬ ‫بسم هللا الرحمن الرحيم‬ ‫إن السمم والبصمم والفؤاد كل‬ ‫أولئك كان عنه سئوال‬ ‫صدق هللا العظيم‬ ‫‪ii‬‬  ‫‪Optics‬‬ ‫مقدمة‬ ‫ثل ى علم خري ثيبب ى زدهاته تبثلي إزدهاته‬ ‫علم البصريات‬ ‫عصر ر رري‬ ‫ح‬ ‫حضر ر ررتها اوتخر ر ررتذ تب ل ت ث عصر ر ررضه ال ضر ر ررت اه الر‬ ‫ررت علم‬ ‫ثي‬ ‫ال ك ضلضجيررت ال ر ل ر ن تا عر ز ر ر ر ر ر ر ر ثيار ىام ال ايح ر ال‬ ‫عل ال ض ال تل ‪:‬‬ ‫البصيات‬ ‫أوال‪ :‬آراء فالسفة اإلغريق‬ ‫ىال ر ر اوهياأ ام ىتح ثر حتتلضا ب خ ر رربي حت ر ر اوإص ر ررته تى م‬ ‫م إت ر ررم اا ت يرتا ‪Optics‬‬ ‫طبيث الضر ررضو ترضااي تاض ثت عي ى ل‬ ‫م ك ت ىذ إ ضث م ى ا ا ال جتح‬ ‫لك م لم ث رضا ى ه اه‬ ‫ى البصيات‬ ‫لم بكر تافي ر ن حبررت زع ر ر ىىالطضذ ىذ اوإص ر ر ر ر ر ر ررته م إ يت ال ضه ثر‬ ‫ر ر ر ر ررضه‬ ‫الثبر لكر ىه ر ر ر ر ر ض رتل ع ال ىي تدعم ىذ اوإصر ر ر ر ررته م إتت بت‬ ‫زل ال جيا ن‬ ‫ا ش ر ر رريتو ى الثبر تاع ا كتتآ كل ت خهاو ىلخر ر ر ر ي جبخر ر ر ر‬ ‫كتذ‬ ‫ىذ خر ب ر ا الثلم ذ ال بلخرض‬ ‫ج ال لخر‬ ‫ل لك لم عر ل ا ال‬ ‫زليع إثرلع اض الصضابن‬ ‫عل ىذ ثت‬ ‫ث‬ ‫ثانيا‪ :‬أراء العلماء فى العصر اإلسالمي‬ ‫ث بي ال خ ررر ر ال بلم تاح ا ثر ىعالا ال ض ررتها او ررالثي ال ر‬ ‫علم‬ ‫زليع ب رري‬ ‫ج ال جياب ن ل لك خر ر‬ ‫كتذ ل م الخ رربأ ى تا رره ال‬ ‫إر ك لررع تكررتتررآ ك ررتإررتبررع ال يجه ا ام للثر ر ثر الجررتثثررت‬ ‫البصر ر ر ر ر ر رياررت‬ ‫الضرضو تب خربي حت ر اوإصرته‬ ‫ر ال بلم بثيا‬ ‫ا تهت ي ن حبت ز ر ت‬ ‫‪iii‬‬  ‫‪Optics‬‬ ‫ىهجث ت زل تجضه ذاب تاض الض ررضون ت لك ز ر ر بث ىكيا ريت شر ر‬ ‫تال‬ ‫لب م زإصر ر ررته الجخر ر ررم ال يب إعن تىر ر ال بلم ى بثيا‬ ‫ثر الثبر‬ ‫الض ر ر ر ررضو إ تع أبتها عر ىش ر ر ر ررثع ل ت ىطضاح تعيتر ب بثت ثر ا جخ ر ر ر ررتا‬ ‫تالجخر ررم ال ضاج تإذا ر ررر آ ا ا شر ررث عل جخر ررم‬ ‫ال ضر رربا كتلور ر‬ ‫أ ثه ثت تثيىع اآلذ عر ال تق الضضبي ن‬ ‫ع تاض ثت‬ ‫ى‬ ‫كلي‬ ‫الثبر ثر ال تحي‬ ‫تتتقش ر ال بلم ع لي اوإصته ت بر بيكب‬ ‫رتايا اوتثعتس‬ ‫ك جزو ثر ىجزاب تن ك ت إ ت ى‬ ‫ال ويا ي تتري‬ ‫تاوتكختهن‬ ‫ثالثا‪ :‬عصر النهضة الحديثة‬ ‫اض زر ياع م ال لخ ر ررعضب‬ ‫ثع ا ته بضذ ى البص ر ريات‬ ‫ىام ثت ز ر ر‬ ‫ثل ض ر ى ه اه ر رضاء الضررضو‬ ‫تال يعيت ررعضبن ك لك ب ررآ تجتحت‬ ‫ك ثر ال ت يكي ( ر ر ر ر ر ر ب ( تال يتخ ر ر ر ر ر ررببر (ه عته ( ت( ىبيثت (ن‬ ‫عل‬ ‫ر ر زل (ال خر ررر ر‬ ‫زح اا ت ى ا‬ ‫تر ي تظيا تذ ث تىخ ر ر تذ ب خ ر ر‬ ‫بث بي ىشر ررث الضر ررضو ثعضتع ثر جخ ر ري ت‬ ‫ال بلم( تثر إث (تبضبر( تال‬ ‫بث بي الضرضو ذت‬ ‫ا ري زل هيج ز‪ ،‬تال‬ ‫هقير ث تهي الصر ي تب خر‬ ‫طبيث ثضجي )‪(Huygen’s wave theory‬ن‬ ‫ى ب خبي رتايا اجتثعتس تاجتكخته‬ ‫تق تج آ رتايا الجخي ت‬ ‫( رب م ه اه ر ت إتل صرب ى ال صر ا تح(‬ ‫تاجت ورته ى ر ض ثخر مي‬ ‫تلك ت لم ب جب ى ب خ ر ر ر ر رربي رضااي ال ار تال بضه تاو ر ر ر ر ر ر تب (تال‬ ‫ال تث ( تال‬ ‫ر ر ررب م ه اه ر ر ر ر ت إتل ص ر ر ررب ى ال صر ر ر رضح ثر اللتلت ح‬ ‫‪iv‬‬  ‫‪Optics‬‬ ‫ز رر ع‬ ‫زى يا ر ررآ ت ر ر تا‬ ‫تج آ ى ب خر رربيات الظتايا ال ضجي ن تال‬ ‫الضضو ى ال ياغن‬ ‫ثضجت‬ ‫ا ثبي‬ ‫ررل تثير بر ال جتح‬ ‫الطيف الكهرومغناطيسىىى‪ :‬جحظ ثتكخررضا تجضه‬ ‫يك بضل ثجتج ث تطيخ رريت‬ ‫الك ي ي ال‬ ‫تطيخ ر ‪ ،‬ىتلو ر‬ ‫الك ي تال‬ ‫بي ضل ثجتج ك ي يت‪ ،‬ت اه ر ر ر ا ا ال الدا بم‬ ‫تطيخ ر ر ر ال‬ ‫تال جتح ال‬ ‫تاره ثثتهل هاتاري ل ت الصر ال ضجي تث ت ز ر ج ىذ ريع ال ضجت‬ ‫المي‬ ‫تبخررتت (‪ 3x108‬ا‪/‬ث( تا ت‬ ‫الك يتث تطيخرري ى ال ياغ ثت‬ ‫بم ال ض ر ر الب ت ع ليت لميتس ر رريع الضر ررضو ى ال ياغ‪ ،‬ا ا ال ت أ‬ ‫ال‬ ‫طبيث الضضو ال ضجي ن )‪(Electromagnetic theory‬‬ ‫كتذ هلب حت م عل‬ ‫تى عررتا ‪1888‬ا ىثبررآ (ايبز( زثعررتذ زتبثررتث طررتق ر ثر ال ر تابي‬ ‫ررت زل هابيا ىري ثو ر ر ر ر ر ر ررت ر هتذ‬ ‫الك ي ير ال ر ر إر ى ال ياغ ال ي‬ ‫الال ررلكي ( تهع آ‬ ‫ى ررب آ تضاا لالبصررتج‬ ‫تن (تال‬ ‫ب‬ ‫زبصررتح ثته‬ ‫تظيا ثتكخرضا الك يتث تطيخري حبت يار عل ىذ ال تق ال بثل ثر‬ ‫إ ب به ج يه قضاتبر الضر ر ررضو ‪ c = f‬ن( ت ر ر ر ثير‬ ‫ال ابيا الك ي ي ال‬ ‫الك يتث تطيخ ر ر ى ا ا الك تب إو ر ر ثر ال صر ررب ى ال ص ر ر‬ ‫لل ي‬ ‫اللتت (‬ ‫عتا ‪1913‬ا كتذ ث ر ىذ طبيث الضر ررضو ى ر ررب آ ثل ض ر ر‬ ‫ح‬ ‫ذلك ال بر لم عر ث تحت الكلبي عر زتبثتث الضضو‬ ‫إوع كببين لكر ح‬ ‫ضاي ىتح ب خبي ث ر ل ا اوتبثتث عل ى تس‬ ‫ىع‬ ‫ثر ال اه ح‬ ‫تع لآ ث تهيم حيث إعتبر الضىىىذء طا طبيعة‬ ‫تظيا الكم تق ح ثآ زار ررتىت‬ ‫‪v‬‬  ‫‪Optics‬‬ ‫يتل (‬ ‫حر ه (ه‬ ‫مزدوجىة فهذ جزييىا يبىدو يمسىىىىىىىىىيمىاا وجزييىا يى مذا ح‬ ‫بر ال ت ر ر ر ررب بر تكتذ ع ي تق ت ‪ 31‬عتثت‪ ،‬تث عتا‬ ‫بي‬ ‫الثالق ال‬ ‫‪(Quantum‬‬ ‫رتو تظيار ىبزارتبير‬ ‫‪ 1925‬ر ى اجع رته عل تظيار الكم ى‬ ‫تالر قررتبأ ا تلير تتور ر ر ر ر ر ر ر ر لررك علم‬ ‫تالجزااررت‬ ‫)‪ theory‬عر عررتلم الر اه‬ ‫ر ثير ل ت إتل صررب ى ثياح‬ ‫ثيعتتيعت الكم تال يعتتيعت ال ضجي تال‬ ‫ه اه ي جحر ن‬ ‫زل ثالث ىقخ ر ررتا هبيخ ر رري ا‬ ‫عر برخ ر ررم البص ر ريات‬ ‫وبشىىىعا عا‬ ‫ي )‪(Geometric optics‬‬ ‫ال‬ ‫البصيات‬ ‫ال ضجي )‪(Wave optics‬‬ ‫البصيات‬ ‫الك ي )‪(Quantum optics‬‬ ‫تالبصيات‬ ‫ا ا الك تب زل ث تذ ىص ررضح ل لر الض ررضو‬ ‫تق بم برخ رريم ث ضات‬ ‫إ يتع ت اللالث ا ثر طبيث الضر ر ر ررضو‬ ‫إور ر ر ررع ثضجز عل البص ر ر ر ريات‬ ‫البصر ر ر ر ريا تىش ر ر ر ررث‬ ‫تثيت اه ضاح ا ثر ىام الثلضا ال ثت ر ر ر ر ريا تا ا ليت‬ ‫اللبزه تات توا إتل ظيا ال خبي الثتث ن‬ ‫إج يه ب صصتب م‬ ‫تبث ثضاضعت ا ا الك تب ايتها ل الب ال‬ ‫ر رربب ال لتح‪ -:‬ىزذ ى م ال تل لظتايا او ر ر ر تب ع ع ثر ال م‬ ‫ىثل‬ ‫ضح تالك ببضبي الو ر ص ر تك لك‬ ‫الجب لل يق بر شررتشررت الك ببضبي ال‬ ‫ث ى‬ ‫البصر ر ر ر ر ريا تىش ر ر ر ر ررث اللبزه ال خر ر ر ر ر ر‬ ‫زل تا ال تل إ بتهئ ا ليت‬ ‫الضضب بضىي لع رل ي عياض ى ثجتج الوبعت تاوبصتج‬ ‫ال اي‬ ‫البصيا تهبياتن‬ ‫‪vi‬‬  ‫‪Optics‬‬ ‫أرجو أن تعم الفائدة من هذا الكتاب‪ ،‬متمنيا لكم جميعا كل توفيق‪.‬‬ ‫عت تصي‬ ‫ت ي ‪2025‬‬ ‫‪vii‬‬  Optics Contents 1 THE NATURE OF LIGHT AND THE LAWS OF GEOMETRIC OPTICS 1.1 Introduction ▪ Newton’s corpuscular theory (particle theory). ▪ Huygen’s wave theory ▪ Electromagnetic theory ▪ Quantum theory 1.2 Fermat's Principle 1.3 Reflection and Refraction 1.3.1 Reflection 1.3.2 Refraction ▪ Refractive index ▪ Reversibility of light ▪ Change in wavelength 1.4 Critical Angle and Total Internal Reflection ▪ Right–angled isosceles prism ▪ Optical fibers Multiple Choice Questions (MCQ) Exercises Summary viii  Optics 2 LIGHT DISPERSION AND ELECTRO- MAGNETIC SPECTRUM 2.1 Dispersion and Prisms 2.2 Electromagnetic Spectrum Gamma rays X-rays Ultraviolet radiation Visible light Infrared radiation Microwaves Radio waves 2.3 Properties of Electromagnetic Radiation 2.4 Power of light Multiple Choice Questions (MCQ) Summary 3 INTERFERENCE OF LIGHT WAVES 3.1 Introduction 3.2 Young's Double-Slit Experiment 3.3 Theory of Interference Fringes 3.4 The Intensity Pattern 3.5 Change of Phase Due to Reflection 3.6 Interference in Thin Films ix  Optics 3.7 Methods of Getting Interference Pattern 3.8 Fresnel's Bi-prism 3.9 Newton's Rings 3.10 Michelson Interferometer Construction and working Application of Michelson interferometer Multiple Choice Questions (MCQ) Exercises Summary 4 DIFRACTION 4.1 Introduction 4.2 Fraunhofer Diffraction at a Single Slit 4.3 Intensity Distribution in the Diffraction Pattern Due do to a Single slit 4.4 Fraunhofer Diffraction at a Double Slit 4.5 The Diffraction Grating Multiple Choice Questions (MCQ) Exercises Summary x  Optics 5 POLARIZATION 5.1 Introduction 5.2 Polarization by Reflection 5.3 Polarization by Selective Absorption. 5.4 Polarization by Double Refraction 5.5 Polarization by Scattering 5.6 Optical Activity 5.7 Laptop Screen Multiple Choice Questions (MCQ) Exercises Summary 6 LASERS 6.1 Introduction 6.2 Absorption 6.3 Spontaneous Emission 6.4 Stimulated Emission 6.5 Population Inversion 6.6 Laser Operation 6.7 Three-Level Laser System 6.8 Four-Level Laser System xi  Optics 6.9 Types of Lasers 6.10 Applications of Lasers Multiple Choice Questions (MCQ) Summary 7 OPTICAL FIBRES 7.1 Introduction 7.2 The Structure of a Fibre 7.3 History 7.4 Advantages of Using Optical Fibres 7.5 Disadvantages 6.6 Areas of Application 7.7 Types of Optical Fibres 7.8 Optical Path in Fibres 7.9 Maximum Angle of Incidence 7.10 Pulse Spreading 7.10-1 Cause of Pulse Spreading Chromatic Dispersion Modal Dispersion 7.10-2 Consequence of Pulse Spreading Frequency Limit Distance Limit xii  Optics Bandwidth Distance Product (BDP) 7.11 Choice of Fibre Single Mode Fibre Multimode Step-index Fibre Multimode Graded-index Fibre 7.12 Cause of Attenuation 7.13 Numerical Aperture Multiple Choice Questions (MCQ) Summary 8 RELATIVITY 8.1 Introduction 8.2 Newtonian Mechanics 8.3 Einstein's Postulates 8.4 Lorentz - Einstein's Transformation Equations 8.5 Relativity of Length 8.6 Length Contraction 8.7 Relativity of Time 8.8 Relativity of Mass 8.9 Relativity of Momentum 8.10 Relativity of Energy Summary References xiii  A.M.Nasr Nature of light 1 CHAPTER (1) NATURE OF LIGHT AND LAWS OF GEOMETRIC OPTICS 1.1 INTRODUCTION When we say that light is travels from one point to another point, the question is “what is the thing that constitutes light?” Many theories were put forward to explain the nature of light, the most important theories are: Particle theory (Newton’s corpuscular theory) Huygen’s wave theory Electromagnetic wave theory Quantum theory Particle theory This theory states that light consists of very small particles called Corpuscles, which are emitted, from a luminous source and travel with very large velocities. When they reach the retina of the eye they produce the sensation of vision. 2 Nature of light A.M.Nasr This theory was able to explain the rectilinear propagation of light, because size of the particles is considered to be so small that Earth’s gravitation has a little effect on them. This theory provided simple explanations of some known experimental facts concerning the nature of light, namely the laws of reflection and refraction. But this theory was not able to explain the phenomenon of interference, diffraction, and polarization. Huygens’s wave theory Huygens supposed that light is a form of longitudinal waves and he was able to explain the phenomenon of reflection, refraction, interference, and diffraction. But the phenomenon of polarization could not be properly explained. Huygens’s Principle: Every point on a given wave front can be considered as a point source for a secondary wavelet. At some alter time, the new position of the wave front is determined by the surface tangent to the set of secondary wavelets (see Fig. 1.1).  A.M.Nasr Nature of light 3 Electromagnetic wave theory Fresnel and Young overcame the above defects by supposing the light waves to be transverse in nature. They assumed that, if V is the velocity of a wave through an elastic medium of elasticity E and density ρ, then 𝑉 = √𝐸/𝜌 which means that ether must have very high elasticity and low density. These two properties are opposite to each other. wave front Primary source Primary source secondary source secondary source Plan wavefront Spherical wavefront Fig. 1.1: Huygens' Principle As gravitational forces can act across space without a medium, electric and magnetic fields can also act without a medium, so light can also travel without any medium. The velocity of 4 Nature of light A.M.Nasr electromagnetic (e.m.) waves is given by 𝑉 = 1/√𝜀𝜇 were ε is dielectric constant and μ, the permeability of the medium. Since these quantities have definite values for vacuum, so there is no necessity of assuming a hypothetical medium like ether. Particle and/or Wave !!!!! Quantum theory Up to the beginning of 20th Century, it was thought that all the characteristics of light could be explained by e.m. theory. But the emission spectrum of light and why light from a particular source gave same set of wavelengths remained a problem. In addition to photo-  A.M.Nasr Nature of light 5 electric effect and compton effect. So we conclude that light has a dual character: the wave theory of light and quantum theory of light compliment each other. Either of the two theories by themselves cannot give the complete nature of light. Stand outside in the sun. The shadow your body makes in sunlight suggests that light travels in straight lines from the sun and is blocked by your body. In this, light behaves like a collection of particles fired from the sun. Place two sheets of glass together with a little water will see fringes. These are formed by the interference of waves. We should simply chose whichever theory worked better in solving our problem. 6 Nature of light A.M.Nasr Summary for theories Old theories : (1) Particle Theory (2) Wave Theory Modern theories : (1) Electromagnetic wave theory Light doesn’t need a medium to propagate. 1 Speed of light in medium = √𝜀𝜇 Light consists of range of wavelengths. (2) Quantum Theory Explained: Compton, Photoelectric effects. Explained: the emission spectrum So, light nature is best explained by : Electromagnetic wave and Quantum theories.  A.M.Nasr Nature of light 7 1.2 FERMAT'S Principle Fermat's principle states that: A light ray traveling between any two points will follow a path which requires the least time. (Fermat's principle is sometimes called the principle of least time). Let us illustrate how to use Fermat's principle to drive the laws of reflection and refraction. 1.3 REFLECTION AND REFRACTION 1 2 φ1 φ1 n1 n2 1 incident ray φ2 2 reflected ray n2 >n1 3 3 refracted ray Fig. 1.2: Ray diagram for light crossing the boundary between two media Whenever a ray of light is incident on the boundary separating two different media, part of the ray is reflected back into the first medium and the remainder is refracted as it enters the second medium, as shown in Fig. 1.2. The laws that describe the directions taken by these rays are given below. 8 Nature of light A.M.Nasr 1.3.1 Reflection The incident and reflected light obey two simple laws: i) The incident ray, the reflected ray, and the normal all lie in the same plane, which is perpendicular to the interface separating the two media. ii) The angle of incidence is equal to the angle of reflection as may be concluded from Figure Fig. 1.3: The relation between the angle of incidence and the angle of reflection.  A.M.Nasr Nature of light 9 1.2.2 Refraction The refraction of light occurs because light travels at different speeds in the different media. (see Fig. 1.2). Snell's law Snell's law states that: the value n1 sin 1 = n2 sin 2 , where n1, n2 are the refractive indices of the two media, 1 , 2 are the angles of incidence and refraction, and C C n1 = , n2 = ( How ???? ) v1 2 Where C is the speed of light in a vacuum, and 1, 2 are the speeds of light in the two media respectively. Hint: Angles of incidence, reflection and Refraction are measured relative to the normal ❑ Refractive index When a ray of light (or wavefront) is incident normally (at 90o) to the interface between two media, no refraction is observed. The wavefronts are closer together in the water and even closer together in the glass. That is, 10 Nature of light A.M.Nasr the light travels slower in the water than in the air and even slower in the glass (see Fig. 1.4a). Fig. 1.4b shows ray and wavefront diagrams for light crossing the boundary between air and water, and between air and glass. The ray of light is refracted towards the normal in water because water is an ‘optically’ denser medium than air (water has a grater refractive index than air). The ray is refracted even more towards the normal in the glass because the glass is even more ‘optically’ dense than water (glass has a grater refractive index than water). Optically denser means that the light travels more slowly in the medium. ❑ Reversibility of light Figure 1.5 illustrates the principle of reversibility of light. This principle states that “light traveling from A to B through any system can reverse its path by traveling from B to A through the same system ( x = i , y = r ). A.M.Nasr Nature of light 11 nair=1.0 nair=1.0 nwater=1.3 nglass=1.5 (a) 1- light is not deviated 2- shorter wavelength parallel waves in water than in air and in glass than in water (shorter wavelength in denser medium) nair=1.0 nair=1.0 nwater=1.3 nglass=1.5 (b) 1- refracted ray bends towards the normal 2- refracted ray bends closer to the normal in denser medium Fig. 1.4: Ray and wavefront diagrams for light crossing the boundary between two media 12 Nature of light A.M.Nasr 1 2 n1 i n1 x n2 n2 >n1 n2 n2 >n1 r y 2 1 1 incident 1 incident 2 refracted 2 refracted Fig. 1.5: Principle of reversibility of light When a ray of light passes through a parallel sided block the emergent ray is parallel to the incident ray. The ray is displaced sideways but not deviated. The sideways displacement d is shown in Fig. 1.6. i n1 n2 r n2 >n1 y x d original emergent path ray Fig. 1.6: Refraction through a glass block  A.M.Nasr Nature of light 13 ❑ Change in wavelength In Fig. 1.7 a wave front xx` is traveling in air towards the boundary between air and a medium. In time t the point x has reached A and x` has reached B. Clearly the wave front xx` has advanced to the position AB in time t. In the same amount of time, t, the point B will have reached the point C (X`B=BC), while the point A will have advanced to the point D, where ADn1 Y` D Y Fig. 1.7: Refraction of wave fronts from Snell’s law n1 sin i = n2 sin r, for air n1=1 14 Nature of light A.M.Nasr then, for any medium of refractive index n sin i n= sin r sin x n= , since x=i and y=r sin y BC AD sin x = , sin y = AC AC BC AD n= / AC AC BC speed of light in air x time t C n= = = AD speed of light in medium x time t v wave speed = f λ; the frequency of the light remains the same regardless of the medium in which the light is traveling. It follows that different speeds of light must be due to different wavelengths. So c fair  n= = = air v fmedium medium The wavelength of the wave in the medium will be shorter than the wavelength in the air. n2 sin 1 1 v1 In general = = = (1.1) n1 sin  2 2 v2  A.M.Nasr Nature of light 15 1.4 CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION The angles the ray makes with the normal in air is always larger than the angle it makes with the normal in the medium, i.e. the angle x is always grater than the angle y (Fig. 1.8). X n2 90o n1 n1>n2 i i Y C (a) (b) (c) Fig. 1.8: Critical angle and total internal reflection The angle of incidence in the denser medium for which the angle of refraction in the air is 90o is called the critical angle c. For angles of incidence greater than c, no light can escape from the denser medium and the ray is said to be totally internally reflected. This ray obeys the law of reflection and we can use Snell's law to find the critical angle. n1 sin c = n2 sin 90 = n2 n Sin c = 2 (n1 > n2) n1 16 Nature of light A.M.Nasr Q: Define the two conditions necessary for TIR to occur. A: 1) The refractive index of the first medium is greater than the refractive index of the second one. 2) The angle of incidence θ1, is greater than the critical angle, θc. The phenomenon of TIR causes 100% reflection. In no other situation in nature, where light is reflected, does 100% reflection occur. So TIR is unique and very useful. Example 1.1 An underwater scuba 1 n1=1.0 diver sees the sun at an apparent n2=1.33  2 angle of 45 from the vertical. Where is the Sun? Solution: Refraction happens as sunlight in air crosses into water. The interface is horizontal, so the normal is vertical: n1 sin 1 = n2 sin 2 gives sin 1 = 1.33 sin (45) sin 1 = (1.33)(0.707) = 0.940 The sunlight is at 1 = 70.5 to the vertical  A.M.Nasr Nature of light 17 Example 1.2 A ray of light strikes a flat block of glass (n = 1 1.50) of thickness 2.0 cm at an 2 angle of 30.0 with the normal. 2 cm 3 Trace the light beam through the 4 glass, and find the angles of incidence and refraction at each surface. Solution: At entry: n1 sin 1 = n2 sin 2 1.00 sin 30 = 1.50 sin 2  0.500  and 2 = sin-1    1.50  = 19.5 To do geometrical optics, you must remember some geometry. The surfaces of entry and exit are parallel so their normals are parallel. Then angle 2 of refraction at entry and the angle 3 of incidence at exit are alternate interior angles formed by the ray as a transversal cutting parallel lines. So 3 = 2 = 29.5 At the exit, n2 sin 3 = n1 sin 4 18 Nature of light A.M.Nasr 1.5 sin 19.5 = 1 sin 4 4 = 30.0 The exiting ray in air is parallel to the original ray in air. Thus, a car windshield of uniform thickness will not distort, but shows the driver the actual direction to every object outside. ========== Example 1.3 A ray of light of wavelength 550 nm traveling in air is incident on a slab of transparent material. The incident ray makes an angle of 40 with the normal, and the refracted beam makes an angle of 26 with the normal. a) Find the index of refraction of the material. b) What is the wavelength of light in the material? Solution: a) At entry n1 sin1= n2 sin2 (Snell's law of refraction) 1 = 40, n1 = 1.0 (for air), and 2 = 26 n sin 1 n2 = 1 = 1.47 sin 2  A.M.Nasr Nature of light 19 1 n 2 b) =  2 n1 1n1 550 2 = = = 374 nm n 2 1.47 ❑ Right–angled isosceles prism A glass (or perspex) prism with angles of 90o, 45o, and 45o is a very useful device for reflecting light. Its main advantage is that it totally reflects light were as mirrors absorb some of the incident energy. This type of prism is used in binoculars, periscopes and certain types of telescopes and microscopes (see Fig. 1.8). original direction 45 original direction final 45 direction Angle of deviation 90o final direction Fig. 1.8: Totally internally reflecting prism and deviation of ray though 180o 20 Nature of light A.M.Nasr ❑ Optical fibers An interesting application of total internal reflection is the use of good quality glass (or transparent plastic) rods to “pipe” light from one place to another. A ray of light entering the fiber at an angle greater than the critical angle will be totally internally reflected along the whole length of the fiber as may be seen from Fig.1.9. Fig. 1.9: Light guiding through optical fibers.  A.M.Nasr Nature of light 21 Exercises 1.1 Find the speed of the light in a) water n = 1.33 b) flint glass n = 1.66 c) zircon n = 2.2 1.2 The speed of light in air is 3.0x108 ms-1. Calculate (a) the speed of light in glass of refractive index 1.5. Calculate the wavelength of the light (b) in air, (c) in glass, if the frequency of the light is 6x1014 Hz. 90o 1.3 Diamond has a refractive C index of 2.4, calculate the critical angle c for the diamond. diamond 1.4 (a) What is meant by (i) total internal reflection. (ii) refractive index (b) calculate the least value of the refractive index of a material which produces total internal reflection with an angle of incidence of 45o. 1.5 A beam of light enters a plastic cube air of side 3 cm from the center of the left Plastic side as shown in Fig.. The refractive index of the plastic is 1.5. 75o From which side of the cube will the ray emerge? Draw a diagram to support your answer. B) At what angle will the ray emerge? 22 Nature of light A.M.Nasr o 1.6 Two mirrors make an angle of 120 with each other as shown in fig.. A ray is incident on mirror M1 at an angle of 55o o 65 to the normal. Find the direction of M2 the ray after it is reflected from the mirror M2. 65o 120o M1 1.7 The following are top view diagrams of a beaker of water. Some of the diagrams represent qualitatively correct paths for light through a beaker of water and some do not. Which diagrams have “flaws”? Briefly explain the flaw in each case.  A.M.Nasr Nature of light 23 MCQ 1. Angle between incident ray and normal ray is called angle of a. Reflection b. Refraction c. Transmission d. Incidence 2. Angle of incidence is equal to angle of a. Reflection b. Refraction c. Transmission d. None of the above 3. In swimming pools they appear shallower than they are actual because of a. Reflection b. Refraction c. both a and b d. None 4. Total internal reflection can occur: a. Only when the incident medium is less dense than the transmitting medium. b. Only when the incident medium is denser than the transmitting medium. c. Only when the media are of approximately equal density. d. Irrespective of the density of the media. 5. A light wave in air enters a substance in which light travels at 2.64×108 m/s at an angle of 48o to the surface. The angle of refraction is: 24 Nature of light A.M.Nasr a. 36.1o b. 49.5 c. 57.6 d. 40.8 6. Light can travel in both air and vacuum ( T / F ) 7. Huygens’s Principle is when a light travels between any two points, its actual path will be the one that requires the least time ( T / F ) 8. The incident ray, normal ray and reflected ray all lie in the same plane ( T / F ) 9. Light is said to have a dual nature because it depends upon wave theory only( T / F ) 10. The optical fiber is an application of total internal reflection ( T / F ) A.M.Nasr Nature of light 25 Summary 26 Nature of light A.M.Nasr  A.M.Nasr Light dispersion 27 CHAPTER (2) LIGHT DISPERSION AND ELECTROMAGNETIC SPECTRUM 2.1 DISPERSION AND PRISMS One of the complicating factors in the design of optical instruments is that the glass used for making lenses does not have a constant refractive index. For a given material, the refractive index is a function of the wavelength of light passing through the material. This effect is called dispersion. In particular, when light passes through a prism, a given ray is refracted at two surfaces and emerges bent away from its original direction by an angle of deviation, . Due to dispersion,  is different for different wavelengths. Sources of light can be described in terms of the frequencies of light they produce. Fig. 2.1 shows the different available sources of light including incandescent (continuous), gas discharge (line) and laser (monochromatic) sources. In addition, Fig. 2.2 shows the dispersion of light from different sources, (a) white light, 28 Light dispersion A.M.Nasr (b) light from a hydrogen gas discharge tube, and (c) monochromatic light (laser source). Note: The most common types of illumination sources are incandescent lights and fluorescent lights, both of which produce a continuous spectrum of light somewhat like the solar spectrum. Gas discharge tube, on the other hand, are sources that produce narrow bands of color. Neon light and sodium, mercury, hydrogen vapor lamps are examples of gas discharge tubes. Fig. 2.1: Different sources of light A.M.Nasr Light dispersion 29 60o red orange yellow white spectrum of green light white light blue 60o 60o indigo violet (a) Continuous spectrum 60o Hydrogen source 60o 60o (b) Line spectrum 60o  angle of deviation Laser 60o 60o (c) Monochromatic light Fig. 2.2: Dispersion of light from different sources 30 Light dispersion A.M.Nasr 2.2 ELECTROMAGNETIC SPECTRUM Seven Forms of Light !!!!! Gamma Rays X-rays Ultraviolet Visible Light Infrared Microwaves Radio Waves Light is just one portion of the various electromagnetic waves flying through space. The electromagnetic spectrum covers an extremely broad range, includes wavelengths other than those visible to the human eye. Infrared radiation, visible light and ultraviolet radiation are all members of the same electromagnetic spectrum. All are transverse waves travelling at the speed of light c=3x108 m/s and can travel through a vacuum. Unlike longitudinal sound waves,  A.M.Nasr Light dispersion 31 which must have a material medium for their transmission, electromagnetic waves can travel through a medium or a vacuum. They are characterized by their different frequencies (or wavelengths); ultraviolet radiation has a greater frequency (shorter wavelength) than visible light and infrared a lower frequency (longer wavelength) see Fig.2.2. At shorter wavelengths electromagnetic radiation behaves like particles, whereas toward the long wavelength end of the spectrum the behavior is mostly wavelike. The visible portion occupies an intermediate position, exhibiting both wave and particle properties in varying degrees. 32 Light dispersion A.M.Nasr Q: Write down the properties of electromagnetic radiation A: All forms of electromagnetic radiation have the following properties: 1. they are all transverse waves travelling in free space at c=3x108 ms-1 and obey the inverse square law for intensity; 2. The wave equation c = f λ can be applied to any known wavelength (or frequency f) to determine its frequency f (or wavelength λ); 3. they are all, under suitable conditions, can be reflected and exhibit the phenomenon of interference; 4. they do not necessarily need a medium through which to travel and thus can travel in a vacuum; 5. they can be plane polarized (longitudinal waves such as sound waves cannot be polarized) A.M.Nasr Light dispersion 33 f (Hz) λ (m) 10 -15 10 23 10 -14 γ rays from 10 22 cosmic rays 10 -13 10 21 10 -12 20 γ rays from 10 radioactive source 10 -11 10 19 10 -10 X rays 10 18 10 -9 UV-C Violet UV-B 10 17 10 -8 Ultra-violet Blue UV-A 10 16 10 -7 Green Visible Yellow 10 15 10 -6 Near IR Orange Far IR 10 14 10 -5 Infra-red Red 10 13 10 -4 Radar and 10 12 10 -3 Microwave 10 11 10 -2 10 10 10 -1 Radio waves 9 10 1 TV channels 10 8 10 FM radio 10 7 10 2 10 6 10 3 AM radio 10 5 10 4 10 4 10 5 long radio 10 3 10 6 waves 10 2 10 7 10 1 10 8 10 9 Fig. 2.2 Electromagnetic spectrum 34 Light dispersion A.M.Nasr ❑ Gamma rays Gamma rays have a range of wavelengths from about 10-11 m or below, and in terms of wavelengths overlap the X-ray region. These rays are emitted from within the nucleus of a radioactive nucleus, such as cobalt-60. They can be detected by Geiger-Muller tube and photographic plates and film. They are also used to kill cancerous cells in humans but car is needed in their use because they also attack and kill healthy cells. ❑ X-rays X-rays have a wide range of wavelengths from about 10-11 to 10-8 m and hence overlap into the ultraviolet region at one end and into the gamma-ray region at the other. They are produced when very fast moving electrons are stooped by a heavy metal target. They can be detected by photographic plates and film, producing photo-electric effects on metals and ionization of gases. X-rays are used in radiology, in medicine to produce pictures of internal organs in the body. They are also used to kill dangerous cells and tumors in the body (healthy cells are also killed). X-rays can be used for detecting cracks in metal castings.  A.M.Nasr Light dispersion 35 ❑ Ultraviolet radiation Ultraviolet radiation is just beyond the violet end of the visible spectrum and has wavelengths in the range 10-8 to 4x10-7 m. (there is no sharp distinction between any pair of radiation, as illustrated in the electromagnetic spectrum chart). Ultraviolet radiation is produced by such objects as a carbon–arc lamp, electric spark, discharge tube, mercury vapor lamp, hot bodies and the Sun. The radiation can be detected by photographic film. Ultraviolet radiation is used in burglar alarms, automatic door openers and counters, for detecting real pearls and forged banknote, for photofinishes in races and discriminating between the values of postage stamps which have been phosphor-coated. Ultraviolet light is arbitrarily broken down into three bands, according to its effects. UV-A (315 - 400 nm) is the least harmful and most commonly found type of UV light, because it has the least energy. UV-A light is often called black light, and is used for its relative harmlessness and its ability to cause fluorescent materials to emit visible light - thus appearing 36 Light dispersion A.M.Nasr to glow in the dark. Most phototherapy and tanning booths use UV-A lamps. UV-B (280 - 315 nm) is typically the most destructive form of UV light, because it has enough energy to damage biological tissues, yet not quite enough to be completely absorbed by the atmosphere. UV-B is known to cause skin cancer. Since most of the extraterrestrial UV-B light is blocked by the atmosphere, a small change in the ozone layer could dramatically increase the danger of skin cancer. Short wavelength UV-C (100 - 280 nm) is almost completely absorbed in air within a few hundred meters. When UV-C photons collide with oxygen atoms, the energy exchange causes the formation of ozone. UV-C is almost never observed in nature, since it is absorbed so quickly. Germicidal UV-C lamps are often used to purify air and water, because of their ability to kill bacteria. ❑ Visible light A very narrow band of radiation of wavelength from about 4x10-7 to 7x10-7 m, i.e. from red to violet, is known as the visible spectrum. It is produced by a flame  A.M.Nasr Light dispersion 37 or any incandescent object and can be detected by the human eye, photographic film or a photo-electric cell. ❑ Infrared radiation A band of radiation of wavelength from about 7x10-7 to 10-3 m is known as the infrared region of the spectrum. Infrared light contains the least amount of energy per photon of any other band. Because of this, an infrared photon often lacks the energy required to pass the detection threshold of a quantum detector. Infrared is usually measured using a thermal detector such as a thermopile, which measures temperature change due to absorbed energy. Since heat is a form of infrared light, far infrared detectors are sensitive to environmental changes - such as a person moving in the field of view. Night vision equipment takes advantage of this effect, amplifying infrared to distinguish people and machinery that are concealed in the darkness. Infrared is unique in that it exhibits primarily wave properties. This can make it much more difficult to manipulate than ultraviolet and visible light. Infrared is 38 Light dispersion A.M.Nasr more difficult to focus with lenses, refracts less, diffracts more, and is difficult to diffuse. Most radiometric IR measurements are made without lenses, filters, or diffusers, relying on just the bare detector to measure incident irradiance. This heat radiation is produced from hot bodies such as Sun, electric fires and furnaces. Infrared radiation is used in cooking, heating, drying, infrared photography, and thermal photocopiers. ❑ Radio waves Beyond the infrared region is a wide range of wavelengths from about 10-3 to 105 m, known as radio waves. This region can be subdivided into particular types of wave, e.g. microwaves, radar and television waves. Radio waves are produced by electrical oscillations in circuits containing inductance and capacitance and can be detected by diodes fitted to similar tuned circuits in the receiver. Radio waves as light waves can be totally internally reflect, e.g. by the Earth’s upper atmosphere. The shorter wavelength T.V. waves are not totally internally reflect by the upper atmosphere and hence can only be relayed around the world by  A.M.Nasr Light dispersion 39 satellites. (In fact, all electromagnetic waves can experience total internal reflection in suitable circumstances. Radio waves are used in communications and microwave cooking. Fig. 2.3: Summary for the electromagnetic spectrum characteristics 40 Light dispersion A.M.Nasr Q: Fill the blanks in the table. Production Detection Application Gamma 1- 1- 1- Medicine rays 2- 2- 2- X-rays 1- 1- 1- 2- 2- 2- Ultra- 1- 1- 1- violet 2- 2- 2- Visible 1- 1- human eye 1- waves 2- 2- 2- Infrared 1- Sun 1- 1- waves 2- electric fires 2- 2- Radio 1- 1- 1- Communications waves 2- 2- 2- 2.4 Power of Light The watt (W), the fundamental unit of optical power, is defined as a rate of energy of one joule (J) per second. Optical power is a function of both the number  A.M.Nasr Light dispersion 41 of photons and the wavelength. Each photon carries an energy that is described by Planck's equation: E = hc / λ where E is the photon energy (joules), h is Planck's -34 constant (6.623 x 10 J s), c is the speed of light (2.998 8 x 10 m s-1), and λ is the wavelength of radiation (meters). All light measurement units are spectral, spatial, or temporal distributions of optical energy. As you can see in Fig. 2.3 and Fig. 2.4, short wavelength ultraviolet light has much more energy per photon than either visible or long wavelength infrared. The lumen (1m) is the photometric equivalent of the watt, weighted to match the eye response of the "standard observer". Yellowish-green light receives the greatest weight because it stimulates the eye more than blue or red light of equal radiometric power: 1 watt at 555 nm = 683.0 lumens To put this into perspective: the human eye can detect a flux of about 10 photons per second at a wavelength of 555 nm; this corresponds to a radiant power of 3.58 x 10- 18 W (or J s-1 ). Similarly, the eye can detect a minimum 42 Light dispersion A.M.Nasr flux of 214 and 126 photons per second at 450 and 650 nm, respectively. Photon energy E(J) E = hc / λ Ultraviolet Visible Infrared Wavelength, (meters) Fig. 2.3 Planck's equation showing photon energy vs. wavelength Fig. 2.4 Penetration power for different types of electromagnetic radiation  A.M.Nasr Light dispersion 43 Example 2.1 Light beam containing two wavelengths (red-blue) is incident from air to glass, which angle of refraction is greater, for the red or blue light? (where the wavelength for red is greater than the wavelength for blue) Solution: 44 Light dispersion A.M.Nasr Example 2.2 Light beam containing two wavelengths (red-blue) is incident from glass to air, which angle of refraction is greater, for the red or blue light? (where the wavelength for red is greater than the wavelength for blue) Solution:  A.M.Nasr Light dispersion 45 MCQ 1. By which optical phenomenon does the splitting of white light into seven constituent colors occur? a. Refraction b. Reflection c. Dispersion d. Interference 2. What is the Electromagnetic Spectrum? a. The range of EM waves according to their frequencies b. The vibration of electrically charged particles that surround the electric field c. A And B d. None of the above 3. Arranging in the ascending order of wavelength, which one is true ? a. Gamma-ray, Visible light, Microwaves b. Radio waves, X-ray, Infra-red c. Infra-red, Microwave, Ultraviolet d. Microwaves, visible light, x-ray 4. The visible spectrum is a. > 700 nm b. 100-400 nm c. 400-700 nm d. < 400 5. Which of the following is wrong about the electromagnetic radiation? a. They can be plane polarized b. They are transverse waves travelling in free space at C=3*10^8 m/s c. They can travel in both vacuum and mediums d. They doesn’t obey the inverse square law for intensity 46 Light dispersion A.M.Nasr 6. The electromagnetic spectrum can be used to diagnose and treat illnesses ( T / F ) 7. The index of refraction for a material usually increases with increasing the wavelength ( T / F ) 8. Type of the spectrum in figure is monochromatic spectrum ( T / F ) Sun light 9. Hydrogen vapor lamps are one of examples of line spectrum ( T / F ) 10. If the energy of photon1 is E and the energy of photon 2 is 4E, then λ2 = 4 λ1 ( T / F ) A.M.Nasr Light dispersion 47 Summary 48 Light dispersion A.M.Nasr  ‫‪Interference of light waves‬‬ ‫‪49‬‬ ‫)‪CHAPTER (3‬‬ ‫‪INTERFERENCE OF LIGHT WAVES‬‬ ‫‪3.1 INTRODUCTION‬‬ ‫ال ت ت ت ت‬ ‫مى ت ت ت ت ة ع‬ ‫تعد ظاهرة التداخل ظاهرة عامة‪ ،‬وليست ت ت ت‬ ‫تتتت‬ ‫وحده‪ ،‬بل ه مش ت ت تتترعة لحريا اللرعاذ التوبوبية والر مية‪ ،‬وعرا‬ ‫ف الف تتل الد ا ت افون فاس طستتة لرشىة لرشتتاهدة التداخل ه م ماذ‬ ‫وعولك ف ال ت ذ عددما ص تد م تد اس يت تياس لرة ل ا‬ ‫تح الرا‬ ‫ماصو صف س ال ت ت ت ت ت ذ ف ا شا و خر‬ ‫فس التر والشت ت ت ت تتدة أمحك وم‬ ‫صف س ف ا ضعيف (تداخل بدا وتداخل وهدام)‬ ‫و‬ ‫هداع ي تتع إة ف مصحا م تتد شو ضت ت‬ ‫فتا‬ ‫ما ف ال ت ت‬ ‫هوه ال ت ت ت ت تتع إة ط ت ت ت ت تتت دام‬ ‫مال ه مفو التل ب ع‬ ‫بدفس التر والح‬ ‫م د واحد وي ته و ي ت و لر د واحد‬ ‫ستر و يست وج م ةة مةدية ع‬ ‫وشرفو تىستي م ةة التداخل مل‬ ‫لفر ل) و م ةة خر‬ ‫الثدا‬ ‫تتان م ىس تتام الحة ة الر مية (مثل الردش ت‬ ‫ان م ىسام السعة ( مثل مقيان التداخل لراصف س س) والسرة‬ ‫مةدية ع‬ ‫ته صحتب س ت متد‬ ‫ه‬ ‫الرر ةة ف تحرإتة مو تحتا ا التتداختل ال ت ت ت ت ت ت ت‬ ‫ىحت و متداظرت و ف الر تتد شو ‪ ،‬والر تتا‬ ‫عم ة ل شة ثابتة ب و‬ ‫الرترا فة‬ ‫ية الت تلىق هوه العم ة تسر الر ا‬ ‫ال‬ 50 Interference of light waves In chapter one on geometric optics, we used light rays to examine what happens when light passes through the interface between two different mediums (reflection, refraction, dispersion) The next three chapters are concerned with wave optics, which deals with the interference, diffraction, and polarization of light. These phenomena cannot be explained with the ray optics, but we describe how treating light as waves. If two beams of light can cross without producing any modification of each other, the resultant amplitude and intensity may be very different from the sum of those contributed by two beams acting separately. This modification of intensity obtained by the superposition of two or more beams of light is called interference. If the resultant intensity is greater than that is expected from the separate intensities, this is constructive interference, while if it is zero or in general less, this is destructive interference.  Interference of light waves 51 The interference is constructive if the waves reinforce each other + = Constructive interference (Waves almost in phase) The interference is destructive if the waves tend to cancel each other + = Destructive interference (Waves out of phase) ❑ Conditions of Interference In order to observe interference in light waves, the following conditions must be met: 1. The two sources must have equal intensities. 2. The two sources should have equal amplitudes and frequencies. 52 Interference of light waves 3. The light from two sources must have either zero phase or a constant difference of phase. This condition is very well observed in the case of coherent sources. 4. The common original source must be monochromatic i.e. emitting light of single wavelength. 5. The two sources must be very narrow because a broad source is equivalent to number of sources and so positions of darkness and brightness cannot be well defined.  Interference of light waves 53 ❑ COHERENT SOURCES Two narrow light sources are said to be coherent if they emit light waves of the same wavelength and frequency and also there is a constant phase relation between them. Two independent sources can never act as coherent sources as a source of light consists of large number of atoms and from the atomic theory, each atom consists of a central nucleus around which revolve electrons in various definite orbits. In an excited state an electron jumps to a higher orbit and so the atom becomes unstable. The electron spontaneously fall back to inner orbit within 10-8 seconds and in doing so will give light pulses. The emission of light pulses from various atoms is random and so there is no constant phase relationship from two pulses. So independent sources or different parts of the same sources cannot act as coherent sources. A real source and its virtual image or two virtual images of a single source can act as coherent sources. So in the interference experiments a double slit is used to produce two virtual sources from a single slit, so that the conditions of coherence is satisfied. 54 Interference of light waves The interference of light waves produces an array of bright and dark fringes called an interference pattern. This pattern is difficult to observe, because of the very short wavelength of light. 3.2 YOUNG'S DOUBLE-SLIT EXPERIMENT Young is the first who demonstrate the experiment on the interference of light. A schematic diagram illustrating the geometry used in Young's double-slit experiment is shown in the figure 3.1. Light from a point source was allowed to pass through two small slits S1 and S2, which serve as coherent monochromatic sources. A series of light and dark fringes on a viewing screen are observed.  Interference of light waves 55 Fig. 3.1 Young's double-slit experiment 3.3 THEORY OF INTERFERENCE FRINGES In actual practice, it is not possible to have two independent sources, which are coherent. But for experimental purposes, two virtual sources formed due to a single source can act as coherent sources. Consider a narrow monochromatic source and two slits S1 and S2, equidistant from the source and separated by a distance d. S1 and S2 act as two monochromatic coherent sources. Suppose a screen is placed at a distance L from the coherent sources. The light intensity at any point on the screen is the resultant of light reaching the 56 Interference of light waves screen from both slits. Also, light from the two slits reaching any point on the screen (except the center) travel unequal path lengths. This difference in length of path is called the path difference. The path difference  between the two waves r2 and r1 can be obtained as follows:  = r2 - r1 = d sin. y =d ( ) L Bright fringes (constructive interference) will appear at points on the screen for which the path difference is equal to an integral multiple of the wavelength. The positions of bright fringes can also be located by calculating their vertical distance from the center of the screen (y). In each case, the number m is called the order number of the fringe. The central bright fringe ( = 0, m = 0) is called the zeroth-order maximum. y  = d sin  = d ( ) =  m, m =0, 1, 2, … L λL ybright =  m (for small ) (3.1) d  Interference of light waves 57 Dark fringes (destructive interference) will appear at points on the screen which correspond to path differences of an odd multiple of half wavelengths. For these points of destructive interference, waves which leave the two slits in phase arrive at the screen 180 out of phase. y 1  = d sin = d ( ) =  (m + ), m = 0, 1, 2, … L 2 λL 1 ydark =  (m + 2) (for small ) (3.2) d Unlike bright fringes which start from the zeroth fringe, dark fringes start from the 1st fringe. As a result, the value m in equation 3.2 is equal to the fringe order -1. The distance between any two consecutive bright or dark fringes is known as fringe width  y. 2L L L y = y2- y1 = − = (3.3) d d d Example 3.1 Two monochromatic coherent sources are placed 0.18 mm apart and the fringes are observed on a screen 80 cm away. It is found that the fourth bright fringe is situated at a distance of 10.8 mm from the central fringe. Find the wavelength of light. Solution: L = 80 cm, d = 0.18 mm = 0.018 cm, 58 Interference of light waves m = 4, y = 10.8 mm = 1.08 cm yd 1.08 × 0.018 λ= = = 6.075x10−5 cm = 60.75 μm mL 4 × 80 Example 3.2 In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.6 mm and the screen is 2 m from the slits. Find the distance along the screen between adjacent bright fringes (fringe width)? Solution:  = 486 x 10-9 m, d = 0.6 mm = 0.6 x 10-3 m, L = 2m L y = d 486 x10−9 x 2 y = −3 = 162 x 10-5 m = 1.62 mm 0.6 x10 ========== 3.4 THE INTENSITY PATTERN Consider the double slit experiment in figure 3.1. any two waves leave the slits initially in phase arrive at the screen out of phase by an amount (  ) depends on the path difference . 2 2 The phase difference is = d sin =     Interference of light waves 59 Two waves initially in phase and of equal amplitude (E0) will produce a resultant amplitude at some point on the screen which depends on the phase difference (and therefore on the path difference). Let E1 and E 2 are the displacements of the two waves, such that: E 1 = Eo sin t E 2 = Eo sin (t + φ) E = E 1 + E 2 = Eo sin t + Eo sin (t + φ)   E = 2 Eo cos sin( wt + ) 2 2 The average light intensity (Iav) at any point P on the screen is proportional to the square of the amplitude of the resultant wave. The average intensity can be written:  ❑ as a function of phase difference ; I av. = (2 Eo cos ) 2 2  Iav = 4 Eo 2 cos2 2 ❑ as a function of the angle () subtended by the screen point at the source midpoint d sin   Iav = Io cos2      60 Interference of light waves 2 Y Where Io= 4Eo2 , = d sin  , sin  =  L ❑ as a function of the vertical distance (y) from the 2πdy center of the screen Iav = Io cos2( ) λL The intensity pattern observed on the screen will vary as the number of equally spaced sources is increased; however, the positions of the principle maxima remain the same as shown in Fig. 3.2. (i) when the phase difference =0, 2, 2(2), m(2) or the path difference  = 0, , 2,…..m, then Iav = 4 Eo2 (ii) when the phase difference  = , 3, (m-1/2) 2 or the path difference  =  , 3 ,....(m − 1/ 2) , then 2 2 Iav = 0  Interference of light waves 61 Fig. 3.2 Intensity distribution Example 3.3 Two radio antennas separated by 300 m as in Figure 3.3 simultaneously transmit identical signals at the same wavelength. A radio in a car traveling due north receives the signals. (a) If the car is at the position of the second maximum, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next minimum in reception? Solution: Note, with the conditions given, the small angle approximation does not work well. That is sin , tan , and  are significantly different. The approach to be used is outlined below. (a) At the m = 2 maximum, 400 m 1000 m 300 m Fig. 3.3 62 Interference of light waves 400m tan  = = 0.400 1000m and  = 21.8 so d sin  (300m) sin 21.8) = = m 2  = 55.7 m (b) The next minimum encountered is the third minimum m = 2, and at that point, 1 5 d sin = (m + )  which becomes d sin =  2 2 or sin  = 5 = 5(55.7m) = 0.464 and  = 27.7 2d 2(300m) so y = (1000 m)tan 27.7 = 524 m Therefore, the car must travel an additional 124 m. Example 3.4 In figure 3.1, let L = 120 cm and d = 0.25 cm. The slits are illuminated with coherent 600-nm light. Calculate the distance y above the central maximum for which the average intensity on the screen is 75% of the maximum.  Interference of light waves 63  d sin   Solution: For small , Iav = I0 cos2      y From the drawing, we see that sin  L y = L cos−1 I avg Substituting, we have d I0 In addition, we are given Iav = 0.75I0 (6.0 x10 −7 m)(1.20m) y= −3 cos−1 0.75  (2.5 x10 m) y = 4.8x10-5m = 0.048 mm Example 3.5 In figure 3.1, let L = 1.2 m and d = 0.12 mm, and assume that the slit system is illuminated with monochromatic 500 nm light. Calculate the phase difference between the two wave fronts arriving at P when (a)  = 0.50 and (b) y = 5.0 mm, (c) What is the value of  for which (1) the phase difference is 0.333 rad and (2) the path difference is /4? Solution (a) The path difference is  = d sin (0.12x10-3m) sin 0.5 = 1.05x10-6m The phase difference is 64 Interference of light waves 2 (1.05 x10−6 m) 2 = = = 13.2 rad  (500 x10− 9 m) This is equivalent to 34 y 5.0x10−3 m (b) tan  = =  sin  L 1.2 2d sin  2(0.12x10−3 m)(4.17x10−3 ) = = = 6.28rad,  500x10− 9 m or (equivalently) =0. 2d sin  (c)  = 0.333 =  −9   -1  −1  (500x10 m )( 0.333)  =sin   = sin  −4  = 1.3x10− 2 deg  2 d   2(1.2 x10 m)   = d sin  ; 4 −9 -1    −1  (500x10 m)  −2 =sin   = sin  −  = 5.8 x10 deg  4d  4  4(1.2 x10 m)  Example 3.6 Monochromatic light ( = 632.8 nm) is incident on two parallel slits separated by 0.20 mm. What is the distance to the first maximum and its intensity (relative to the central maximum) on a screen 2.0 m beyond the slits?  Interference of light waves 65 Solution d sin  = m y because sin , and m = 1, L L (632.8x10−9 m)(2.0m) ybright = = d 2.0x10− 4 m ybright = 6.33 mm The relative intensity can be written as I1  d  = cos 2  ybright  I0  L  L I1 But since ybright = , = cos2() = 1 d I0 So the intensity will be approximately the same as at the central maximum. 3.5 CHANGE OF PHASE DUE TO REFLECTION An electromagnetic wave undergoes a phase change of 180 upon reflection from a medium that is optically more dense than the one in which it was travelling. There is also a 180 phase change upon reflection from a conducting surface. 66 Interference of light waves 3.6 INTERFERENCE IN THIN FILMS Interference effects in thin films depend on the difference in length of path traveled by the interfering waves as well as any phase changes, which may occur due to reflection. It can be shown that when light is reflected from a surface of index of refraction greater than the index of the medium in which it is initially traveling the reflected ray undergoes a 180 (or  radians) phase change. In analyzing interference effects, this can be considered equivalent to the gain or loss of a half wavelength in path difference. Therefore, there are two different cases to consider: (a) a film surrounded by a common medium and (b) a thin film located between two different media. These cases are illustrated in figure 3.4. 180o phase 180o phase change change 1 2 1 2 no phase 180o phase n1 < n change n1 < n change n t n t n1 < n n2 > n (a) (b) Fig. 3.4: Interference of light resulting from reflections at two surfaces of a thin film of thickness t, refractive index n.  Interference of light waves 67 In case (a), where a phase change occurs at the top surface, the reflected rays will be in phase (constructive interference) if the thickness of the film is an odd number of quarter wavelengths-so that the path lengths will differ by an odd number of half wavelengths. In case (b), the phase changes due to reflection at both the top and bottom surfaces are offsetting and, therefore, constructive interference for the reflected rays will occur when the film thickness is an integer number of half wavelengths-and, therefore, the path difference will be a whole number of wavelengths. In thin film interference (Fig. 3.4), the wavelength of light in the film, n, is not the same as the wavelength in the surrounding medium.  n = n The conditions for interference in thin films can be stated in terms of the thickness (t) and index of refraction of the film. 2nt=  m +  , m = 0,1,2,… (constructive interference) 1  2 2nt = m, m = 0,1,2,… (destructive interference) 68 Interference of light waves These conditions for constructive and destructive interference are valid only when the film is surrounded by a common medium. Thin film interference conditions summary: Consider the following thin film interference case where the ray reflected from the upper layer (1) interferes with the ray reflected from the lower layer (2). The conditions for constructive and destructive interference can be realized based on the relation between the two rays (1) , (2) as follows: Rays (1) , (2) are in phase : Constructive:2nt = mλ, Destructive:2nt = (m + 0.5)λ Rays (1) , (2) have 180 phase difference :  Interference of light waves 69 Constructive:2nt = (m + 0.5)λ, Destructive:2nt = mλ m=0, 1 , 2 , …… Example 3.7 Find the minimum thickness of a layer of magnesium fluoride (n = 1.38) on flat glass (n = 1.8) that will cause destructive interference of reflected light of wavelength  = 550 nm near the middle of the visible spectrum. Solution: In Fig. (3.5) both rays reflect from a medium of higher refractive index than the medium they are traveling in, so both undergo a phase shift of  rad upon reflection. Air Magnesium n=1.38 t Glass n=1.8 Fig. 3.5 So for destructive interference 1 2nt = (m + ) (m = 0, 1, 2,……) 2 70 Interference of light waves 1 2( 1.38 t ) = ( 550 x 10-9 ) 2 550 x10 −9 t= = 99.6 nm. 4(1.38) Example 3.8 A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat parallel plates of glass (n = 1.5). What must be the thickness of the liquid layer if normally incident 600-nm light is to be strongly reflected? Solution: A phase change of  occurs upon reflection at the first interface but not at the second. The total phase shift of the second reflected wave relative to the first is then:  =  = 2  = 2 (2nt)    (600 x10 −9 m) t= = = 85.4x10-9m 4n 4(1.756 ) Example 3.9 An oil film (n = 1.45) floating on water (n=1.33) is illuminated by white light at normal incidence. The film is 280 nm thick. Find the dominant observed color in the reflected light. (visible 400nm – 700 nm) Solution: The light reflected from the top of the oil film undergoes phase reversal. Since 1.45 > 1.33, the light  Interference of light waves 71 reflected from the bottom undergoes no reversal. For constructive interference of reflected light, we then have 2nt =  m +   or 1 2 nt 2(1.45)( 280 nm) = =  2 (m + ) 1 (m + ) 1 2 2 (a) for m = 0,  = 1624 nm (infrared) m = 1,  = 541 nm (green) m = 2,  = 325 nm (ultraviolet) Infrared and ultraviolet are invisible to the human eye, so the dominant color is green One of the very important applications of thin film interference is the use of nonreflecting coatings for camera lenses. These coatings reduce the loss of light at various surfaces of multiple lens system, as well as prevent internal reflections, which might mar the image. 72 Interference of light waves 3.7 METHODS OF GETTING INTERFERENCE PATTERN 3.7.1 FRESNEL'S BIPRISM Fresnel used a biprism to show interference phenomenon. The biprism abc consists of two acute angled prisms placed base to base, (Fig. 3.6). Fig. 3.6 Interference by Fresnel’s biprism Actually, it is constructed as a single prism of obtuse angle of about 179. The acute angle  on both sides is about 30` to 1o. The prism is placed as shown in figure. When light falls from S on the lower portion of the prism it is bent upwards and appears to come from the virtual source B. Similarly, light falling from S on the upper  Interference of light waves 73 portion of the prism is bent downwards and appears to come from the virtual source A. Therefore, A and B act as two coherent sources. Interference fringes of equal width are produced between E and F. To observe the fringes, the screen can be replaced by an eyepiece or a low power microscope. Theory: The point O is equidistant from A and B. Therefore, it has maximum intensity. On both sides of O, alternately bright and dark fringes are produced. As in double slit experiment: L The fringe width  y = , the central fringe is bright d and is called the zeroth order fringe. The condition for bright fringe L y bright = m where m = 0, 1, 2, …… d The condition for dark fringe (m+0.5)λ.L y= where m = 0, 1, 2, ……. d Unlike bright fringes which start from the zeroth fringe, dark fringes start from the 1st fringe. So, the value m for the dark fringes condition is equal to the fringe order -1. 74 Interference of light waves Determination of wavelength of light: Fresnel's biprism can be used to determine the wavelength of a given source of monochromatic light. The eyepiece is moved horizontally to determine the fringe width. Suppose for crossing 2 bright fringes from the field of view, the eyepiece has moved through a distance . ℓ Then the fringe width,  y = 2 L But the fringe width,  y = d y d = L  Interference of light waves 75 Example 3.10 A biprism is placed at a distance of 5 cm in front of a narrow slit, illuminated by sodium light ( = 5890 x 10-8 cm) and the distance between the virtual sources is found to be 0.05 cm. Find the width of the fringes observed in an eyepiece placed at a distance of 75 cm, from the biprism. Solution:  = 5890 x 10-8 cm, d = 0.05 cm. L = 5 + 75 = 80 cm L width of the fringe,  y = d 5890 x10−8 x80 y = = 9424 x 10-5 cm. 0.05 Example 3.11 In a biprism experiment the eyepiece was placed at a distance of 120 cm from the source. the distance between the two virtual sources was found to be equal to 0.075 cm. Find the wavelength of light of the source if the eyepiece has to be moved through a distance 1.888 cm for 20 fringes to cross the field of view. Solution n = 20,  = 1.888 cm  1.888  Fringe width  y = = cm n 20 76 Interference of light waves d = 0.075 cm, L = 120 cm y.d 1.888 0.075 = = x = 5900 x 10-8 cm = 590 nm L 20 120 Determination of the thickness of a thin film of transparent material The biprism experiment can be used to determine the thickness of a given thin film of transparent material e.g. glass or mica. F P A y d O B L Fig 3.7 Effect of thin transparent sheet  Interference of light waves 77 Suppose A and B are two virtual coherent sources. The point O is equidistant from A and B. When a transparent film F of thickness t and refractive index n is introduced in the path of one of the beams Fig. 3.7, the fringe which was originally at O shifts to P. The time taken by the wave from B to P in air is the same as the time taken by the wave from A to P partly through air and partly through the film. BP The time taken from B to P = C where C the velocity of light in air AP − t t The time taken from A to P = + C v where v the velocity of light in the medium BP AP − t t  = + C C v C But =n v C  BP = AP + t + t n BP = AP - t + nt BP-AP = nt - t = (n-1)t 78 Interference of light waves If P is the point originally occupied by the mth fringe, then the path difference BP - AP = m  (n-1) t = m Also, the distance through which the fringe shifted is: m L y= d L where = ∆y, the fringe width. d yd  m = L y.d and (n - 1)t = L (𝑛 − 1)𝑡 = 𝑚𝜆 Example 3.12 A thin film of transparent material (n = 1.6) is placed in the path of one of the interfering beams in a biprism experiment using sodium light,  = 589nm. The central fringe shifts to a position normally occupied by the 12th bright fringe. Calculate the thickness of the film. Solution n = 1.6,  = 589 nm, m = 12  (n - 1) t = m  Interference of light waves 79 m 12 (589 x10 −7 ) t= = = 0.01178 mm (n − 1) (1.6 − 1) Example 3.13 When a thin film of transparent material of thickness 6.3x10-3 mm is introduced in the path of one of the interfering beams, the central fringe shifts to a position occupied by the sixth bright fringe. If  = 546 nm, find the refractive index of the film. Solution t = 6.3 x 10-3 cm,  = 546 x 10-7 cm, m=6  (n-1) t = m m 6(546x10− 7 )  n= +1 = + 1 = 1.52 t 6.3x10− 4 Example 3.14 Consider the double-slit arrangement, where the separation d of the slits is 0.3 mm and the distance L to the screen is 1 m. A very thin film of transparent plastic, with a thickness of t = 0.05 mm and a refractive index of n = 1.5, is placed over only the upper slit. Find the distance by which the central maximum of the interference pattern moves upward? 80 Interference of light waves Solution y.d (n - 1) t = L (n − 1)t L (1.5 − 1)(0.005)(100) y= = = 8.33 cm d 0.03 3.7.2 NEWTON'S RINGS When a plano-convex lens of long focal length is placed on a plane glass plate, a thin film of air (or some other transparent medium) is enclosed between the lower surface of the lens and the upper surface of the plate. The thickness of the air film is very small at the point of contact and gradually increases from the center outwards. The fringes produced with monochromatic light are circular. The fringes are concentric circles, uniform in thickness and with the point of contact as the center. Suppose the radius of curvature of the lens is R and the air film of thickness t is at a distance r, from the point of contact o as shown in figure 3.8.  Interference of light waves 81 Fig.3.8 Newton Rings Experiment Here, interference is due to reflected light. Therefore, for the dark ring 2nt = m (m = 0, 1, 2,…) (i) for the bright ring 1 2nt = (m + ) (m = 0, 1, 2,……) (ii) 2 As the central fringe (zeroth) is dark, then the value of m in equation (ii) is equal to the fringe order -1. for air n=1 From Fig. (3-7) r2 = R2 - (R - t)2 = R2 – (R2 - 2Rt – t2 ) Since t is very small, we may drop terms in t2 to obtain r2  2Rt 82 Interference of light waves r2 r2 1 t , Sub. in (i), then 2n = (m + ) 2R 2R 2 For bright rings 1 r2 = (m + ) R (n=1) 2 1 r = ( m + ) R (m = 0, 1, 2, …..) 2 For dark rings r2 = m  R  r = m R (m = 0, 1, 2, ….) when m = 0, the radius of the dark ring is zero and the R radius of the bright ring is. Therefore, the center is 2 dark. Alternately dark and bright rings are produced Fig. 3.9.  Interference of light waves 83 Fig. 3.9 Newton's rings The radii of the dark ring and bright ring depend on m, , R and n. Example 3.15 In a Newton's rings experiment, the radius of the 12th, dark ring changes from 1.4 cm to 1.27 cm when a liquid is introduced between the lens and the plate. Calculate the refractive index of the liquid Solution For air : rd = √mλ0 R/n → 1.4 × 10−2 = √12λ0 R/1 1.96 × 10−4 = 12λ0 R (1) For liquid : rd = √mλ0 R/n → 1.27 × 10−2 = √12λ0 R/n 1.61 × 10−4 = 12λ0 R/n (2) −4 1.96 × 10−4 from (1)into (2) → 1.61 × 10 = n n = 1.215 84 Interference of light waves 3.7.3 MICHELSON INTERFEROMETR ❑ Principle The amplitude of light beam from a source is divided into two parts of equal intensities by partial reflection and transmission. These beams are then sent in two directions at right angles and are brought together after they suffer reflection from plane mirrors to produce interference fringes. ❑ Construction and working It consists of two highly polished mirrors M1 and M2 and two plane glass plates A and C parallel to each other. The rear side of the glass plate A is half silvered so that light coming from the source S is equally reflected and transmitted by it. Light from a monochromatic source S falls on the plate A. The plate A is inclined at an angle of 45. One half of the energy of the incident beam is reflected by the plate A towards the mirror M1 and the other half is transmitted towards the mirror M2. These  Interference of light waves 85 two beams (reflected and transmitted) travel along two mutually perpendicular paths and are reflected back by the mirrors M1 and M2. These two beams return to the plate A. The beam reflected back by M1 is transmitted through the glass plate A and the beam reflected back by the mirror M2 is reflected by the glass plate A towards the eye (Fig. 3.10). The beam going towards the mirror M1 and reflected back, has to pass twice through the glass plate A. Therefore, to compensate for the path the plate C is used between the mirror M2 and A. The light beam going towards the mirror M2 and reflected back towards A also passes twice through the compensating plate C. Therefore, the paths of the two rays in glass are the same. The mirror M1 can be moved. The image of M2 is at M`2 parallel to M1. Hence, M`2 and M1 form the equivalent of a parallel air film. The effective thickness of the air film is varied by moving mirror M1 parallel to itself. Under these conditions, the interference pattern is a series of bright and dark circular rings which resemble newton's rings. If a dark circle appears at the center of the pattern, the two rays interfere destructively. If the mirror M1 is then moved a distance of /4, the path difference changes by /2. The two rays will now interfere constructively, 86 Interference of light waves giving a bright circle in the middle. As M1 is moved an additional distance of /4, a dark circle will appear once again. the wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1. Fig. 3.10 The basic component of Michelson interferometer ❑ Application of Michelson interferometer Some of the applications of Michelson interferometer are given as follows: a) Optical testing of various transparent materials.  Interference of light waves 87 A Michelson interferometer uses a beam splitter to create two different optical paths. This can be used for optical testing. If the plate distorts the waves, this will show up in the interference fringes as shown in figure 3.10. b) The refractive index or thickness of various transparent materials. In the path of the rays going towards M1 a thin transparent film is introduced and the fringes are obtained in the center of the field of view. In that case, if the refractive index of the film is n and the thickness is t, and the number of fringes crossing the center of the field of 88 Interference of light waves view is m, the path difference introduced between the two interfering beams will be 2 (n-1) t  2 (n - 1) t = m  m n= +1 2t If t is known we can find n, or if n is known we can find t. c) The wavelength of a given monochromatic light. If M1 and M2 are equidistant from the glass plate A, the field of view will be perfectly dark. The Mirror M2 is kept fixed and the mirror M1 is moved and the number of  Interference of light waves 89 fringes that cross the field of view is counted. Suppose, for the monochromatic light of wavelength , the distance through which the mirror is moved equal d and the number of fringes that cross the center of the field of view = m. Then, m d= 2 because for one fringe shift, the mirror moves through a  distance equal to. Hence  can be determined. 2 90 Interference of light waves Example 3.16 In an experiment for determining the refractive index of air with Michelson interferometer a shift of 150 fringes is observed when all the air was removed from the tube. If the wavelength of the light used is 4000 A in air and the length of the tube is 20 cm. Calculate the refractive index of air. Solution Here, m = 150,  = 4000 A = 400 x 10-7 cm L = 20 cm, n=? 2 (n - 1) L= m  m 150 x 400 x10 −7. n= +1 = +1 2L 2 x 20 = 1.00015  Interference of light waves 91 Example 3.18 In moving one mirror in a Michelson interferometer through a distance of 0.1474 mm, 500 fringes cross the center of the field of view. What is the wavelength of light? Solution Here, m = 500, d = 0.1474 mm = 0.01474 cm m  d= 2 2d 2 x0.01474  = m 500 = 5896 x 10-8 cm = 589.6 nm 92 Interference of light waves Exercises 1- A pair of narrow, parallel slits separated by 0.25 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.2 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands. 2- A laser beam (λ = 632.8 nm) is incident on two slits 0.20 mm apart. Approximately how far apart are the bright interference lines on a screen 5.0 m away from the double slits? 3- Young's double-slit experiment is performed with 589-nm light and a slit-to-screen distance of 2.00 m. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. 4- Two narrow parallel slits separated by 0.85 mm are illuminated by λ = 600 nm light, and the viewing screen is 2.80 m away from the slits. (a) What is the phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe? (b) What is the ratio of the intensity at the center of a bright fringe?  Interference of light waves 93 5- A thin film of oil ( n= 1.25 ) covers a smooth wet pavement. When viewed perpendicular to the pavement, the film appears to be predominantly red (λ = 640 nm ) and has no blue (λ = 512 nm ). How thick is it? 6- Mirror M1 is displaced a distance L during this displacement, 250 fringes ( formation of successive dark or bright bands) are counted. The light being used has a wavelength of 632.8 nm. Calculate the displacement L. 7- Calculate the minimum thickness of a soap bubble film ( n= 1.33) that results in constructive inter