Engineering Physics PDF - SIT HUB

FOREWORD This is the second iteration of preparing our own courseware material, after successful completion of a similar task undertaken a few years ago. These contents have been carefully prepared and should serve as excellent auxiliary material for both instructors and students. The Special Acad...

FOREWORD This is the second iteration of preparing our own courseware material, after successful completion of a similar task undertaken a few years ago. These contents have been carefully prepared and should serve as excellent auxiliary material for both instructors and students. The Special Academic Group, Autonomy (SAGA) was formed for the sole purpose of preparing courseware contents primarily in the first-year theory subjects; a few second-year subjects were also included. The subjects for the first year were - Basic Electrical Engineering, Basic Electronics Engineering, Computer Programming, Data Structures and Algorithms, Engineering Mathematics I and II, Engineering Physics, Engineering Chemistry, Constitution of India, Environmental Science and Engineering and Communicative and Technical English. For the second year, the subjects for which courseware material was prepared were Analog Electronic Circuits, Digital Systems Design, Circuit Theory and Measurements and Instruments. Faculty members from all the departments contributed to the task. They were, in no particular order, Nalini Singh, Bimal Meher, Saumyaranjan Dash, Mukti Routray, Susmita Biswal, Manasa Dash, Bipin Tripathy, Sibasankar Nayak, Janmejay Senapati, Subrat Sahu, Pradeep Moharana, Rupambika Pattanaik, Dhananjay Tripathy, Jagadish Patra, Sachin Das, Deepak Ranjan Nayak, Amulya Roul, Bodhisattva Dash, Sanghamitra Das, Gyana Ranjan Biswal, Nibedita Swain and Rajan Mishra. The entire group worked diligently to successfully complete the task which included a peer review of the material. I take this opportunity to thank all the members of the SAGA group for a job well done. I sincerely hope that this courseware material comes in handy and is utilized to the fullest extent. These are readily available additional resources prepared in accordance with the Silicon autonomy syllabus, to complement textbooks and classroom lectures. If there are any errors, I would be grateful if they are brought to my notice so that we can correct them in subsequent versions. Dr. Jaideep Talukdar, Principal Silicon Institute of Technology Bhubaneswar December 2020  Module – 1: WAVES 1.0 Wave: a wave can be described as a disturbance that travels through a medium, transporting energy from one location to another location without transporting matter. As a disturbance moves through a medium from one particle to its adjacent particle, energy is being transported from one end of the medium to the other. The general name of this disturbance or wave is called progressive wave. Types of waves: Waves are of two types Transverse wave Longitudinal wave If the disturbance takes place perpendicular to the direction of propagation of wave, the wave is called transverse wave. If the disturbance takes place along the direction of propagation of wave, then it is called longitudinal wave. Wave as a periodic variation in space and time A wave is a disturbance that travels through a medium from one location to another. A wave is the motion of a disturbance that moves through a medium in such a manner that at any position, the displacement of the particle of the medium is a function of time, and at any instant, the displacement of the particles of the medium is a function of the position at that point. The disturbance that is propagated is called a progressive or a travelling wave. During its propagation, energy is transferred through the medium  At a given instant, the wave periodically varies with position and at a given position, wave varies periodically with time. Example: In sound wave, the disturbance is a pressure variation in a medium. In light wave, the disturbance is the variation of the strength of electric and magnetic fields. 1.1 Wave in one dimension:  ( x, t ) = A sin(kx  wt ) = A sin[k ( x  vt)] If we consider the initial phase factor Ø 𝜓(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛( 𝑘𝑥 ± 𝑤𝑡 + Ø) Wave Characteristics: ❖ Amplitude (A) – Maximum displacement of particle of the medium from its equilibrium point. The bigger the amplitude, the more energy the wave carries. ❖ Wavelength ( ) – Distance between two consecutive crests (max positive displacement) or between two consecutive trough (max negative displacement). ❖ Wave vector or propagation vector (k) – 2 /  ❖ Period (T) –The time required for one complete oscillation. It is the reciprocal of frequency: T = 1 / f. ❖ Frequency (f) – The number of oscillations/vibrations per second. The SI unit for frequency is the Hertz (Hz). ❖ Wave speed (v) – How fast the wave is moving (the disturbance). Speed depends on the medium. It is expressed as v =  f. 1.2 Wave equation in differential form: Let’s take the equation, 𝜓 (𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛( 𝑘𝑥 − 𝑤𝑡 + 𝜙) 𝜕𝜓 = −𝑤𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝑤𝑡 + 𝜙) 𝜕𝑡 𝜕2𝜓 = −𝑤 2 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝑤𝑡 + 𝜙) = −𝑤 2 𝜓 𝜕𝑡2 Again, 𝜕𝜓 = 𝑘𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝑤𝑡 + 𝜙) 𝜕𝑥 𝜕2𝜓 = −𝑘 2 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝑤𝑡 + 𝜙) == −𝑘 2 𝜓 𝜕𝑥 2  𝑘2 𝜕2 𝜓 1 𝜕2 𝜓 = 𝑤2 𝜕𝑡2 = 𝑣2 𝜕𝑡2 𝝏 𝟐𝝍 𝟏 𝝏 𝟐𝝍 So, 𝟐 = 𝝏𝒙 𝒗 𝟐 𝝏𝒕 𝟐 It is the general wave equation, in second order differential form with wave velocity/ phase 𝒘 velocity 𝒗 =. 𝒌 1.3 Reflection and transmission of a transverse wave at boundary of two media When a wave strikes an obstacle, or comes to the end of the medium in which it is traveling, part of it is reflected and a part is transmitted ✓ Incident wave – the wave that strikes the boundary ✓ Reflected wave – the wave that is returned as a result of reflection ✓ Transmitted wave – the wave that is propagated to the second medium 1.3.1 Rarer to Denser medium:  Parameters Reflected Component Transmitted Component Amplitude Decreases Decreases Velocity No change Decreases Frequency No Change No change Wavelength No change Decreases Phase Changes by π No change 1.3.2 Denser to Rarer medium Parameters Reflected Component Transmitted Component Amplitude Decreases Decreases Velocity No change Increases Frequency No Change No change Wavelength No change Increases Phase No change No change Superposition 2.1 Principle of Superposition: The principle of superposition states that the resultant displacement of a particle of a medium acted upon by two or more waves simultaneously is the algebraic sum of the displacement of the same particle due to individual waves. Let the displacement of the particle due to individual waves are 1 and 2 Then the resultant displacement is 𝜳 = 𝜳 𝟏 + 𝜳 𝟐 (If both are in same direction) and 𝜳 = 𝜳 𝟏 − 𝜳𝟐 (If both are in opposite direction) 2.2 Types of Superposition: Coherent Superposition Incoherent Superposition Coherent Superposition Phase difference remains constant. The resultant intensity differs from sum of the individual intensity. 𝐼 ≠ 𝐼1 + 𝐼2. It gives an interference (fringe) pattern. Incoherent Superposition Phase difference changes frequently and randomly. The intensity of the resultant wave is equal to the sum of the intensities of the component waves. 𝐼 = 𝐼1 + 𝐼2. Cannot form interference pattern because their intensity is constant. 2.3 Two beam Superposition: Let’s take two beams or two waves having different amplitude, same frequency and different phases superpose 1 = 𝐴1 sin (𝑘𝑥 − 𝑤𝑡 + 𝜙1 ) 2 = 𝐴 2 sin(𝑘𝑥 − 𝑤𝑡 + 𝜙2 ) The resultant wave is  = 1 + 2  = 𝐴1 sin (𝑘𝑥 − 𝑤𝑡 + 𝜙1 ) + 𝐴 2 sin(𝑘𝑥 − 𝑤𝑡 + 𝜙2 )  = 𝐴1 [sin (𝑘𝑥 − 𝑤𝑡)𝑐𝑜𝑠𝜙1 + cos(𝑘𝑥 − 𝑤𝑡) 𝑠𝑖𝑛𝜙1 ] + 𝐴 2 [sin (𝑘𝑥 − 𝑤𝑡)𝑐𝑜𝑠𝜙2 + cos(𝑘𝑥 − 𝑤𝑡) 𝑠𝑖𝑛𝜙2 ] = sin(𝑘𝑥 − 𝑤𝑡)[𝐴1 𝑐𝑜𝑠𝜙1 + 𝐴 2 𝑐𝑜𝑠𝜙2 ]+ cos(𝑘𝑥 − 𝑤𝑡)[𝐴1 𝑠𝑖𝑛𝜙1 + 𝐴 2 𝑠𝑖𝑛𝜙2 ] Let 𝐴𝑐𝑜𝑠𝜙 = [𝐴1 𝑐𝑜𝑠𝜙1 + 𝐴 2 𝑐𝑜𝑠𝜙2 ] 𝐴𝑠𝑖𝑛𝜙 = [𝐴1 𝑠𝑖𝑛𝜙1 + 𝐴 2 𝑠𝑖𝑛𝜙2 ] Then  = sin(𝑘𝑥 − 𝑤𝑡)𝐴𝑐𝑜𝑠𝜙 + cos(𝑘𝑥 − 𝑤𝑡)𝐴𝑠𝑖𝑛𝜙 = 𝐴sin (𝑘𝑥 − 𝑤𝑡 + 𝜙) So the resultant wave is of same nature as that of the incident wave. Where A is the resultant amplitude and 𝜙 is the resultant phase. So 𝐴 2 = (𝐴𝑠𝑖𝑛𝜙) 2 + (𝐴𝑐𝑜𝑠𝜙) 2 = (𝐴1 𝑠𝑖𝑛𝜙1 + 𝐴 2 𝑠𝑖𝑛𝜙2 )2 + (𝐴1 𝑐𝑜𝑠𝜙1 + 𝐴 2 𝑐𝑜𝑠𝜙2 ) 2 = 𝐴1 2+𝐴 2 2 + 2𝐴1 𝐴 2 cos(𝜙1 − 𝜙2 ) The third term of above equation is the interference term. The resultant intensity depends upon the interference term. 2.3.1 Coherent superposition: In this case phase difference remains constant, so (𝜙1 − 𝜙2 ) is constant Case I If(𝜙1 − 𝜙2 ) = 2𝑛𝜋, where 𝑛 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑒 = 0, ±1, ±2, ±3, … … … … Then cos(𝜙1 − 𝜙2 )= 1, 𝐴 2 = (𝐴1 + 𝐴 2 )2 So intensity 𝐼 = 𝐼𝑚𝑎𝑥 = (𝐴1 + 𝐴 2 ) 2 Special case If 𝐴1 = 𝐴 2 = 𝑎, Then 𝐼𝑚𝑎𝑥 = 4𝑎 2 , that means the resultant intensity (intensity of bright fringe) is four times the intensity of the individual wave. Case II If (𝜙1 − 𝜙2 ) = (2𝑛 + 1)𝜋, where 𝑛 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑒 = 0, ±1, ±2, ±3, … … … … Then cos(𝜙1 − 𝜙2 )= -1, 𝐴 2 = (𝐴1 − 𝐴 2 ) 2 So intensity 𝐼 = 𝐼𝑚𝑖𝑛 = (𝐴1 − 𝐴 2 )2 Special case If 𝐴1 = 𝐴 2 = 𝑎 , Then 𝐼𝑚𝑖𝑛 = 0, that means the resultant intensity is zero. Dark fringe is formed. So, in case of coherent superposition alternate bright and dark fringes appear. 2.3.2 Incoherent superposition: In this case phase difference changes frequently and randomly, so any particular value cannot be taken. So, time average value of cos(Ø1 − Ø2 ) is considered. But< cos( 𝜙1 − 𝜙2 ) >= 0, as its value varies in between +1 to -1. So, 𝐴 2 = 𝐴1 2+𝐴 2 2 Special case If 𝐴1 = 𝐴 2 = 𝑎 , then 𝐴 2 = 2𝑎 2 , which is constant always. So, no fringe pattern is observed. Intensity distribution curve 2.4 Multiple beams Superposition: Let’s take a number of beams or waves having different amplitude, same frequency and different phases superpose 1 = 𝐴1 sin (𝑘𝑥 − 𝑤𝑡 + 𝜙1 ) 2 = 𝐴 2 sin(𝑘𝑥 − 𝑤𝑡 + 𝜙2 ) 3 = 𝐴 3 sin(𝑘𝑥 − 𝑤𝑡 + 𝜙3 ) n = 𝐴 n sin(𝑘𝑥 − 𝑤𝑡 + 𝜙𝑛 ) The resultant wave is  = 1 + 2 + 3 + ………+n n 𝑛  = ∑ i = ∑ 𝐴 𝑖 𝑠 𝑖𝑛(𝑘𝑥 − 𝑤𝑡 + 𝜙𝑖 ) i=1 𝑖=1 Where 𝐴 𝑖and Ø𝑖 represents amplitude and phase of the i th component wave. By comparison with two beam superposition the resultant wave is  𝛹 = 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝑤𝑡 + Ø) where 𝐴 𝑠𝑖𝑛Ø = [𝐴1 𝑠𝑖𝑛𝜙1 + 𝐴 2 𝑠𝑖𝑛𝜙2 + 𝐴 3 𝑠𝑖𝑛𝜙3 + ⋯ + 𝐴 𝑛 𝑠𝑖𝑛𝜙𝑛 ]=∑𝑛𝑖=1 𝐴 𝑖 𝑠𝑖𝑛𝜙𝑖 and 𝐴 𝑐𝑜𝑠Ø = [𝐴1 𝑐𝑜𝑠𝜙1 + 𝐴 2 𝑐𝑜𝑠𝜙2 + 𝐴 3 𝑐𝑜𝑠𝜙3 + ⋯ + 𝐴 𝑛 𝑐𝑜𝑠𝜙𝑛]=∑𝑛𝑖=1 𝐴 𝑖 𝑐𝑜𝑠𝜙𝑖 So 𝐴 2 = (∑𝑛𝑖=1 𝐴 𝑖 𝑠𝑖𝑛Ø𝑖 ) 2 + (∑𝑛𝑖=1 𝐴 𝑖 𝑐𝑜𝑠𝜙𝑖 ) 2 =∑𝑛𝑖=1 𝐴 2𝑖 + 2 ∑𝑛𝑖,𝑗=1 𝐴 𝑖 𝐴𝑗 cos(𝜙𝑖 − 𝜙𝑗 ) 𝑖≠𝑗 Discussion: 2.4.1 Coherent superposition: (𝜙𝑖 − 𝜙𝑗 ) remains constant (𝜙𝑖 − 𝜙𝑗 ) =2nπ where 𝑛 = 0, ±1, ±2, ±3, … … … … So, cos(𝜙𝑖 − 𝜙𝑗 ) =1 Then 𝐴 2 = =∑𝑛𝑖=1 𝐴 2𝑖 + 2 ∑𝑛𝑖,𝑗=1 𝐴 𝑖 𝐴𝑗 𝑖≠𝑗 Example Let n=3 so 𝐴 2 = 𝐴12 + 𝐴 22 + 𝐴 23 + 2𝐴1 𝐴 2 + 2𝐴 2 𝐴 3 + 2𝐴1 𝐴 3 =(𝐴1 + 𝐴 2 + 𝐴 3 )2 Special Case: If 𝐴1 = 𝐴 2 = 𝐴 3 = 𝑎 Then 𝐴 2 = (3𝑎) 2=9𝑎 2 So, in general 𝑰 𝒄𝒐𝒉 = (𝒏𝒂) 𝟐 where ‘n’ is the number of beams superpose. 2.4.2. Incoherent superposition: (𝜙𝑖 − 𝜙𝑗 ) changes frequently and randomly So < cos (𝜙𝑖 − 𝜙𝑗 ) >= 0 Then 𝐴 2 =∑𝑛𝑖=1 𝐴 2𝑖 Example Let n=3 so 𝐴 2 = 𝐴12 + 𝐴 22 + 𝐴 23 Then 𝐴 2 = 3𝑎 2 So, in general 𝑰 𝒊𝒏𝒄𝒐𝒉 = 𝒏𝒂𝟐 where ‘n’ is the number of beams superpose. 𝑰 𝒄𝒐𝒉 = 𝒏𝑰 𝒊𝒏𝒄𝒐𝒉 2.5. Superposition of multiple waves with a constant and successive phase difference: Consider ‘N’ number of waves having same amplitudes, frequencies but phase varies with a constant amount (Ф, say) and are represented as follows; 𝜓1 = 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) 𝜓2 = 𝐴𝑠𝑖𝑛 (𝑘𝑥 − 𝜔𝑡 + 𝜙) 𝜓3 = 𝐴𝑠𝑖𝑛 (𝑘𝑥 − 𝜔𝑡 + 2𝜙) 𝜓𝑁 = 𝐴𝑠𝑖𝑛 (𝑘𝑥 − 𝜔𝑡 + [𝑁 − 1]𝜙) Phase difference between successive waves is Ø. The resultant superposed wave can be written as; 𝜓 = 𝜓1 + 𝜓 2 + ⋯ + 𝜓𝑁 In order to get the resultant wave in a convenient way we can express the wave function in exponential form 𝜓1 = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) 𝜓2 = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡+𝜙) 𝜓3 = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡+2𝜙) Similarly, 𝜓𝑁 = 𝐴𝑒 𝑖( 𝑘𝑥−𝜔𝑡+( 𝑁−1)𝜙) So the resultant superposed wave can be written as; 𝜓 = 𝜓1 + 𝜓2 + ⋯ + 𝜓 𝑁 𝜓(𝑥, 𝑡) = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) {1 + 𝑒 𝑖𝜙 + 𝑒 𝑖2𝜙 + 𝑒 𝑖3𝜙 + ⋯ + 𝑒 𝑖(𝑁−1) 𝜙 }……………(i) Using the Algebraic Formulation; 1−𝑥 𝑚+1 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑚 = 1−𝑥 Now, equation (i) can be written as; 1−𝑒𝑖𝑁𝜙 𝑒𝑖𝑁𝜙−1 𝜓 = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) { } = 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) { } 1−𝑒𝑖𝜙 𝑒𝑖𝜙 −1 With suitable arrangement of the exponent power terms; 𝑒 𝑖( 𝑁⁄2)𝜙 [𝑒𝑖 (𝑁⁄2) 𝜙−𝑒−𝑖(𝑁⁄2)𝜙] 𝜓= 𝐴𝑒 𝑖(𝑘𝑥−𝜔𝑡) 𝑒𝑖 (𝜙⁄2)[𝑒𝑖(𝜙⁄2)−𝑒−𝑖(𝜙⁄2)]  𝑁𝜙 𝑁𝜙 𝑖(𝑘𝑥−𝜔𝑡+ 𝑁−1 𝜙) (2𝑖𝑠𝑖𝑛 ) 𝑁−1 𝑖(𝑘𝑥−𝜔𝑡+ 𝜙) 𝑠𝑖𝑛 2 2 𝜓 = 𝐴𝑒 2 𝜙 = 𝐴𝑒 2 𝜙 ………….(ii) 2𝑖𝑠𝑖𝑛 𝑠𝑖𝑛 2 2 If we represent the resultant wave function as 𝜓 = 𝑅𝑒 𝑖(𝑘𝑥−𝑤𝑡+𝜃)…………(iii) where R------- Resultant amplitude 𝜃 ⋯ ⋯ ⋯ 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑝ℎ𝑎𝑠𝑒 Then comparing equations (ii) and (iii), we have 𝑁𝜙 𝑠𝑖𝑛 𝑁−1 2 𝑅=𝐴 𝜙 and 𝜃 = 𝜙 𝑠𝑖𝑛 2 2 We will apply this resultant amplitude expression in diffraction phenomena like, single slit and multiple slit (diffraction grating). 3.1 Interference: We have already discussed that in case of coherent superposition, there is redistribution of light intensity leading to interference. Definition: When two light waves from different coherent sources meet together (superpose), then the resultant amplitude (or intensity) in the region of superposition is different than that of the amplitude (or intensity) of individual waves. This modification in the distribution of intensity (light energy) due to superposition of two or more light waves is called "Interference of light. Examples ✓ One of the best examples of interference is demonstrated by the light reflected from a film of oil floating on water. ✓ The thin film of a soap bubble, which reflects a spectrum of beautiful colors when illuminated by natural or artificial light sources. Interference is of two types: ❖ Constructive interference ❖ Destructive interference Constructive interference occurs when the wave amplitudes reinforce each other, building a wave of even greater amplitude. It occurs when the crests of one wave overlap with the crests of the other wave, and troughs of one wave with the troughs of the other wave, causing an increase in wave amplitude/intensity. Or the resultant amplitude is the sum of the amplitudes of the waves Condition: 𝜆 Path difference = nλ or 2n ( ) 2 Phase difference = 2nπ, n is an integer = 0, ±1, ±2, ±3, … … … … Destructive interference occurs when the wave amplitudes oppose each other, resulting in waves of reduced amplitude. It occurs when the crests of one wave overlap with the troughs of the other wave, causing a decrease in wave amplitude/intensity. Or if we take two waves then the resultant amplitude is the difference of two amplitudes, then that is called destructive interference. Condition: 𝜆 Path difference = (2n + 1) 2 Phase difference = (2n + 1)π, n is an integer = 0, ±1, ±2, ±3, … … … … 𝟐𝝅 (Phase difference = x path difference) 𝝀 3.2 Methods of obtaining interference: i. Division of Amplitude: In this method, the amplitude of the incident beam is divided into two or more parts either by partial reflection or refraction. Thus, we have coherent beams produced by division of amplitude. These beams travel different optical paths and are finally united together to produce interference. Examples are Newton’s rings, Michelson interferometer.  ii. Division of Wave front: Under this category, the coherent sources are obtained by dividing the wave front, originating from a common source, by employing mirrors, biprisms or lenses. These beams travel unequal optical path and fin ally reunite to produce interference. Examples are Fresnel’ biprism, Lloyds single mirror. 3.2.1 Condition for Interference of light: To obtain a permanent or stationary interference pattern the conditions are classified into following three categories ▪ Condition for sustained interference ▪ Condition for clear observation ▪ Condition for good contrast between maxima and minima 3.2.2 Condition for Sustained Interference Pattern: ▪ The two interfering wave should be coherent i.e., both light waves are in same pha se or maintain constant phase difference between them. ▪ The source should have same frequency. ▪ Both the wave must be in same set of polarization. 3.2.3 Condition for clear observation of fringes ✓ The distance between the coherent sources should small. ✓ The distance between the source and screen must be large. ✓ The background should be dark. 3.2.4 Condition for Good Contrast between maxima and minima ✓ The amplitude of the both interfering wave be the equal or very nearly equal. ✓ The sources must be narrow. ✓ The sources must be monochromatic. Newton’s Rings (Fringes due to Division of Amplitude) 4.1: Description: When a plane-convex lens of large focal length is placed as a plane glass plate, a thin film of air is formed between the plano-convex lens and plane glass plate. The thickness of the air film is very small at the point of contact and gradually increased from the centre upwards. If a monochromatic light is allowed to fall normally on this film, a set of alternate dark and bright fringes will be seen in the film. The fringes are concentric circle. These circles or rings are called Newton’s rings. Normal incidence of monochromatic light on the plano- convex lens will produces bright and dark concentric rings around the point of contact between the lens and the glass plate. The phenomenon first explained by Newton so called Newton’s rings. Experimental arrangement: S – Source, producing monochromatic light like, sodium light L1 - Convergent lens, used to produce parallel rays G - A glass plate, making an angle 450 with the direction of incident light, so that vertically reflected downward reflected rays are produced L - A plano-convex lens of large radius of curvature, placed on a plane glass plate P, with its convex surface facing upward M - A travelling microscope, which is used to see the fringe patterns 4.1.1 Explanation of formation of Fringes: Newton’s rings are formed as a result of interference between light wave reflected from the upper and lower surfaces of the air film. Fig. below shows that from a single incident ray (A), two reflected rays (E&F) are produced by division of amplitude. These two reflected rays are coherence and superimpose to produce interference pattern. Suppose, the radius of curvature of the plano-convex lens is R and the thickness of the air film is ‘t’. The ray E and F are coherent as they produced from same incident ray A. The effective path difference between the interfering rays is 𝜆 ∆= 2µ𝑡𝑐𝑜𝑠𝑟 + 2 For air, μ=1 and for normal incidence, r = 0 𝜆 So effective path difference in this case is ∆= 2𝑡 + ……………………(1) 2 At the centre, 𝑡 = 0 𝜆 So effective path difference ∆= , this is the condition for minimum intensity. So the centre 2 is dark. For the bright fringes ∆= 𝒏𝝀 𝜆 So 2𝑡 + = 𝒏𝝀 2 𝜆 Or, 2𝑡 = (2𝑛 − 1) ……………………………………… (2) 2 𝝀 For the dark rings ∆= (𝟐𝒏 + 𝟏) 𝟐 𝜆 𝝀 So 2𝑡 + = (𝟐𝒏 + 𝟏) 𝒏 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒 …. 2 𝟐 Or, 2𝑡 = 𝒏𝝀 ……………………………………………… (3) It is clear that a bright or dark fringe of any order n depend upon the thickness of the air film. Since t is constant along a circle with its centre at the point of contact, the fringes are in the form of concentric circle. 4.1.2 Diameter of Bright Rings Substituting the value of t in eqn. (2) For bright rings So from eqn. 5 Thus, the diameters of bright rings are proportional to the square root of the odd number The separations between successive rings are 0.732:0.504:0.410. 4.1.3 Diameter of Dark Rings From equation (3) and (4) If D is the diameter of dark ring Thus, the diameters of dark ring are proportional to the square root of natural number 4.1.4 Newton’s rings by transmitted light: The effective path difference between the interfering rays is ∆= 2µ𝑡𝑐𝑜𝑠𝑟 For μ=1 and for normal incidence r=0 So effective path difference in this case is ∆= 2𝑡 At centre 𝑡 = 0 So effective path difference ∆= 0 , this is the condition for maxima so center is bright. For the bright fringes ∆ = 𝒏𝝀 So 2𝑡 = 𝒏𝝀 So, for bright fringe 𝐷𝑛2 = 4𝑅𝑛𝝀 so 𝐷𝑛 = √4𝑅𝑛𝝀 𝝀 For the dark rings ∆ = (𝟐𝒏 − 𝟏) 𝟐 𝝀 So 2𝑡 = (𝟐𝒏 − 𝟏) 𝒏 = 𝟏, 𝟐, 𝟑, 𝟒 …. 𝟐 So 𝐷𝑛2 = 2𝑅𝝀(𝟐𝒏 − 𝟏) So 𝐷𝑛 = √𝟐𝑹𝝀(𝟐𝒏 − 𝟏) Hence, the fringe patterns, in reflected and transmitted light are complimentary. 4.1.5 Application of Newton’s rings: 4.1.5.1 Determination of wavelength of light: By forming Newton’s rings and measuring the radii of the rings formed, we can calculate the wavelength of the light used if the radius of curvature of the lens is known. Let R be the radius of curvature of the lens and 𝜆 is the wavelength of the light used. So the diameter of the nth dark ring can be written as Dn2 = 4 n 𝜆𝑅 (1) Similarly, the diameter of the (n +m) th dark ring is D(2n+m)= 4 (𝑛+m) 𝜆𝑅 (2) Subtracting equation (1) from (2) we get D(2n+m) − Dn2 = 4 (𝑛+m) 𝜆𝑅 – 4 n 𝜆𝑅 = 4mRλ 𝐃𝟐(𝐧+𝐦)−𝐃𝟐𝐧 Or, λ = 𝟒𝒎𝑹 Using the above relation, wavelength of the monochromatic light can be calculated 4.1.5.2 Determination of refractive index of a liquid using Newton’s rings: By forming Newton’s rings and measuring the diameter of the rings formed, we can calculate the refractive index of the liquid. In air film, the diameters of the nth and (n+m)th dark rings are measured with the help of travelling microscope. D(2n+m) − Dn2 = 4 (𝑛+m) 𝜆𝑅 – 4 n 𝜆𝑅 = 4mRλ ………………………… (3) Now, the air film is replaced by liquid film (refractive index µ). In liquid film, the diameters of the same nth and (n+m)th dark rings are 𝐷′𝑛 and 𝐷′(n+m) are measured with the help of travelling microscope. D′2n = 4 n 𝜆𝑅/µ and D′2(n+m)= 4 (𝑛+m) 𝜆𝑅/µ So, D′2( n+m)– D′2n = 4mRλ/µ ………………………………………..(4) Dividing equation (3) by (4) 𝐃𝟐(𝐧+𝐦)−𝐃𝟐𝐧 =µ 𝐃′𝟐(𝐧+𝐦)–𝐃′𝟐𝐧 Using the above relation 𝜇 can be calculated. Numerical: 1. A wave along s string is given by the equation 𝑦 = 0.01 sin (50𝜋𝑡 − 31.4𝑥) 𝑚 calculate the speed with which the wave travels. 𝑤 50𝜋 Sol: V= = = 5𝑚/𝑠 𝑘 31.4 2. Two coherent sources, whose intensity ratio is 9:4, produce interference fringes. Find the ratio of maximum to minimum intensity of the fringe system. Sol: Let I1 and I2 be the intensities, and a1 and a2 the amplitudes of the two interfering beams. We can write 𝐼1 𝑎21 = 𝐼2 𝑎22 𝑎1 𝐼 3 or, =√1 = 𝑎2 𝐼2 2 If 𝑎1 =3x then 𝑎 2 = 2x 𝐼𝑚𝑎𝑥 ( 𝑎1+𝑎2) 2 ( 3𝑥+2𝑥 )2 25𝑥 2 25 Now, =( = = 2 = 𝐼𝑚𝑖𝑛 𝑎1−𝑎2) 2 ( 3𝑥−2𝑥 )2 𝑥 1 3. Find the ratio of the intensity at the center of a bright fringe in an interference pattern to the intensity at a point one-quarter of the distance between two fringes from the center. Sol: Resultant intensity I = 2𝑎 2( 1 + cos ϕ ) Where a is the amplitude of the wave. At the center, ϕ = 0 So, I0 = 2𝑎 2( 1 + cos ϕ ) = 2𝑎 2 ( 1 + cos 0 ) = 4𝑎 2 At one-quarter of the distance between two fringes from the center, ϕ = π/2 I1 = 2𝑎 2 ( 1 + cos ϕ ) = 2𝑎 2 ( 1 + cos π/2)= 2𝑎 2 So, I0 : I1 = 4𝑎 2 : 2𝑎 2 = 2 :1 4. In a two source interference experiment, let ‘p’ be a point on the screen, which is equidistant from both the sources. If one source is closed the intensity at ‘p’ is 0.02 watt/m2 then what will be the intensity at ‘p’ when both sources are opened? Sol: ‘P’ is the point of central principal maxima I1=0.02 watt/m2 Resultant intensity I= 4I 1=0.08 watt/m2 5. Twenty five sinusoidal waves of equal amplitude superpose incoherently to produce a resultant intensity of 0.5 watt/m 2. What would be the maximum possible resultant intensity if the superpose coherently? Sol: In case of incoherent superposition, Iin-coherent = na2 =nI, and in case of coherent superposition, I coherent = n2a2 =n2I  So, Icoherent = n Iin-coherent = 25 × 0.5 = 12.5 watt/m2. 6. Newton’s rings are observed normally in reflected light of wavelength 5893 Å. The diameter of the 10th dark ring is 0.005 m. Find the radius of curvature of the lens and the thickness of the film. Sol: Dn2 = 4 n 𝜆 𝑅 Here, Dn = 0.005 m, n=10, 𝜆 = 5893 × 10-9 m. D2n Now, R = = 1.06 m = 106 cm 4nλ If t be the thickness of the film corresponding to a ring of diameter D, then we have D2n 2t = = 6 × 10-6 m 4R or, t = 3 × 10-6 m 7. If in a Newton’s rings experiment, the air in the interspace is replaced by a liquid of refractive index 1.33, in what proportion would the diameter of a ring change? (D2n) 𝑎𝑖𝑟 Sol: ( =µ D2n)𝑙𝑖𝑞𝑢𝑖𝑑 (𝐷𝑛)𝑙𝑖𝑞𝑢𝑖𝑑 1 or, ( 𝐷𝑛) 𝑎𝑖𝑟 = √ =0.867 µ The rings are contracted to 0.867 their previous diameter. 8. In a Newton’s rings experiment in air, the diameter of 10th dark ring is 0.272 cm and that of 15th dark ring is 0.555 cm. If the radius of curvature of the plano-convex lens is 200 cm, calculate the wavelength of the monochromatic light used. Sol: 𝐷𝑛 = 0.272 cm 𝐷𝑛+𝑝 = 0.555 cm 𝑛 = 10, n + m =15, m = 5 𝑅 = 200 cm D2(n+m)− D2n We know that λ = 4𝑚𝑅 0.5552 − 0.2722 or, λ = = 5851 Å. 4×5×200 9. In a Newton’s rings experiment, the diameter of the 10th bright ring changes from 1.5 cm to 1.27 cm when a liquid is introduced between the plate and the lens. Calculate the refractive index of the liquid. 10. In a Newton’s rings experiment, the diameter of the 5 th ring was 0.336 cm and the diameter of the 15th ring was 0.59 cm. Find the radius of curvature of the plano convex lens if the wavelength of light is 5890 Å. Questions: Short & medium answer type 1. Write the difference between coherent and incoherent superposition. 2. Discuss why two independent sources of light of the same wavelength cannot produce interference fringes 3. Do you expect any fringe pattern in case of incoherent superposition? Give justification to your answer. 4. Find the ratio of intensity at the centre of a bright fringe to the intensity at a point one quarter of the distance between two fringes from the centre. 5. Write down the condition for the destructive interference in terms of phase difference and in terms of path difference. 6. Write the conditions of sustained interference pattern. 7. Write the conditions for good contrast between dark and bright fringes. 8. What are the different methods of obtaining interference pattern? Give example of each. 9. Show the intensity distribution curve in an interference pattern. 10. Why the fringes in Newton’s ring interference are circular? 11. In a Newton’s ring experiment the center is bright, whether it is formed by reflected light or transmitted light. Justify your answer. 12. In a Newton’s ring experiment the center is dark, whether it is formed by reflected light or transmitted light. Justify your answer. 13. Why the planoconvex lens has large radius of curvature in Newton’s ring experiment? 14. In a Newton’s ring experiment, the fringes near to the center are wider than the fringes away from the centre. Justify it why? 15. Write down the formula for the refractive index of a liquid determined by In a Newton’s ring experiment. 16. Explain how refractive index of liquid can be determined by Newton's rings method. Long answer type 1. Show that the resultant wave due to the superposition of two harmonic waves of same frequency, different phases and travelling in same direction is also harmonic. Find the amplitude and phase of the resultant wave. 2. Show that coherent superposition leads to interference pattern, discuss it with two beams having different amplitude, different phases propagating in same direction. 3. Draw the schematic diagram for production of Newton's rings with reflected light. Explain how will u obtained two coherent waves.  4. With a suitable diagram, explain the formation of Newton’s rings. Derive the expression for the diameter of Newton's rings formed by transmitted light. Explain why the fringes in Newton’s rings interference are circular. 5. Explain the formation of fringes in a Newton’s rings in reflected light with a suitable diagram. Prove that in reflected light (i) diameters of the dark rings are proportional to the square root of natural numbers, and (ii) diameters of the bright rings are proportional to the square root of odd natural numbers. DIFFRACTION 5.1 Diffraction is the slight bending of light as it passes around the edge of an object. The amount of bending depends on the relative size of the wavelength of light to the size of the opening. If the opening is much larger than the light's wavelength, the bending will be almost unnoticeable. Diffraction of light Diffraction is the spreading out of waves as they pass through an aperture or around objects. It occurs significantly when the size of the aperture or obstacle is of similar linear dimensions to the wavelength of the incident wave. The essential condition for diffraction to occur is that the wavelength of light should be comparable to that of the size of the object/obstacle. It might also occur if the size of the object is less than the wavelength of light. It is due to the interaction among the secondary wavelets which are not blocked off by the obstacle. Diffraction pattern The most striking examples of diffraction are those that involve light; for example, the closely spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow pattern seen when looking at a disc. Difference between Diffraction and Interference Interference Diffraction Interference may be defined as waves Diffraction on the other hand can be emerging from two different coherent termed as secondary waves that sources, producing different wave emerge from the different parts of the fronts. same wave. In interference the intensity of all the In diffraction, there is a variance of the positions on maxima are equal. intensity of all maxima positions. The width of the fringes in interference The widths of the fringes are not equal may be equal (bi-prism) or may not be in diffraction. equal (Newton’s rings). It is absolutely dark in the region of In the case of diffraction, the position of minimum intensity, in the case of minima is not perfectly dark interference. If the number of sources are few such as If the number of sources is many, that is two sources, then they are referred to as more than two then it is referred to as interference sources. diffraction sources. Types of diffraction (Fresnel diffraction and Fraunhofer diffraction) Characteristics Fresnel diffraction Fraunhofer diffraction Observation distance Either the source or screen or The source and the screen both are at finite distance from are effectively at infinite the obstacle distance from the obstacle Requirement of lens Lens are not required Lens (usually 02) are required Wave fronts Cylindrical or spherical wave Plane wave fronts fronts Diffraction pattern Shape and intensity of diffraction Shape and intensity pattern change as the waves remain constant propagate downstream of the scattering source  Movement of Move along the corresponding Remains in a fixed position diffraction pattern shift in the object 5.1.1 Diffraction due to Single Slit (Fraunhofer diffraction). ‘AB’ is a narrow slit of width ‘e’ perpendicular to the plane of the paper. Consider a plane wavefront ‘ww’ originated from a monochromatic source of wavelength ‘λ’ propagating normal to the slit be incident on it. After diffraction Let the diffracted rays are focused by using a convex lens and the screen placed at the focal plane of the lens. According to Hygen’s theory, each point of the wave front in the plane of the slit is a source of secondary wavelets. The secondary wavelets traveling normal to the slits are brought to focus by the lens at the point P 0 on the screen. So, the point P0 must be the point of maxima called as central principal maxima. In order to consider all the secondary wavelets originated and propagated in different directions, consider the secondary wavelets in the direction of’ θ’ are brought to the focus by the lens to the point P 1 on the screen as shown in the figure. The intensity at the point P1 is either minimum or maximum and depends upon the path difference between the secondary waves originating from the corresponding points of the wave front. Theory: In order to find out the intensity at P1, draw a perpendicular AC on BR. The path difference between secondary wavelets from A and B in directionθ is BC i.e., So, the phase difference, Let us consider that the width of the slit is divided into ‘n’ equal parts and the amplitude of the wave from each part is ‘a’. So, the phase difference between two consecutive points............(2.32) Then the resultant amplitude R is calculated by using the method of vector addition of amplitudes The resultant amplitude of n number of waves having same amplitude 'a' and having common phase difference of ' ' is...................(2.33) Substituting the value of in equation (2.33) 𝑛 2𝜋 sin( ( )𝑒𝑠𝑖𝑛𝜃) 2 𝑛𝜆 𝑅=𝑎 1 1 2𝜋 sin( ) 𝑒𝑠𝑖𝑛𝜃 2 𝑛𝜆 𝜋 sin( ( )𝑒𝑠𝑖𝑛𝜃) 𝜆 So, 𝑅 =𝑎 1𝜋............(2.34) sin 𝑒𝑠𝑖𝑛𝜃 n𝜆 Substituting in equation 2.34 As is small value; and na = A Therefore,..............(2.35) Therefore, the Intensity is given by.........(2.36) Case (i): Principal Maximum: Eqn (2.35) takes maximum value for =0 or The condition The condition means that this maximum is formed by the secondary wavelets which travel normally to the slit along OPo and focus at Po. This maximum is known as “Principal maximum”. Intensity of Principal maxima Therefore Case (ii): Minimum Intensity positions: For minimum intensity, sin =0. The values of ' ' which satisfy are where...(2.37) In the above eq. (2.37) n = 0 is not applicable because corresponds to principal maximum. Therefore, the positions according to eq. (2.37) are on either side of the principal maximum. Case (iii): Secondary maximum: In addition to principal maximum at = 0, there are weak secondary maxima between minima positions. The positions of these weak secondary maxima can be obtained with the rule of finding maxima and minima of a given function in calculus. So, differentiating eq n (2.35) and equating to zero, we have Because correspond to minima positions so, 𝛼𝑐𝑜𝑠𝛼 − 𝑠𝑖𝑛𝛼 = 0...........(2.38) The values of ' ' satisfying the eqn (2.38) are obtained graphically by plotting the curves and on the same graph. The points of intersection of the two curves gives the values of ' ’ for which satisfy eqn (2.38). The points of intersections are But , gives principal maximum, substituting the values of ' ' in eqn(2.36), we get and so on. From the above expressions, Imax, I1, I2, I3… it is evident that most of the incident light is concentrated at the principal maximum. 5.1.2 Intensity distribution graph: A graph showing the variation of intensity with ' ' is as shown in the adjacent figure 5.1.3 Plane diffraction grating [Diffraction due to N-Slits (Grating)]: An arrangement consisting of large number of parallel slits of the same width and separated by equal opaque spaces is known as Diffraction grating. Plane diffraction grating consists of a number of parallel and equidistant lines ruled on an optically plane and parallel glass plate by a fine diamond point. Each ruled line behaves as an opaque line while the transparent portion between two consecutive ruled lines behaves as a slit. If a is the width of transparency and b is width of opacity (width of ruling lines) then a+b is called grating element. Theory: A section of a plane transmission grating AB placed perpendicular to the plane of the paper is as shown in the figure. Let ‘e’ be the width of each slit and‘d’ the width of each opaque space. Then (e+d) is known as grating element and XY is the screen. Suppose a parallel beam of monochromatic light of wavelength ' ' be incident normally on the grating. By Huygen’s principle, each of the slit sends secondary wavelets in all directions. Now, the secondary wavelets travelling in the direction of incident light will focus at a point Po on the screen. This point Po will be a central maximum. Now consider the secondary waves travelling in a direction inclined at an angle ' ' with the incident light will reach point P1 in different phases. As a result dark and bright bands on both sides of central maximum are obtained. The intensity at point P1 may be considered by applying the theory of Fraunhofer diffraction at a single slit. The wavelets proceeding from all points in a slit along their direction are equivalent to a single wave of amplitude starting from the middle point of the slit, Where If there are N slits, then we have N diffracted waves. The path diffe rence between two consecutive slits is Therefore, the phase difference......(2.39) Hence the intensity in a direction can be found by finding the resultant amplitude of N vibrations each of amplitude and a phase difference of '' Since in the previous case Substituting these in equation The resultant amplitude on screen at P 1 becomes..........(2.40) Thus, Intensity at P1 will be...(2.41) The factor gives the distribution of Intensity due to a single slit while the factor gives the distribution of intensity as a combined effect of all the slits 5.1.4 Intensity Distribution: Case (i): Principal maxima: The eqn (2.40) will take a maximum value if.......(2.42) n = 0 corresponds to zero order maximum. For n = 1,2,3,… we obtain first, second, third,… principal maxima respectively. The ± sign indicates that there are two principal maxima of the same order lying on either side of zero order maximum. Case(ii): Minima Positions: The eq. (2.40) takes minimum value if but...........(2.43) where m has all integral values except m = 0, N, 2N, …, nN, because for these values becomes zero and we get principal maxima. Thus, m = 1, 2, 3, …, (N-1). Hence where gives the minima positions which are adjacent to the principal maxima. Case(iii): Secondary maxima: As there are (N-1) minima between two adjacent principal maxima there must be (N-2) other maxima between two principal maxima. These are known as secondary maxima. To find their positions Only............(2.44) The roots of the above equation other than those for which give the positions of secondary maxima The eqn (2.44) can be written as From the triangle we have Since intensity of principal maxima is proportional to N2, Hence if the value of N is larger, then the secondary maxima will be weaker and becomes negligible when N becomes infinity. 5.1.5 Formation of Spectra with Grating The principle maxima in a grating are formed in direction given by (𝒆 + 𝒅)𝒔𝒊𝒏𝜽 = 𝒏𝝀. where (𝒆 + 𝒅) is the grating element, ‘n’ the order of the maxima and ‘ ’ the wavelength of the incident light. 1) For a given wavelength ‘ ’ the angle of diffraction’ ’ is different for principal maxima of different orders. 2) For white light and for a particular order n, the light of different wavelengths will be diffracted in different directions. The longer the wavelength, greater is the angle of diffraction. So in each order, we will get the spectra having as many lines as the wavelength in the light source. At centre (n = 0, zero order) gives the maxima of all wavelengths. So here different wavelengths coincide to form the central image of the same color as that the light source. Similarly, the principal maxima of all wavelengths corresponding to n = 1 will form the first order spectrum, the principal maxima of all wavelengths corresponding to n = 2, will form the second order spectrum and so on. Important characteristics of grating spectra: ✓ Spectra of different orders are situated symmetrically on both sides of zero order. ✓ Spectral lines are almost straight and quite sharp. ✓ Spectral colors are in the order from Violet to Red. ✓ Spectral lines are more dispersed as we go to higher orders. ✓ Most of the incident intensity goes to zero order and rest is distributed among the other orders. 5.1.6 Maximum number of orders formed by a Grating: The principal maxima in a grating satisfy the condition Or The maximum angle of diffraction is 90o, hence the maximum possible order is given by Ex: Consider a grating having grating element which is less than twice the wavelength of the incident light, then i.e., only the first order is possible. 5.1.7 Absent spectrum with a diffraction grating: We know the intensity maxima at a point due to grating can be observed when that satisfies the maxima for single slit s well as for ‘N’ no. of slits. But when the condition for principal maxima due to ‘N’ no. of slits and the condition for minima due to single slit are satisfied simultaneously for the same angle θ, then the principal maxima due to grating for that particular order will be missed or absent, which is called missing spectra. We know that, in case of a grating the principal maxima are obtained in the directions given by 𝑛 = 0,1,2,3,4 … … Also, in case of a single slit, the minima are obtained in the directions given by 𝑚 = 1,2,3,4 … ….. If both the conditions are satisfied simultaneously for same angle ‘θ’, a particular maximum of order n will be missing in the grating spectrum. Dividing above equations we have which is the condition of absent spectra. If the width of the ruling is equal to the width of the slit, the second order spectrum will be missed. 5.1.8 Dispersive power of grating Dispersive power of grating is defined as the ratio of the difference in the angle of diffraction of any two neighboring spectral lines to the difference in the wavelength between the two spectral lines. It can also be defined as the difference in the angle of diffraction per unit change in wave length. The diffraction grating of the nth order principal maxima for wavelength given by equation (𝑒 + 𝑑)𝑠𝑖𝑛𝜃 = 𝑛𝜆 Differentiating this equation, we have , (𝑒 + 𝑑)𝑐𝑜𝑠𝜃 𝑑𝜃 = 𝑛 𝑑𝜆 (e+d) is constant and ‘n’ is constant for given order. 𝑑𝜃 𝑑𝜃 𝑛 So, is the dispersive power of grating = ( 𝑒+𝑑)𝑐𝑜𝑠𝜃 𝑑𝜆 𝑑𝜆 𝑑𝜃 So dispersive power of grating α n (order of spectrum) 𝑑𝜆 1 α (𝑒+𝑑) 1 α 𝑐𝑜𝑠𝜃 5.1.9 Determination of Wavelength Using Grating In laboratory, the grating spectrum can be obtained by using a spectrometer. Adjustments: Before performing the experiment, the following adjustments are made. (1) The spectrometer is adjusted for parallel rays by Schuster’s method. (2) The grating is adjusted for normal incidence. Measurement of : When a white light is incident on the grating normally, the beam gets dispersed and in each order of the spectrum we can observe constituent wavelengths ( i.e., VIBGYOR) 1) The telescope is now turned to get the first order spectrum in the field of view on left. 2) The cross-wire is adjusted on the line whose wavelength is to be determined (say on RED line) 3) Now, the readings of the two verniers are noted. 4) The telescope is then turned to the right side to receive the first order spectrum and repeat steps (2) & (3). 5) The difference between readings of the same vernier gives twice the angle of diffraction θ for that line in first order. 6) By substituting the values of ‘θ’, (𝑒 + 𝑑) and n in we can determine the wavelength of light. The same procedure from step (1) through (6) is repeated for second order and even in higher orders. Numerical: 1. In Fraunhoffer diffraction due to a narrow slit, a screen is placed 2 meters away from the lens to obtain the pattern. If the slit width is 0.2 mm and the first minima lie 5 mm on either side of the central maximum, find the wavelength of light. Sol: In the Fraunhoffer diffraction due to a single slit of width e, the direction of minima are given by e sin θ = ±m λ  For first minima on either side of central maximum, m =1. Also, when θ is very small sin θ = θ. Hence, e θ = λ or, θ = λ/e = λ/0.02 θ is also given by 0.5/200 So, λ/0.02 = 0.5/200 or, λ = 5000 Å 2. A parallel beam of light is normally incident on a plane transmission grating having 4250 lines per cm and a second order spectral line is observed at an angle of 30 0. Calculate the wavelength of light. Sol: We know that (e +d) sinθ = nλ, n is an integer Here, (e +d) is the grating element, n is the order of the spectrum There are 4250 lines per cm. therefore 1 (e +d) = cm 4250 Also, n =2, θ = 300 (e +d) sinθ So, λ = = 5882 Å 𝑛 3. In a grating spectrum, which spectral line in 4th order will overlap with 3rd order line of 5600 Å? Sol: The grating equation is (e +d) sinθ = nλ If nth order of λ1 coincides with (n+1)th order of λ 2, then (e +d) sinθ = nλ1 = (n+1) λ2 nλ1 3×5600 or, λ2 = = = 4200 Å 𝑛+1 4 4. What is the highest order spectrum which may be seen with light of wavelength 5000 Å by means of a grating with 3000 lines/cm? Sol: The grating equation is (e +d) sinθ = nλ The maximum value of sinθ is 1. Therefore, the highest order visible with grating spectrum is given by 𝑒+𝑑 1 nmax = = =6.6 λ 3000×5 ×10−5 So, the highest order, which may be seen, is 6.  5. A plane transmission grating having 5000 lines per cm is being used under normal incidence of light. If the width of opaque parts be double than that of the transparent parts of the grating, then which orders of spectra will be absent? Sol: The condition for absent spectra is 𝑒+𝑑 𝑛 = where m = 1,2,3……………… e 𝑚 If d = 2e, then n = 3m = 3, 6, 9, …………… Hence the 3rd, 6th, 9th ………..order spectra would be absent. 6. For a grating with grating element 18000 Å, obtain the dispersions in the first order spectrum around wavelength 5000 Å, assuming normal incidence. Sol: The expression for angular dispersion is given by 𝑑𝜃 𝑛 1 =( = 2 dλ 𝑒+𝑑) 𝑐𝑜𝑠𝜃 √(𝑒+𝑑) −λ2 𝑛 (𝑒 + 𝑑)= 18000 Å, λ = 5000 Å, n=1 Substituting, we get 𝑑𝜃 = 5.78 × 10-5 rad/ Å. dλ 7. In a plane transmission grating, the angle of diffraction for the second order principal maxima for wavelength 5000 Å is 300. Calculate the number of lines in one cm of the grating surface. 8. A plane transmission grating produces an angular separation of 0.01 radian between two wavelengths observed at an angle of 300. If the mean value of the wavelength is 5000 Å and the spectrum is observed in the second order, calculate the difference in the two wavelengths. Questions: Short & medium answer type 1. Distinguish between Fresnel’s and Frounhofer’s diffraction. 2. Distinguish between diffraction and interference. 3. Diffraction can occur if slit width is greater than wavelength of light. Justify. 4. Find the approximate expression for the secondary maxima in case of Fraunhofer’s single slit diffraction pattern. 5. Explain the meaning of missing spectra in the diffraction pattern of a plane transmission grating. Obtain the expression for missing spectra. 6. In a plane diffraction grating, if the width of opaque space is equal to the slit width, find the orders of missing spectra. 7. Define dispersive power of grating. Write the factors on which it depends.  8. A glass piece and a student grating both are identical in size and shape is given to you. How can you distinguish the students grating from the glass piece without using any optical instrument? 9. Find the position of maxima, minima and secondary maxima in a single slit diffraction pattern. 10. Draw the intensity distribution curve in case of a single slit diffraction. 11. Why the secondary maxima are not visible in case of grating? 12. Write the characteristics of grating spectra. Long answer type 1. Derive an expression for intensity incase of Fraunhofer’s single slit diffraction with a neat diagram. 2. In Fraunhofer’s diffraction due to single slit, obtain the conditions for principal maximum, secondary maxima and minima. Find its expression, and then show it graphically. 3. How grating spectra are formed? Write its important characteristics. Explain the meaning of missing spectra in the diffraction pattern of a plane transmission grating. Obtain the expression for missing spectra. 4. Define dispersive power of grating. Obtain the expression for it Write the factors on which it depends. Show that the dispersive power of a grating is 𝑛 ⁄{(𝑎 + 𝑏) 2 − 𝑛2 𝜆2 }12 Where (𝑎 + 𝑏) is the grating element and ′𝑛’ is the order of the spectrum.  Module – 2: ELECTROMAGNETISM Scalar and Vector Function: A scalar function f(x, y, z) specifies a magnitude but independent of directi ons. A vector function F(x, y, z) specifies a magnitude and a direction at every point (x, y, z) in some region of space. We can picture a vector function as a collection of arrows (Fig. 1), one for each point (x, y, z). Fig. 1: Vector function The direction of the arrow at any point is the direction specified by the vector function, and its length is proportional to the magnitude of the function. A vector function, like any vector, can be resolved into components as in fig. 2. Fig. 2: Vector function components Letting, i, j and k unit vectors along the x-, y-, and z-axes, respectively, we write 𝐹⃗ (x, y, z) = 𝑖̂𝐹𝑥 +𝑗̂𝐹𝑦 +𝑘̂𝐹𝑧 (1) Fx, Fy and Fz are themselves scalar functions of x, y and z. The magnitude or the direction (or both) of the vector quantities may change with position and time. The various laws of electromagnetism describe how the different scalar and the vector functions change with position and time. The laws are expressed mathematically in terms of derivatives of the fields with respect to space coordinates and time. The derivatives w.r.t. space coordinates are expressed in terms of gradient, divergence and curl operators. Gradient(⃗𝛁⃗ ): Suppose, we have a function of three variables—say, the temperature T (x, y,z) in this room. (Start out in one corner, and set up a system of axes; then for each point (x, y,z) in the room, T gives the temperature at that spot.) We want to generalize the notion of “derivative” to functions like T , which depend not on one but on three variables. A derivative is supposed to tell us how fast the function varies, if we move a little distance. But this time the situation is more complicated, because it depends on what direction we move. In fact, the question “how fast does T vary?” Taking the total derivatives on the temperature (T), we have; 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝑑𝑇 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 (2) 𝜕𝑥 𝜕𝑦 𝜕𝑧 This tells us how T changes when we alter all three variables by the infinitesimal amounts dx, dy, dz. Interestingly, the eq. (2) is a reminiscent of the dot product i.e., 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝑑𝑇 = (𝑖̂ + 𝑗̂ + 𝑘̂ ) ∙ (𝑖̂𝑑𝑥 + 𝑗̂𝑑𝑦 + 𝑘̂ 𝑑𝑧) (3) 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑑𝑇 = (∇𝑇) ∙ (𝑑𝑟) where 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝛁𝑇 ≡ 𝑖̂ + 𝑗̂ + 𝑘̂ (4) 𝜕𝑥 𝜕𝑦 𝜕𝑧 is the gradient of T. Note that ⃗∇⃗ 𝑇 is the generalized derivative and is a vector quantity i.e., like any vector, the gradient has magnitude and direction. To determine its geometrical meaning, we can rewrite the total derivative as; 𝑑𝑇 = (∇𝑇) ∙ (𝑑𝑟) = |𝑑𝑇||𝑑𝑟| cos 𝜃 (5) If we fix the magnitude |dr| and search around in various directions (that is, vary θ), the maximum change in T evidentally occurs when θ = 0. That is, for a fixed distance |dr|, dT is greatest when I move in the same direction as ⃗∇⃗ 𝑇. Thus, The gradient points in the direction of maximum increase of the scalar function (f say). In other words, ⃗∇⃗ 𝑓is normal to the surface of constant f. The Del (𝛁) Operator: ⃗⃗ , operates a scalar function as seen from The gradient has the formal appearance of a vector, ∇ the eq (4). Generally, we can represent the del operator in vector notation as; 𝜕 𝜕 𝜕 ⃗∇⃗ = 𝑖̂ + 𝑗̂ + 𝑘̂ (6) 𝜕𝑥 𝜕𝑦 𝜕𝑧 It is also known as the nabla operator. Operation with the Del/Nabla operator: Just like vector operations, Del operates with scalar and vector functions in three different ways. These are ⃗⃗. ): The Divergence (∇ From the definition of ⃗∇⃗ we construct the divergence: 𝜕 𝜕 𝜕 ⃗⃗. 𝑓⃗ = (𝑖̂ ∇ + 𝑗̂ + 𝑘̂ ). (𝑖̂𝑓𝑥 + 𝑗̂ 𝑓𝑦 + 𝑘̂ 𝑓𝑧 ) 𝜕𝑥 𝜕𝑦 𝜕𝑧  𝜕𝑓𝑥 𝜕𝑓𝑦 𝜕𝑓𝑧 =( + + ) (7) 𝜕𝑥 𝜕𝑦 𝜕𝑧 Where fx, f y and fz are the components of the vector function f. The divergence operates on a vector function f yields a scalar quantity. The divergence of a vector function (F) is, like any dot product of a vector. Geometrical Interpretation: The name divergence is well chosen, for ⃗∇⃗ ∙ ⃗f is a measure of how much the vector f spreads out (diverges) from the point in question. For example, the vector function in Fig. 3 has a large (positive) divergence (if the arrows pointed in, it would be a negative divergence). Fig. 3: positive divergence Example 2. Here, 𝑖̂, 𝑗̂ and 𝑘̂ are replaced by 𝑥̂, 𝑦̂ and 𝑧̂ respectively. Solenoidal Field: If divergence of a vector field/function (say 𝑓⃗) vanishes (∇ ⃗⃗. 𝑓⃗ = 0) then the field is known as a solenoidal field. The Curl (⃗𝛁⃗ ×): From the definition of ∇ we construct the curl of a function as;  𝜕 𝜕 𝜕 ⃗⃗ × 𝐹⃗ = (𝑖̂ ∇ + 𝑗̂ + 𝑘̂ ) × (𝑖̂𝐹𝑥 + 𝑗̂ 𝐹𝑦 + 𝑘̂ 𝐹𝑧 ) 𝜕𝑥 𝜕𝑦 𝜕𝑧 ⃗∇⃗ × 𝐹⃗ = 𝑖̂ (𝜕𝐹𝑦 − 𝜕𝐹𝑧 ) + 𝑗̂ (𝜕𝐹𝑥 − 𝜕𝐹𝑧 ) + 𝑘̂ (𝜕𝐹𝑦 − 𝜕𝐹𝑥) (8) 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑧 Notice that the curl of a vector function (F) is, like any cross product of a vector. Geometrical Interpretation: The name curl is also well chosen, for 𝛻⃗⃗ × 𝐹⃗ is a measure of how much the vector F swirls around the point. Example: Irrotational Field: ⃗⃗ × 𝐹⃗ = 0 The (vector) field (F say) is said to be irrottaional if curl of the field vanishes i.e., ∇ Operations using the Del:  (vii) Divergence of gradient of a scalar function (say T): 𝜕2𝑇 𝜕2 𝑇 𝜕2𝑇 i.e., ∇ · (∇T ) = ∇2T = 2 + 2 + = Laplacian of T 𝜕𝑥 𝜕𝑦 𝜕𝑧2 (viii) Curl of gradient of a scalar function (say T) is always zero i.e., ∇ × (∇T) = 0 (ix) Gradient of divergence ‘∇(∇ · v)’ seldom occurs in physical applications, and it has not been given any special name of its own. (x) The divergence of a curl, like the curl of a gradient, is always zero: ∇ · (∇ × v) = 0 (xi) Curl of curl: ∇ × (∇ × v) = ∇(∇ · v) − ∇2 v (For readers perspective 𝛻⃗⃗ is here replaced by ∇ in the above formulas) Integrals: Line, Surface and the Volume Integrals: In electrodynamics, we encounter several different kinds of integrals, among which the most important are line (or path) integrals, surface integrals (or flux), and volume integrals. a. Line Integrals of a vector field: A line integral is an expression of the form (9) where v is a vector function, dl is the infinitesimal displacement vector, and the integral is to be carried out along a prescribed path, from point a to point b as shown in the fig. 4. Fig. 4: Line integral between two points a and b. If the path in question forms a closed loop (that is, if b = a), we shall put a circle on the integral sign then, the line integral becomes;  (10) Ordinarily, the value of a line integral depends critically on the path taken from a to b, but there is an important special class of vector functions for which the line integral is independent of path and is determined entirely by the end points. In this case, the vector field is known as a conservative field. b. Surface Integrals: A surface integral is an expression of the form (11) where v is again some vector function, and the integral is over a specified surface (see Fig. 5). Fig. 5: surface integral Here da is an infinitesimal patch of area, with direction perpendicular to the surface. For a closed surface, the surface integral can be written as; If v describes the flow of a fluid (mass per unit area per unit time), then represents the total mass per unit time passing through the surface— hence the alternative name, ‘flux’. c. Volume Integrals: A volume integral is an expression of the form (12) where T is a scalar function and dτ is an infinitesimal volume element. In Cartesian coordinates, dτ = dx dy dz For example, if T is the density of a substance (which may vary from point to point), then the volume integral would give the total mass. Theorems: Gauss Divergence Theorem: The volume integral of divergence of a vector field (say V) over a given volume is equal to the surface integral of the vector over a closed surface (a) enclosing the volume. Mathematically, (13) Stoke’s Theorem: The surface integral of the curl of a vector field over a given region (say S) is equal to the line integral of the vector along the boundary of the area (P). (14) Maxwell’s Equations: The various laws of electromagnetism were pulled together and were cast into four equations involving time and space derivatives of electric and magneti c fields. These equations are known as Maxwell’s electromagnetic equations. The equations are as follows; 1). Gauss’ Law in Electrostatics: Statement: The total electric flux ‘ɸ E’ over a closed surface is equal to 1/ϵ0 times the net charge enclosed by the surface. We have, the electric flux (ɸE) associated with the charge (as seen from the fig.6) enclosed by a surface‘s’ is given by 𝑞 𝑒𝑛𝑐 ɸE = (15) 𝜖0 Fig.6: Flux emanating from the elementary surface ‘ds’. Here, q enc = +q where ϵ0 is permittivity of the free space. Also, the electric flux is defined as the surface integral of the electric field i.e., ɸE = 𝐸⃗⃗. 𝑑𝑠 ⃗⃗⃗⃗⃗ (16) Equating both the equations, finally the Gauss’ law can be expressed mathematically as; 𝑞 ɸE = 𝐸⃗⃗. ⃗⃗⃗⃗⃗ 𝑑𝑠 = 𝑒𝑛𝑐 (17) 𝜖0 Notes: The charge enclosed (q enc ) by the surface may be point charges or continuously distributed charges. The net enclosed charges may be positive, negative or zero. Accordingly, the net electric flux may be outward, inward or zero. The surface enclosing the charges is termed as a Gaussian surface. The electric flux doesn’t depend on the shape or size of the Gaussian surface as long as the charges are enclosed by the surface. Symmetry is crucial w.r.t. application of the Gauss’ law. The eq. (17) is known as the integral form of the Gauss’ law 1.1: Gauss’ Law in terms of Electric Displacement (D): We have the electric displacement vector (D) in vaccum/free space is defined as; ⃗⃗ = 𝜖0 𝐸⃗⃗ 𝐷 Putting this value in the equation (17), we get ɸE = ⃗⃗. ⃗⃗⃗⃗⃗ 𝐷 𝑑𝑠 = 𝑞𝑒𝑛𝑐 (18) 1.2: Gauss’ Law in differential form: Let, ρ is the volume charge distributions (dq/dV). The net enclosed charge by the Gaussian surface will be, qenc = ρ dV (19) where dV is the elementary volume encloses by the Gaussian surface. Putting the equation (19) in eq (17), we get; 𝐸⃗⃗. ⃗⃗⃗⃗⃗ 𝑑𝑠 = ( ρ dV)/ ϵ0 (20) Applying the Gauss’ divergence theorem to the left hand side of eq.(20), we have  𝐸⃗⃗. 𝑑𝑠 ⃗⃗⃗⃗⃗ = ⃗⃗. 𝐸⃗⃗ )𝑑𝑉 (∇ (21) Equating eq. (21) and (22), we get 1 (⃗∇⃗. 𝐸⃗⃗ )𝑑𝑉 = ρ dV 𝜖0 Or ⃗⃗. 𝐸⃗⃗ − 𝜌 )dV =0 (∇ 𝜖0 The integrand must vanish, leading to 𝜌 ⃗⃗. 𝐸⃗⃗ = ∇ (22) 𝜖0 This is the differential form of the Gauss’ law. In terms of Displacement Vector, eq. (22) can be written as; ⃗⃗. 𝐷 ∇ ⃗⃗ = 𝜌 (23) 2. Gauss’ law in Magnetostatics: Statement: The magnetic flux (ɸ B ) over a closed surface enclosing the volume is always zero. Mathematically, ɸB = ⃗⃗. ⃗⃗⃗⃗⃗ 𝐵 𝑑𝑠 = 0 (24) where B, is the magnetic flux density (magnetic flux per unit area) with unit is Weber/m 2. Since the magnetic poles always occur in pairs, isolated magnetic poles don’t exi st. Therefore, within any macroscopic volume, the net magnetic pole is always zero. Applying the Gauss’ divergence theorem to the equation (24), we have ɸB = ⃗⃗. ⃗⃗⃗⃗⃗ 𝐵 𝑑𝑠 = (⃗∇⃗. 𝐵 ⃗⃗)𝑑𝑉 = 0 or (⃗∇⃗. 𝐵 ⃗⃗) = 0 (25) Equation (25) describes the differential form of the Gauss’ law in Magnetostatics. In other words, it shows that non-existence of magnetic monopoles. 3. Faraday’s Law of Electromagnetic Induction: In 1831 Michael Faraday reported that A changing magnetic field induces an electric field or simply an emf ‘ε’. The magnitude of the emf induced in the loop is equal to the negative rate of change of magnetic flux through the surface enclosed by the loop. Mathematically, 𝑑ɸ𝐵 𝜀= − (28) 𝑑𝑡 The emf (ε) is defined as the line integral of the electric field i.e., ε = ∮ 𝐸⃗⃗.𝑑𝑙 ⃗⃗⃗⃗ (29) C Equating eq. (28) and (29) respectively, 𝑑ɸ𝐵 ∮ 𝐸⃗⃗.𝑑𝑙 ⃗⃗⃗⃗ = − 𝑑𝑡 C ⃗⃗ ⃗⃗⃗⃗⃗ or ⃗⃗⃗⃗ = − 𝑑 ∮ 𝐵.𝑑𝑠 ∮ 𝐸⃗⃗.𝑑𝑙 ⃗⃗. ⃗⃗⃗⃗⃗ (ɸB = 𝐵 𝑑𝑠) 𝑑𝑡 This is known as the Faraday’s law of Electromagnetic Induction. 3.1: Differential form of the Faraday’s Law: Applying Stoke’s theorm to the lhs and equating both side, we get 𝑑 ∮ 𝐵.𝑑𝑠 ⃗⃗ ⃗⃗⃗⃗⃗ (⃗∇⃗ × 𝐸⃗⃗ ). ⃗⃗⃗⃗⃗ 𝑑𝑠 = − 𝑑𝑡 ⃗⃗ 𝑑𝐵 or, (⃗∇⃗ × 𝐸⃗⃗ ) = − (30) 𝑑𝑡 The eq. (30) is known as the diferential form of the Faraday’s Law of Electromagnetic Induction. 4. Ampere’s Circuital Law: Statement: The line integral of magnetic field along a closed loop (C) is equal to μ0 times the net electric current enclosed by the loop. (26) C Here I = Ienc= algebraic sum of the currents enclosed by the loop C (Amperian loop). Let us considered an Amperian loop, which encloses the current I 1, I2 respectively as shown in fig.7. The Ampere’s circuital law becomes  Fig. 7: Amper’s loop carrying current in two straight wires. ⃗⃗.𝑑𝑙 ∮𝐵 ⃗⃗⃗⃗ = 𝜇 0 (𝐼1 + 𝐼2 ) C Alternatively says, the integral of magnetic field intensity (H) along an imaginary closed path is equal to the current enclosed by the path. Example: Use Ampere’s circuital law, find the magnitude of the magnetic field B, at a point due to a straight, infinitely long conductor carrying a steady current. Solution: Let us consider an infinitesimal length dl of the Amperian loop at the same point. At each point on the Amperian loop, the value of B is constant since the perpendicular distance of that point from the axis of conductor is fixed, but the direction will be along the tangent on the loop at that point. Now, according to Ampere’s Circuital Law (B, dl are in the same direction) Therefore, or B = μ0I/2πr (here, B indicates a vector quantity) 4.1 Ampere’s Law in Differential Form: We have, From the current distribution, I = ⃗⃗⃗⃗⃗𝐽. ⃗⃗⃗⃗⃗ 𝑑𝑠 where J = current density (current per unit area). ⃗⃗. ⃗⃗⃗⃗ 𝐵 𝑑𝑙 =μ0 ⃗⃗⃗⃗⃗⃗𝐽. ⃗⃗⃗⃗⃗ 𝑑𝑠 Applying the Stoke’s theorm to the lhs, ⃗⃗ × 𝐵 (∇ ⃗⃗). 𝑑𝑠 ⃗⃗⃗⃗⃗ = μ0 ⃗⃗⃗⃗𝐽. 𝑑𝑠 ⃗⃗⃗⃗⃗ Equating both sides, we have (⃗∇⃗ × 𝐵 ⃗⃗) = μ0 ⃗⃗𝐽 (27) The eq. (27) is known as the diferential form of the Ampere’s Circuital Law. 5.0: Maxwell’s Equations: 5.1: Electrodynamics before Maxwell So far, we have encountered the following laws, specifying the divergence and curl of electric and magnetic fields: These equations represent the state of electromagnetic theory, when Maxwell began his work. 5.2: Maxwell’s Correction to the Ampere’s Law: Just as a changing magnetic field induces an electric field (Faraday’ law) , Maxwell proposed that ‘a changing electric field induces a magnetic field’. The real confirmation of Maxwell’s theory came in 1888 with Hertz’s experiments on electromagnetic waves. Maxwell called his extra term the displacement current (Id): 𝑑𝐸⃗⃗ 𝐼𝑑 = 𝜖0 𝐴 (28) 𝑑𝑡 The displacement current exists as long as the electric field changes with time. When the electric field attains a constant value (just like the plates of the capacitor gets fully charged), displacement current vanishes and the filed is established by the conduction current (I = due to flow of charge particles). Maxwell introduced this displacement current in the Amper’s Circuital Law for a time varying electric field and modified the equation as; ⃗⃗.𝑑𝑙 ∮𝐵 ⃗⃗⃗⃗ = 𝜇 0 (𝐼 + 𝐼𝑑 ) (29) This is known as the Modified Amper’s Circuital Law with Maxwell’s correction. The corresponding differentional form will be, ⃗⃗ × 𝐵 (∇ ⃗⃗) = μ0 (𝐽⃗ + ⃗⃗⃗⃗ 𝐽𝑑 ) (30) Where, ⃗⃗⃗⃗ 𝐽𝑑 is known as the displacement current density (Id/A). or, ⃗⃗ ⃗⃗) = μ0 (𝐽⃗ + 𝜖0 𝑑𝐸) (⃗∇⃗ × 𝐵 (31) 𝑑𝑡 5.3: Distinction between the displacement and the conduction currents. Displacement Current Conduction Current The displacement current evolves due The Conduction current originates from the to time varying electric field. flow of charges in a conducting medium or in a metal. It is a ficticious current which can exist It is a real current. in vaccum or in any medium. It depends on the electric permitivity of It obeys the Ohm’s Law and depends on the the medium and the rate at which the resistance and potential difference of the electric field changes with time. conductor. Now, the finishing touches on Maxwell’s equations are: (32) This is known as the differential form of Maxwell’s Equations. The final differential and Integral form of the Maxwell’s equations can be qritten as; Gauss’ Law Electric flux for through a closed Electrostatic surface is s proportional to the charged enclosed Gauss’ Law No magnetic for monopole; The total Magnetism magnetic flux through a closed surface is zero Changing magnetic Faraday’s Law flux gives emf. of Induction Electric current and Ampere’s Law changing electric flux produces a magnetic field For a charge and current free medium (free space), the Maxwell’s equations become; (33) Corresponding the Integral forms are;  E dA = 0  B dA = 0 (34) d B  E dl = − dt d E  B dl =   0 0 dt 5.4: Physical Significance of Maxwell’s Equations: Maxwell’s equations incorporate all the laws of electromagnetism, which were developed from experimental observations and were expressed in the form of various emprical laws. Maxwell’s equations lead to the existence of electromagnetic waves, which has been amply confirmed by experimental observations. These equations are consistence with all the observed properties of e.m. waves. Maxwell’s equations are consistence with the special theory of relativity. Maxwell’s equations are used to describe the classical electromagnetic field as well as the quatum theory of interaction of charged particles with electromagnetic field. Maxwell’s equations provided a unified description of the electric and magnetic phenomena which were treated independely. Electromagnetic Waves 6: Maxwell’s Equations in Free space/vacuum: In free space, the charge density 𝜌 and the current density 𝐽⃗ are zero. So the four Maxwell’s equations in free space are (i) ⃗∇⃗. 𝐸⃗⃗ = 0 (ii) ⃗⃗. 𝐵 ∇ ⃗⃗ = ∇ ⃗⃗. 𝐻 ⃗⃗ = 0 𝜕𝐵 ⃗⃗ (iii) ⃗∇⃗ × 𝐸⃗⃗ = − 𝜕𝑡 ⃗⃗ 𝜕𝐷 (iv) ⃗∇⃗ × 𝐻 ⃗⃗ = 𝜕𝑡 (with 𝜌 and σ are zero, 𝐽⃗ = σ𝐸⃗⃗ = 0, 𝐷 ⃗⃗ = ε0 𝐸⃗⃗, and 𝐵 ⃗⃗ = 𝜇 0 𝐻 ⃗⃗) 6.1: Electromagnetic Wave Equations in terms of 𝐸⃗⃗ From Maxwell’s 3rd equation ⃗⃗ ⃗∇⃗ × 𝐸⃗⃗ = − 𝜕𝐵 𝜕𝑡 Taking curl on both the sides, we can write 𝜕 𝜕 ⃗⃗ × (∇ ∇ ⃗⃗ × 𝐸⃗⃗ ) = − ⃗⃗ × 𝐵 (∇ ⃗⃗) = −µ0 ⃗⃗ × 𝐻 (∇ ⃗⃗) 𝜕𝑡 𝜕𝑡 Applying Maxwell’s 4th equation to the right hand side of the above equation we get; 2𝐸 ⃗⃗ ⃗⃗ × 𝐸⃗⃗ ) = −ε0µ0 𝜕 Or, ⃗∇⃗ × (∇ …………….(1) 𝜕𝑡2 But, ⃗∇⃗ × (∇ ⃗⃗ × 𝐸⃗⃗ ) = ⃗∇⃗ (∇ ⃗⃗. 𝐸⃗⃗ ) - ∇2 𝐸⃗⃗ Now, equation (1) can be written as 2𝐸 ⃗⃗ ∇ ⃗⃗. 𝐸⃗⃗ ) - ∇2 𝐸⃗⃗ = −ε0µ0 𝜕 ⃗⃗ (∇ ……………………..(2) 𝜕𝑡2 ⃗⃗. 𝐸⃗⃗ = 0 From, Maxwell’s 1st equation, ∇ Equation (2) can be written as 𝜕2𝐸⃗⃗ - ∇2 𝐸⃗⃗ = −ε0µ0 𝜕𝑡2 𝝏𝟐𝑬 ⃗⃗ Or, 𝛁 𝟐⃗𝑬⃗ = ε0µ0 ……………………………..(3) 𝝏𝒕 𝟐 Equation (3) is the required electromagnetic wave equation in free space, in terms of electric vector, 𝐸⃗⃗ (2nd order differential wave equation). 6.2 ⃗⃗ Electromagnetic Wave Equations in terms of 𝐻 We have Maxwell’s 4th equation ⃗⃗ ⃗⃗ = − 𝜕𝐷 ⃗∇⃗ × 𝐻 𝜕𝑡 Taking curl on both the sides, we can write ⃗∇⃗ × (∇ ⃗⃗ ) = − 𝜕 (∇ ⃗⃗ × 𝐻 ⃗⃗ ) = −ε0 𝜕 (∇ ⃗⃗ × 𝐷 ⃗⃗ × 𝐸⃗⃗ ) 𝜕𝑡 𝜕𝑡 2𝐻 ⃗⃗ ⃗⃗ × (∇ ∇ ⃗⃗ ) = −ε0µ0 𝜕 ⃗⃗ × 𝐻 ……………………………(4) 𝜕𝑡2 (Applying Maxwell’s 3rd equation) 2𝐻 ⃗⃗ Or, ⃗∇⃗ (∇ ⃗⃗. 𝐻 ⃗⃗ = −ε0µ0 𝜕 ⃗⃗) - ∇2 𝐻 𝜕𝑡2 𝜕2𝐻 ⃗⃗ ⃗⃗ = −ε0µ0 Or, - ∇2 𝐻 (Applying Maxwell’s 2nd equation) 𝜕𝑡2 𝝏𝟐⃗𝑯 ⃗⃗ ⃗⃗⃗ = ε0µ0 Or, 𝛁 𝟐𝑯 ……………………………(5) 𝝏𝒕 𝟐 Equation (5) is the required electromagnetic wave equation in free space, in terms of magnetic ⃗⃗ (2nd order differential wave equation). vector, 𝐻 6.3 Velocity of electromagnetic wave The general wave equation (2nd order differential wave equation) is 1 𝜕2⃗𝛹 ⃗⃗ ⃗⃗⃗ = ∇2 𝛹 𝑣2 𝜕𝑡2 Comparing this equation with e.m. wave equation (equation 3 or 5), we can write the expression for velocity, 𝑣 1 Hence, 𝑣 = ≅ 2.99 × 108 m/s ≅ 𝑐 (velocity of light) √µ0ε 0 7.0 Maxwell’s equations in a dielectric medium: Maxwell’s equations in a dielectric medium are (i) ⃗∇⃗. 𝐸⃗⃗ = 0 (ii) ⃗⃗. 𝐵 ∇ ⃗⃗ = ∇ ⃗⃗. 𝐻 ⃗⃗ = 0 ⃗⃗ (iii) ⃗∇⃗ × 𝐸⃗⃗ = − 𝜕𝐵 𝜕𝑡 ⃗⃗ 𝜕𝐷 (iv) ⃗⃗ × 𝐻 ∇ ⃗⃗ = 𝜕𝑡 (with 𝜌 and 𝐽⃗ are zero, 𝐷 ⃗⃗ = ε𝐸⃗⃗ , and 𝐵 ⃗⃗ = µ𝐻 ⃗⃗) 7.1 Electromagnetic Wave Equations in terms of 𝐸⃗⃗ : From Maxwell’s 3rd equation  ⃗⃗ ⃗∇⃗ × 𝐸⃗⃗ = − 𝜕𝐵 𝜕𝑡 Taking curl on both the sides, we can write 𝜕 𝜕 ⃗∇⃗ × (∇ ⃗⃗ × 𝐸⃗⃗ ) = − ⃗⃗ × 𝐵 (∇ ⃗⃗) = −µ ⃗⃗ × 𝐻 (∇ ⃗⃗) 𝜕𝑡 𝜕𝑡 2𝐸 ⃗⃗ ⃗⃗ × 𝐸⃗⃗ ) = −εµ 𝜕 Or, ⃗∇⃗ × (∇ …………….(6) (Applying Maxwell’s 4th equation) 𝜕𝑡2 But, ⃗∇⃗ × (∇ ⃗⃗ × 𝐸⃗⃗ ) = ⃗∇⃗ (∇ ⃗⃗. 𝐸⃗⃗ ) - ∇2 𝐸⃗⃗ Now, equation (1) can be written as 2𝐸 ⃗⃗ ∇ ⃗⃗. 𝐸⃗⃗ ) - ∇2 𝐸⃗⃗ = −εµ 𝜕 ⃗⃗ (∇ …………………… (7) 𝜕𝑡2 ⃗⃗. 𝐸⃗⃗ = 0 From, Maxwell’s 1st equation, ∇ Equation (2) can be written as 𝜕2𝐸⃗⃗ - ∇2 𝐸⃗⃗ = −εµ 𝜕𝑡2 𝝏𝟐𝑬 ⃗⃗ Or, 𝛁 𝟐⃗𝑬⃗ = εµ …………………………….. (8) 𝝏𝒕 𝟐 Equation (8) is the required electromagnetic wave equation in a dielectric medium, in terms of electric vector, 𝐸⃗⃗ (2nd order differential wave equation). ⃗⃗ and is represented by Similarly, we can derive for magnetic component 𝐻 𝝏𝟐⃗𝑯 ⃗⃗ ⃗⃗⃗⃗ = εµ 𝛁 𝟐𝑯 ……………………………… (9) 𝝏𝒕𝟐 7.2 Velocity of electromagnetic wave: The general wave equation (2nd order differential wave equation) is 1 𝜕2⃗𝛹 ⃗⃗ ⃗⃗⃗ = ∇2 𝛹 𝑣2 𝜕𝑡2 Comparing this equation with em wave equation (equation 8 or 9), we can write the expression for velocity, 𝑣 1 1 𝑐 𝑣= = = √εµ √ε 0ε r μ0μr √ε r μr If μr =1 (non-magnetic medium), then 𝒄 𝒄 𝒗= = √𝛆 𝐫 𝒏 n is refractive index of the medium So, 𝑛 = √𝛆 𝐫 = √𝑘 (k is dielectric constant of the medium) 8.0 Transverse nature of electromagnetic wave: Suppose the em wave is propagating in any arbitrary direction defined by the propagation vector k. The electric and magnetic vectors are represented by the equations 𝐸⃗⃗ (𝑟, 𝑡) = 𝐸⃗⃗0 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝐻 ⃗⃗ (𝑟, 𝑡) = 𝐻⃗⃗0 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) Here, the vectors E0 and H0 are constant in time, k is propagation vector whose magnitude is 2𝜋 equal to and its direction is the direction of propagation of wave and 𝑖 = √−1. 𝜆 8.1 Transverse nature of 𝐸⃗⃗ From Maxwell’s 1st equation (in free space) ⃗∇⃗. 𝐸⃗⃗ = 0 ⃗⃗. 𝐸⃗⃗0 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) = 0 or, ∇ 𝜕 𝜕 𝜕 or, [𝑖̂ + 𝑗̂ + 𝑘̂ ]. [𝑖̂𝐸0𝑥 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) + 𝑗̂ 𝐸0𝑦 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) + 𝑘̂ 𝐸0𝑧 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) ] = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧 or, 𝑖𝐸0𝑥 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝑘𝑥 + 𝑖𝐸0𝑦 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝑘 𝑦 + 𝑖𝐸0𝑧 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝑘 𝑧 = 0 or, 𝑖 (𝐸𝑥 𝑘𝑥 + 𝐸𝑦 𝑘𝑦 + 𝐸𝑧 𝑘𝑧 ) =0 or, 𝐸⃗⃗. 𝑘⃗⃗ = 0 Two vectors are perpendicular to each other. That means the electric vector 𝐸⃗⃗ is perpendicular to the direction of propagation. 8.2 Transverse nature of 𝐻⃗⃗ From Maxwell’s 2nd equation ⃗∇⃗. 𝐻 ⃗⃗ = 0 or, or, ⃗∇⃗. 𝐻 ⃗⃗0 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) = 0 𝜕 𝜕 𝜕 or, [𝑖̂ + 𝑗̂ + 𝑘̂ ]. [𝑖̂𝐻0𝑥 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) + 𝑗̂𝐻0𝑦 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) + 𝑘̂ 𝐻0𝑧 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) ] = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧 or, 𝑖𝐻0𝑥 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝑘 𝑥 + 𝑖𝐻0𝑦 𝑒 𝑖 (𝑤𝑡−𝑘.𝑟)𝑘 𝑦 + 𝑖𝐻0𝑧 𝑒 𝑖(𝑤𝑡−𝑘.𝑟) 𝑘𝑧 =0 or, 𝑖 (𝐻𝑥 𝑘 𝑥 + 𝐻𝑦 𝑘 𝑦 + 𝐻𝑧 𝑘 𝑧) = 0 ⃗⃗. 𝑘⃗⃗ = 0 or, 𝐻 ⃗⃗ is Two vectors are perpendicular to each other. That means the magnetic vector 𝐻 perpendicular to the direction of propagation. 8.3 Relative orientation of vector ⃗𝑬⃗ and ⃗⃗𝑯 ⃗⃗ From Maxwell’s 3 rd equation,  ⃗⃗ ⃗⃗ ⃗∇⃗ × 𝐸⃗⃗ = − 𝜕𝐵 = −µ 𝜕𝐻