B. Tech. Physics – I (Waves and Oscillations) PDF
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These notes cover various oscillatory systems, including waves and oscillations, simple harmonic motion, compound pendulums, and LC circuits. The document details periodic motion, types of oscillatory motion, oscillatory systems, and characteristics like time period and frequency. It also describes damped oscillations.
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Subject: B. Tech. PHYSICS – I (3 – 1 – 0) Waves and Oscillations Periodic & Oscillatory Motion:- The motion in which repeats after a regular interval of time is called periodic motion. 1. The periodic motion in which there is existence of a restoring force and the body moves along the same p...
Subject: B. Tech. PHYSICS – I (3 – 1 – 0) Waves and Oscillations Periodic & Oscillatory Motion:- The motion in which repeats after a regular interval of time is called periodic motion. 1. The periodic motion in which there is existence of a restoring force and the body moves along the same path to and fro about a definite point called equilibrium position/mean position, is called oscillatory motion. 2. In all type of oscillatory motion one thing is common i.e each body (performing oscillatory motion) is subjected to a restoring force that increases with increase in displacement from mean position. 3. Types of oscillatory motion:- It is of two types such as linear oscillation and circular oscillation. Example of linear oscillation:- 1. Oscillation of mass spring system. 2. Oscillation of fluid column in a U-tube. 3. Oscillation of floating cylinder. 4. Oscillation of body dropped in a tunnel along earth diameter. 5. Oscillation of strings of musical instruments. Example of circular oscillation:- 1. Oscillation of simple pendulum. 2. Oscillation of solid sphere in a cylinder (If solid sphere rolls without slipping). 3. Oscillation of a circular ring suspended on a nail. 4. Oscillation of balance wheel of a clock. 5. Rotation of the earth around the sun. Oscillatory system:- 1. The system in which the object exhibit to & fro motion about the equilibrium position with a restoring force is called oscillatory system. 2. Oscillatory system is of two types such as mechanical and non- mechanical system. 3. Mechanical oscillatory system:- In this type of system body itself changes its position. For mechanical oscillation two things are specially responsible i.e Inertia & Restoring force. E.g oscillation of mass spring system, oscillation of fluid-column in a U-tube, oscillation of simple pendulum, rotation of earth around the sun, oscillation of body dropped in a tunnel along earth diameter, oscillation of floating cylinder, oscillation of a circular ring suspended on a nail, oscillation of atoms and ions of solids, vibration of swings…etc. 4.Non-mechanical oscillatory system:- In this type of system, body itself doesn‟t change its position but its physical property varies periodically. e.g:-The electric current in an oscillatory circuit, the lamp of a body which is heated and cooled periodically, the pressure in a gas through a medium in which sound propagates, the electric and magnetic waves propagates undergoes oscillatory change. Simple Harmonic Motion:- It is the simplest type of oscillatory motion. A particle is said to be execute simple harmonic oscillation is the restoring force is directed towards the equilibrium position and its magnitude is directly proportional to the magnitude and displacement from the equilibrium position. If F is the restoring force on the oscillator when its displacement from the equilibrium position is x, then F –x Here, the negative sign implies that the direction of restoring force is opposite to that of displacement of body i.e towards equilibrium position. F= -kx …………. (1) Where, k= proportionality constant called force constant. Ma=-kx M =-kx M kx=0 + x=0 ω2x=0 ………… (2) Where ω2= Here ω=√ is the angular frequency of the oscillation. Equation (2) is called general differential equation of SHM. By solving these differential equation x= + ……… (3) Where , are two constants which can be determined from the initial condition of a physical system. Appling de-Moiver‟s theorem x= cos +isin )+ cos -isin ) x= + ) cos + ) sin x= Ccos +Dsin ………(4) Where C = + & D= Let assume, C=A D=Acos Putting these value in equation (4) x=A cos +Acos sin x=A ( cos ) x=A sin ) ……….. (5) 2 Where A=√ +D2) & Similarly, the solution of differential equation can be given as x=Acos ) ………(6) Here A denotes amplitude of oscillatory system, ) is called phase and is called epoch/initial phase/phase constant/phase angel. Equation (5) and (6) represents displacement of SHM. Velocity in SHM:- =Asin ) =A cos ) v=A cos ) ………… (7) The minimum value of v is 0(as minimum value of Asin )=0 & maximum value is A. The maximum value of v is called velocity amplitude. Acceleration in SHM:- 2 a= -A sin ) …………. (8) 2 The minimum value of „a‟ is 0 & maximum value is A. The maximum valueof „a‟ is called acceleration amplitude. 2 Also, a= x (from equation (5)) a –y It is also the condition for SHM. Time period in SHM:- The time required for one complete oscillation is called the time period (T). It is related to the angular frequency( ) by. T= ……………… (9) Frequency in SHM:- The number of oscillation per time is called frequency or it is the reciprocal of time period. ʋ= = ……………(10) Potential energy in SHM:- The potential energy of oscillator at any instant of time is, U=-∫ =-∫ 2 = 2 = sin2 ) ………… (11) (By using equation (5)). Kinetic energy in SHM:- The kinetic energy of oscillator at any instant of time is, K= )2 = v2 K= A2ω2 cos2 ) ……. (12) (By using equation (7)) Both kinetic and potential energy oscillate with time when the kinetic energy is maximum, the potential energy is minimum and vice versa. Both kinetic and potential energy attain their maximum value twice in one complete oscillation. Total energy in SHM:- Total energy= K.E+P.E = A2ω2 cos2 )+ 2 sin2 ) 2 = cos2 )+ 2 sin2 ) 2 Total energy = Total energy = A 2ω 2 The total energy of an oscillatory system is constant. Graphical relation between different characteristics in SHM. COMPOUND PENDULUM (Physical pendulum):- Compound /physical pendulum is a rigid body of any arbitrary shape capable of rotating in a vertical plane about an axis passing through the pendulum but not through the pendulum but not through centre of gravity of pendulum. The distance between the point of suspension the centre of gravity is called the length of length of the pendulum &denoted by When the pendulum is displaced through a angle θ from the mean position,a restoring torque come to play which tries to bring the pendulum back to the mean position.But the oscillation continues due to the inertia of restoring force. Here the restoring force is -mgsinθ. So the restoring torque about the point of suspension “O” is τ=-mg sinθ. If the moment of inertia of the body about “OA” is “I”, the angular acceleration becomes, α=τ/I α= ……………….(1) For very small angular displace “θ “, we assume that Sin θ~θ. So, α=-mglθ/I. α=-(mg /I) θ………. (2) Also α=d2θ/dt2 Now we can write d2θ/dt2+ ( mg /I) θ =0……………..(3) d2θ/dt2+ω2θ=0……………..(4) Where, ω2= mg /I. And eqn(4) is the general equation of simple harmonic. T=2π(I/mg )1/2 T=2π( M(k2 +L2)/Mg )1/2. T=2π( (K2/l+l)/g)1/2……………………………(5). Here + =L, Called as equivalent length of pendulum.. If a line which is drawn along the line joining the point of suspension & Centre of gravity by the distance “ k 2/l”.we have another Point on the line called centre of Oscillation is equivalent Length of pendulum. So,the distance between centre of suspension & centre of Oscillation is equivalent length of pendulum.If these two points are interchanged then “time period” will be constant. L.C CIRCUIT(NON MECHANICAL OSCILLATION ):- In this region,it is combination “L” &”C” with the DC source through the key.If we Press the Key for a while then capacitor get charged & restores the charge as “+Q” and”-Q” with the potential “v=q/c” between the plates.When the switch is off the capacitor gets discharged. As capacitor gets discharged, q also decreases. So, current at that situation is given by I=dq/dt. As q decreases, electric field energy (Energy stored in electric field ) gradually decreases.This energy is transferred to magnetic field that appears around the inductor. At a time,all the charge on the capacitor becomes zero,the energy of capacitor is also Zero. Even though q equals to zero,the current is zero at this time. Mathematically, Let the potential difference across the two plates of capacitor at any instance” V” is given by V=q/c…………………(1) In the inductor due to increases in the value of flow of current, the strength of magnetic field ultimately the magnetic lines of force cut/ link with inductor changes. So a back emfdevelops which is given by ε =-L …………….(2) Now applying KVL to this LC circuit, +v-ε=0 +L =0 + =0 + =0………………………..(3). This represents the general equation of SHM, Here there is periodic execution of energy between electric field of capacitor & magnetic field of inductor. Here this LC oscillation act as an source of electromagnetic wave. Here, ω 2= ⁄ ω= ⁄ √ T=2π√ Damped oscillation:- For a free oscillation the energy remains constant. Hence oscillation continues indefinitely. However in real fact, the amplitude of the oscillatory system gradually decreases due to experiences of damping force like friction and resistance of the media. The oscillators whose amplitude, in successive oscillations goes on decreasing due to the presence of resistive forces are called damped oscillators, and oscillation called damping oscillation. The damping force always acts in a opposite directions to that of motion of oscillatory body and velocity dependent. Fdam –v Fdam=-bv b= damping constant which is a positive quantity defined as damping force/velocity, Fnet = Fres+ Fdam Fnet= -kx –bv Fnet= -kx– b M +kx+ b =0 + + x=0 +2β +ω02x = 0 …………. (2) Where β= is the damping co-efficient & ω0=√ is called the natural frequency of oscillating body. The above equation is second degree linear homogeneous equation. The general solution of above equation is found out by assuming x(t), a function which is given by x(t) = A =A = x = Aα2 = α2x Putting these values in equation α2x + 2α2βx + ω02x =0 α2 + 2α2β + ω02 =0 ………… (3) α = -β±√ ω02, is the general solution of above quadratic equation. As we know, x(t) =A1 + A2 ( √ ) ( √ ) x(t) = A1 + A2 x(t) = (A1 √ + A2 √ ) … (4) Depending upon the strength of damping force the quantity (β 2-ω02) can be positive /negative /zero giving rise to three different cases. Case-1:- if β ω02 underdamping (oscillatory) Case-2:- if β ω02 overdamping (non-oscillatory) Case-3:- if β ω02 critical damping (non-oscillatory) Case-1: [Under damping ω02 β2] If β2 ω02, then β2- ω02= -ve let β2-ω02=-ω21 √ =i ω1 whereω1= √ ω02- β2 = Real quantity So the general equation of damped oscillation/equation (IV) becomes X (t) = e-βt (A1eiω1t +A2e-iω1t) By setting A1=r/2eiθ and A2= r/2e-iθ, X(t)= e-βt[r/2ei(θ+ω1t)+ r/2 e-i(θ+ω1t)] =re-βt[ei(θ+ω1t)+ e-i(θ+ω1t)]/2 X(t)= re-βt cos(θ+ ω1t)………(v) Here cos(θ+ ω1t) represents the motion is oscillatory having angular frequency „ω1‟.The constant „r‟ and ‟ θ‟ are determined from initial potion & velocity of oscillatior T1=2π/ ω1 T1=2π/√ ω02- β2……(vi) (time period of damped oscillator) T1 T (where T= time period of undamped oscillator Implies f1 f Frequency of damped oscillator is less than that of the undamped oscillator. In under damped condition amplitude is no more constant and decreases exponentially with time, till the oscillation dies out. Mean life time:The time interval in which the oscillation falls to 1/e of its initial value is called mean life time of the oscillator. (τ) 1/e a= a e-βτm = , -β =loge1/e = Velocity of underdamped oscillation: X(t)=r cos(ω1t+ θ) r[-βe-βt cos(ω1t+ θ)-e-βtω1 sin( ω1t+ θ) = v=-re-βt[βcos(ω1t+ θ)+ω1 sin( ω1t+ θ)…(vi) Now , x=0& t=0, X(t)= re-βt cos(ω1t+ θ) 0 = re0 cos(0+ θ) 0 = cosθ Using the value of θ & t=0 in the equation (vii) we have = -r ω1 Where value of V0 in …………… Calculation of Energy(instantaneous): K.E = mv2 K.E = mv2 [β2cos2(ω1t+ θ)+ ω12sin2(ω1t+ θ)+ βω1sin2(ω1t+ θ)] Potential Enegy: P.E= kx2 = kr2 cos2 (ω1t+ θ) Total Energy: T.E=K.E+P.E = [( mv2+ kr2)cos2(ω1t+θ)+ mr2ω12sin2(ω1t+θ) mv2 βω1sin2(ω1t+ θ)] Total average energy: = mr2 ω02 =E0 Where, E0 =Total energy of free oscillation The average energy decipated during one cycle =Rate of energy = = Decrement The decrement measures the rate at which amplitude dies away. The ratio between amplitude of two successive maxima, is the decrement of the oscillator. re-βt/ re-β(t+T) = re+βt The logarithmic decrement of oscillator is „λ‟ loga π /√ ω02- β2 logaa0/a1=loga1/a2=..........=eβt= Rate of two amplitudes of oscillation whichare separated by one period Relaxation time( : It is the time taken by damped oscillation by decaying of its energy 1/e of its initial energy. ε0=ε0 = Loge-1=log -1= =1/2 =m/b Case-II:(over damping oscillation) Here β2 ω02 √ β2-ω02=+ve quantity = α (say) X (t) = e-βt (A1eαt +A2e-αt)…………(viii) Depending upon the relative values of α, β ,A1 , A2& initial position and velocity the oscillator comes back to equilibrium position. The motion of simple pendulum in a highly viscous medium is an example of over damped oscillation. Quality factor: Q= π =. Critical damping: β2 = ω02 The general solution of equation (ii) in this case, X(t) = (Ct+D) e-βt …………………………………(ix) Here the displacement approaches to zero asymptotically for given value of initial position and velocity a critically damped oscillator approaches equilibrium position faster than other two cases. Example: The springs of automobiles or the springs of dead beat galvanometer. Curves of three Cases: Forced Oscillation The oscillation of a oscillator is said to be forced oscillator or driven oscillation if the oscillator is subjected to external periodic force. If an external periodic sinusoidal force „Fcosωt‟ acts on a damped oscillator, its equation of motion is written as, Fnet= -kx- b +Fcosωt m + = -kx –b +Fcosωt ++ + x = cosωt + 2β + ω02x = f0cosωt ----------------------------------------------------- (i) , Where β= ω02= and f0 = , and β and ω02 respectively called as damping coefficient, natural frequency. Equation (i) is also represented as ̈+ ̇+ 0 2 x= f0cosωt Equation (i) represents the general equation of forced oscillation. Equation (i) is a non-homogenous differential equation with constant co-efficient. For weak damping (ω02 >β 2) , the general equation contains, x(t) = xc(t) + xp(t) Where xc(t) is called complementary solution and its value is √ √ xc(t)= A1 +A2 ) ………………(ii) Now xp(t) is called the particular integral part. Let us choose xp(t) = P cos (ωt-δ) ̇ (t)= -Pωsin(ωt-δ) ̈ (t)=-Pω2cos(ωt-δ)………………………(iii) Putting xp(t) , ̇ , ̈ (t) in eqn (i) we get - Pω2cos (ωt-δ)-2βPω sin (ωt-δ) + ω02Pcos(ωt-δ) = f0cosωt - Pω2cos (ωt-δ)-2βPω sin (ωt-δ) + ω02Pcos(ωt-δ) = f0cos (ωt-δ + δ ) - Pω2cos(ωt-δ)-2βPω sin(ωt-δ) + ω02Pcos(ωt-δ) = f0 [ cos (ωt-δ ).cos δ – sin(ωt-δ ).sinδ ] Now, compairing the coefficient of cos(ωt-δ) and sin (ωt-δ) on both sides, (ω02-ω2)P = f0cosδ ………………………………(iv) 2βPω = f0sinδ ………………………………………(v) Squaring and adding eqn (iv) & (v) {(ω02-ω2)P}2 +4 β2 P 2ω2 =f02 P= …………………(vi) √( ) Now dividing eqn (v) by (iv) δ= ( ) …………………..(vii) xp = cos(ωt-δ) (steady state solution) √( ) Now, x(t) = xc(t) + xp(t) √ √ x(t) = A1 + A2 )+ cos(ωt- √( ) δ) Steady state behavior: Frequency:-The Oscillator oscillates with the same frequency as that of the periodic force. ω0and ω are very close to each other then beats will be produced and these beats are transient as it lasts as long as the steady state lasts. The duration between transient beats is determined by the damping coefficient „β‟. Phase: The phase difference „δ‟ between the oscillator and the driving force or between the displacement and driving is δ= ( ) This shows that there is a delay between the action of the driving force and response of the oscillator. (In the above figure fQ= ω0 and fp= ω ) At ω=ω0 , φ₌ , the displacement of the oscillator lags behind the driving force by. At ω ω0 then δ =- → -0= Amplitude: The amplitude of driven oscillator , in the steady state , is given by A= = √( ) √( ) It depends upon (ω02-ω2). If it is very small, then the amplitude of forced oscillation increases. Case-1: At very high driving force i.e ω>>ω0 and damping is small (β is small) or ( β→0) A= √ A= A= Amplitude is inversely proportional to the mass of the oscillator & hence the motion is mass controlled motion. Case-2: At very low driving force (ω = (( ) ) ( ) =2 ( ) 2 ⁄ | | [ ] | | For a particular A | | INTERFERENCE Coherent Superposition: The superposition is said to be coherent if two waves having constant phase or zero phase difference. In this case, the resultant intensity differs from the sum of intensities of individual waves due to interfering factor. i.e. I I1 I 2 Incoherent Superposition: The superposition is said to be incoherent if phase changes frequently or randomly. In this case, the resultant intensity is equal to the sum of the intensities of the individual waves. i.e. I I1 I 2 Two Beam Superposition: When two beam having same frequency, wavelength and different in amplitude and phase propagates in a medium, they undergo principle of superposition which is known as two beam superposition. Let us consider two waves having different amplitude and phase are propagated in a medium is given as (1) (2) Applying the principle of superposition (3) Let (4) and (5) (6) Squaring and adding equation (4) and (5) A √ (7) We know, √ √ (8) Dividing equation (5) by (4), we get, Coherent Superposition: In coherent superposition, the phase difference remains constant between two beams. Now equation (7) and (8) becomes, √ and √ √ √ The two beams having same amplitude, Again, if √ √ √ √ For same amplitude, Incoherent Superposition: In incoherent superposition the phase difference between the waves changes frequently or randomly, so the time average of the interfering term √ ) vanishes as the cos value varies from -1 to 1. Here, √ Multiple beam superpositions: When a number of beams having same frequency, wavelength and different amplitude and phase are undergoing the superposition, such superposition is known as multiple beam superpositions. Let , , ,............. be the number of beams having same frequency, wavelength and different in amplitude and phase are propagating in a medium are given as, : : : According to principle of superposition, , , ,............. ∑ ∑ ∑ (1) where resultant amplitude of the ith component. Phase of the ith component. ∑ (2) ∑ (3) Squaring and adding (2) and (3) we get, ∑ ∑ The phase angle is given as, ∑ ∑ Coherent Superposition: In this case the phase difference between the waves remains constant i.e. ∑ ∑ If all the beams having equal amplitudes. i.e. Now, I coherent N 2 I1 Incoherent Superposition In incoherent superposition, the phase difference between the beams changes frequently or randomly due to which the time average of factor ∑ vanishes as cos value varies from- 1 to +1 ∑ ∑ Now , I incoherent KA 2 ∑ I coherent I incoherent NI 1 N I icoherent Interference: The phenomenon of modification in distribution of energy due to superposition of two or more number of waves is known as interference. To explain the interference, let us consider a monochromatic source of light having wavelength and emitting light in all possible directions. According to Huygens‟s principle, as each point of a given wavefront will act as centre of disturbance they will emit secondary wave front on reaching slit S1 and S2. As a result of which, the secondary wave front emitted from slit S1 and S2 undergo the Principle of superposition. During the propagation, the crest or trough of one wave falls upon the crest and trough of other wave forming constructive interference, while the crest of one wave of trough of other wave producing destructive interference. Thus, the interfering slit consisting of alternate dark and bright fringes, which explain the phenomenon of interference. Mathematical treatment: Let us consider two harmonic waves of same frequency and wavelength and different amplitude and phase are propagating in a medium given as Let Squaring and adding (2) and (3) * √ + As, I [ √ √ ] Dividing equation (3) by (2) we get, Condition for maxima: The intensity will be maximum when the constructive interference takes place i.e. , n=0, 1, 2... [ ] The constructive interference is when difference is even multiple of or integral multiple of 2 and path difference is an integral multiple of. Now, [ √ ] [ √ √ ] If the waves having equal amplitude, ] Condition for minima The intensity will be minimum destructive interference takes place i.e. Where n = 0, 1, 2, 3... [ ] Thus destructive interference takes place when phase difference is odd multiple of and path difference is odd multiple of. Now, [ √ ] [ √ √ ] Intensity distribution curve If we plot a graph between phase difference or path difference along X-axis and intensity along Y-axis, the nature of the graph will be symmetrical on either side. From the graph, it is observed that, 1) The fringes are of equal width 2) Maxima having equal intensities 3) All the minima‟s are perfectly dark The phenomenon of interference tends to conservation of energy i.e. the region where intensity is 0, actually the energy present is maxima. As the minima‟s and maxima position changes alternatively so the disappearance of energy appearing is same as the energy appearing in other energy which leads to the principle of conservation of energy. Sustained Interference The interference phenomenon in which position of the maxima and minima don‟t changes with time is called sustained interference. Condition for Interference 1) The two waves must have same frequency and wavelength. 2) The two source of light should be coherent. 3) The amplitude of wave may be equal or nearly equal. Condition for good Contrast I. The two slit must be narrow. II. The distance between the two slit must be small. III. The background should be perfectly dark. IV. The distribution between the slit and the screen should be large. V. The two waves may have equal or nearly equal amplitude (for sharp superposition). Coherent Sources The two sources are said to be coherent if they have same phase difference, zero phase difference or their relative phase is constant with respect to time. Practical resolution of Coherent Coherent sources from a single source of light can be realised as follows A narrow beam of light can be split into its number of component waves and multiple reflections. Component light waves are allowed to travel different optical path so that they will suffer a path difference and hence phase difference. [ ] Methods for producing coherent sources/Types of interferences Coherent sources can be produced by two methods 1) Division of wave front 2) Division of amplitude Division of Wave front The process of coherent source or interference by dividing the wave front of a given source of light is known as division of wave front. This can be done by method of reflection or refraction. In this case a point source is used. Examples 1. YDSE 2.Lylord‟s single mirror method 3.Fresnel‟s bi-prism 4.Bilet splitting lens method DIVISION OF AMPLITUDE The process of obtaining a coherent source by splitting the amplitude of light waves is called division of amplitude which can be done by multiple reflections. In this case, extended source of light is used. 1.Newton‟s ring method 2.Thin film method 3. Michelson‟s interferometer Young’sDouble Slit Experiment: In 1801 Thomas Young demonstrated the phenomenon of interference in the laboratory with a suitable arrangement. It is based on the principle of division of wavefront of interference. The experiential arrangement consists of two narrow slits, S1 and S2 closely spaced, illuminated by a monochromatic source of light S. A screen is placed at a distance D from the slit to observe the interference pattern. In the figure, d Slit separation D Slit and screen separation Wavelength of light Y distance of interfering point from the centre of slit x Path difference coming from the light S1 and S2 Optical path difference between the rays coming through S1 and S2 Now the path difference, In figure, [ ] (Using binomial theorem) Similarly, * + The alternative dark and bright patches obtained on the interference screen due to superposition of light waves are known as fringe. Condition for bright fringe The bright fringe is obtained when the path difference is integral multiple of i.e. x n From equation (4) and (5), we get Where n = 0, 1, 2 …… Condition for dark fringe It will be obtained when the path difference is an odd multiple of λ/2 i.e. From (4) and (6), we get Where n = 0, 1, 2 …… Fringe Width The separation between two consecutive dark fringes and bright fringes is known as fringe width. If and be the two consecutive bright fringe. Similarly, is and be the two consecutive dark fringes. It is concluded that the separation between the two consecutive bright fringes is equal to the consecutive dark fringes. So Hence bright and dark fringes are equispaced. Discussion: From the expression for If young double slit apparatus is shifted from air to any medium having refractive index (µ), fringe pattern will remain unchanged and the fringe width decreases (1/µ) as λ decreases. If YDSE is shifted from air to water, the fringe width decreases3/4 times width in air. When YDSE is performed with white light instead of monochromatic light we observed, I. Fringe pattern remains unchanged II. Fringe width decreases gradually III. Central fringe is white and others are coloured fringes overlapping When YDSE is performed with red, blue and green light So [ ] Wavelength of light in any given medium, decreases to1/µ times of wavelength in vacuum. [ ] So, it decreases 1/µ times. Newton’s Ring The alternate dark and bright fringe obtained at the point of contact of a Plano convex lens with its convex side placed over a plane glass plate are known as Newton‟s ring as it was first obtained by Newton. The formation of the Newton‟s ring is based on the principle of interference due to division of amplitude. Experimental Arrangement The experimental arrangement consist of a) S: Monochromatic source of monochromatic light b) P: A plane glass plate c) L: A convex lens which is placed at its focal length to make the rays parallel after refraction d) G: A plane glass plate inclined at on 450 to make the parallel rays travel vertically downwards e) L‟: A plane convex lens of long focal length whose convex side kept in contact with plane glass plate f) T: Travelling microscope mounted over the instrument to focus the Newton‟s ring. Formation of Newton’s Ring I. To explain the formation of Newton‟s ring, let us consider a plano-convex lens with its convex side kept in contact with a plane glass plate. II. At the point of contact air film is formed whose thickness gradually goes on increasing towards outside. III. When a beam of monochromatic light is incident on the arrangement, a part of it get reflected from the upward surface of the air film and the part of light get reflected from the lower surface of the air film. IV. The light which reflected from glass to air undergoes a phase change of „π‟ and those are reflected from air glass suffers no phase change. V. As a result of which they super-impose constructively and destructively forming the alternate dark and bright fringe at the point of contact. Condition for bright and dark fringe in Reflected light In Newton‟s ring experiment, the light travels from upper and lower part of the air film suffers a path difference of λ/2 (phase change of π). Again, as the ray of light reflected twice between the air films having thickness„t‟. Then the total path travelled by the light is given as. Now, from the condition for bright ring, we have, From the condition for the dark fringe we have, Newton’s ring in transmitted light The Newton‟s rings obtained in transmitted light are complementary to that of Newton‟s ring obtained in reflected light i.e. In transmitted light, the condition for bright ring is, And for dark ring is, Newton’s ring in Reflected Light and Transmitted Light In reflected light In transmitted light (a)Condition for bright ring; (a)Condition for bright ring; (b) Condition for dark (b) Condition for bright ring; ring; (c)Newton‟s rings are more (c)Newton‟s rings are less intense. intense. DETRMINATION OF DIAMETER OF NEWTON‟S RING LOL‟ is the section of lens placed on glass plate AB. C is the centre of curvature of curved surface LOL‟. R is its radius of curvature and r is the radius of Newton‟s ring corresponding to film if thickness t. From the property of circles, t = thickness of air film ( t From the condition for bright Newton‟s ring, , For the nth ring. Q) Show that diameter of Newton‟s dark or bright fringe is proportional to root of natural numbers. √ √ √ √ , n = 1, 2, 3……. Thus the diameter of Newton‟s bright ring is proportional to square root of odd natural numbers. Similarly from the Newton‟s dark ring, √ √ √ √ √ √ Thus the diameter of Newton‟s dark ring is proportional to square root of natural numbers. Determination of wavelength of light using Newton’s ring method To determine the wavelength of light, let us consider the arrangement which involves a travelling microscope mounted over the Newton‟s ring. Apparatus, on focusing the microscope over the ring system and placing the crosswire of the eye piece on tangent position, the readings are noted. On taking readings on different positions of the crosswire on various rings we are able to calculate the wavelength of light used. Let and be the nth and (n+p)th dark ring, then we have, Subtracting equation (1) from (2) we get, This is the required expression from the wavelength of light for Newton‟s ring method. If we plot a graph between the orders of ring along X-axis and the diameter of the ring along Y-axis, the nature of the graph will be a straight line passing through origin. From the graph the wavelength of light can be calculated the slope of the slope of the graph. Slope of the graph = wavelength of light Determination of refractive index of liquid by Newton’s ring The liquid whose refractive index is to be determined is to be placed between the gap focused between plane convex lens and plane glass plate. Now the optical path travelled by the light is to be 2µt, instead of 2t where µ be the refractive index of the liquid from the condition for the Newton‟s ring we have, For nth ring, Let and be the diameter of the (n+p)th and nth dark ring in presence of liquid then 4n p R 4nR Dn'2 p and Dn'2 Now , 4n p R 4nR Dn'2 p - Dn'2 = - = 4 pR (1) If the same order ring observed in air then Dn2 p Dn2 4 pR (2) Dividing equation (2) by (1) ,we have D 2 n p Dn2 air D '2 n p D '2 n liquid This is the required expression for refractive index of the liquid. DIFFRACTION Fundamental Idea about diffraction: The phenomenon of bending of light around the corner of an aperture or at the edge of an obstacle is known as diffraction The diffraction is possible for all types of waves The diffraction verifies the wave nature of light Diffraction takes place is due to superposition of light waves coming from two different points of a single wave front Diffraction takes place when the dimension of the obstacle is comparable with the wavelength of the incident light. Explanation of diffraction: To explain diffraction, let us consider an obstacle AB is placed on the path of an monochromatic beam of light coming from a source „S‟ which produces the geometrical shadow CD on the screen. This proves the rectilinear propagation of light. If the dimension or size of the obstacle is comparable with the wave length of the incident light, then light bends at the edge of the obstacle and enters in to the geometrical shadow region of the obstacle. According to Fresnel inside a well region, the destructive interference takes place for which we get brightest central maxima, which is associated with the diminishing lights on either side of the shadow as the constructive interference takes place out side the well region. This explains the diffraction phenomena. Types of Diffraction: Depending on the relative position of the obstacle from the source and screen, the diffraction is of 2 types. a. Fresnel Diffraction b. Fraunhoffer Diffraction Fresnel’s Diffraction Fraunhoffer Diffraction (1) The type of diffraction (1) The type of diffraction in which the distance of in which the distance of either source or screen or either source or screen or both from the obstacle is both from the obstacle is finite, such diffraction is infinite, such diffraction is known as Fresnel‟s known as Fraunhoffer diffraction. diffraction. (2) No lenses are used to (2) Lenses are used to make the rays converge or make the rays converge or parallel. parallel. (3) The incident wave (3) The incident wave front is either cylindrical or front is plane. spherical. Ex:. The diffraction at the Ex:The diffraction at the narrow. straight edge. Fraunhoffer Diffraction due to a single slit: Let us consider a parallel beam of monochromatic light inside on a slit „AB‟ having width „e‟. The rays of the light which are incident normally on the convex lens „L2‟, they are converged to a point „P0‟ on the screen forming a central bright image. Fraunhoffer diffraction due to single slit Schematic digram for Fraunhoffer diffraction due to single slit The rays of light which get deviated by an angle „θ‟, they are converged to a point „P1‟, forming an image having lens intensity. As the rays get deviated at the slit „AB‟ they suffer a path difference. Therefore path difference, BK = AB Sinθ = e sinθ 2 Therefore, Phase difference = e sin Let us divide the single slit into ‟n‟ no. of equal holes and a be the amplitude of the light coming from each equal holes. 1 2 Then Avg. phase difference= e sin n Now the resultant amplitude due to superposition of waves is given as nd n 1 2 a sin a sin e sin a sin e sin R 2 = 2 n = d 1 2 1 sin e sin sin e sin n 2 sin n 2 a sin Let e sin ,then R sin n Since is very small and n is very large so is also very n small. Therefore, sin n n na sin A sin Thus, R = a sin whereA an n Now the intensity is given as A sin 2 2 sin 2 Iα R2 I KR I K = I0 where I 0 KA 2 2 2 Condition for Central /principal maxima: When α = 0, e sin 0 sin 0 0 Thus, the condition for principal maxima will be obtained at 0 position for all the rays of light. Position for/Condition for minima: The minimum will be obtained when sin 0 sin(m ) sin sin m m e sin m e sin m where m 1,2,3,4,....... m e Thus, the minimas are obtained at ,2 ,3 ,4 ,........ e e e e Position/Condition for secondary maxima: The maxima‟s occurring in between two consecutive secondary maxima is known as secondary maxima. The positions for secondary maxima will be obtained as dI 0 d d sin 2 I 0 0 d 2 sin cos sin 2I 0 0 2 cos sin =0 2 cos sin 0 tan This is a transdectional equation.It can be solved by graphical method. Taking y and y tan ,where the two plots are interests, this intersection points gives the position for secondary maxima. Thus the secondary maxima‟s are obtained at 3 5 7 , , ......... 2 2 2 From the expression for amplitude we have R = A sin A 3 5 7 ............. 3! 5! 7! = A x 1 2 4 ................ = A, since α ‹‹ 1 3! 5! Thus the intensity at the central principal maxima is I0 3 Sin 2 ( ) Sin 2 For α= 3 , I1 =I0 = I0 2 = I0 2 2 3 2 22 2 5 Sin 2 ( ) For α= 5 , I2 =I0 Sin 2 = I0 2 2 = I0 and so on …… 2 5 2 62 2 Intensity distribution curve: The graph plotted between phase difference and intensity of the fringes is known as intensity distribution curve. The nature of the graph is as follows: Intensity distribution curve From the nature of the graph it is clear that 1. The graph is symmetrical about the central maximum 2. The maxima are not of equal intensity 3. The maxima are of not equal width The minima are of not perfectly dark PLANE TRANSMISSION GRATING: It is an arrangement consisting of large no.of parallel slits of equal width separated by an equal opaque space is known as diffraction grating or plane transmission grating. Diffraction grating Construction: It can be constructed by drawing a large no. of rulings over a plane transparent material or glass plate with a fine diamond point. Thus the space between the two lines act as slit and the opaque space will acts as obstacle. N.B.Though the plane transmission grating and a plane glass piece looks like alike but a plane transmission grating executes rainbow colour when it exposed to sun light where as a plane glass piece does not executes so. Grating element: The space occurring between the midpoints of two consecutive slit in a plane transmission grating is known as Grating element. It can be measured by counting the no. of rulings present in a given length of grating. Let us consider a diffraction grating having e = width of the slit d = width of the opacity If “N” be the no. of rulings present in a given length of grating “x” each having width (e+d), then N (e+d) = x x (e d ) Grating element N For example if a grating contain 15,000 lines per cm in a grating then the grating element of the grating 1 Grating element, (e+d) = =0.00016933 cm 15000 Diffraction due to plane transmission grating /Fraunhoffer diffraction due to N-parallel slit: Let us consider a plane wave front coming from an infinite distance is allowed to incident on a convex lens “L” which is placed at its focal length. The rays of light which are allowed to incident normally on the lens are converged to a point “P o” forming central principal maxima having high intensity and the rays of light which are diffracted through an angle are “θ” are converge to a point “P1” forming a minima having less intensity as compared to central principal maxima. Again those rays of light which are diffracted through an angle “θ” are undergoes a path difference and hence a phase difference producing diffraction. Let AB- be the transverse section of the plane transmission grating WW ' - be a plane wave front coming from infinite distance e = width of the slit d = width of the opacity (e+d) = grating element of the grating N = be the no. of rulings present in the grating Now the path difference between the deviated light rays is S2K = S1S2Sinθ = (e d )Sin Therefore, Phase difference = 2 x S2K = 2 (e d )Sin = 2 (say) where (e d ) Sin Now the resultant amplitude due to superposition of “N” no.of waves coming from “N” parallel slit is given as Sin SinN RA Sin and intensity is given as Sin 2 Sin 2 N Sin 2 Sin 2 N IR 2 I KR 2 KA 2 I 2 Sin 2 2 Sin 2 0 Sin 2 where I0 =this is contributed due to diffraction at single slit 2 Sin 2 N and = this is contributed due to interference at ” N” Sin 2 parallel slit Position for central principal maxima /condition for central principal maxima: The principal maxima will be obtained when Sin o Sin ( m ) m (e d ) Sin m (e d ) Sin m where m 0,1,2,3.......This is called grating equation or condition for central principal maxima. Position for minima /condition for minima: The minima will be obtained when SinN o Sin ( n ) N n N (e d ) Sin n N (e d ) Sin n Where n can take all the values except n 0, N ,2 N ,3N ,.......... This is the condition for minima due to diffraction at N-parallel slit. Position/Condition for secondary maxima: The maxima‟s occurring in between two consecutive secondary maxima is known as secondary maxima. The positions for secondary maxima will be obtained as dI 0 d d Sin 2 Sin 2 N I 0 0 d 2 Sin 2 Sin 2 Sin N N cos N sin sin N cos 2I 0 0 2 Sin sin 2 N cos N sin sin N cos =0 sin 2 N cos N sin sin N cos =0 N cos N sin sin N cos N tan N tan N This is a transdectional equation. It can be solved by graphical method. Taking y tan N and y N tan N ,where the two plots are interests, this intersection points give the position for secondary maxima.Thus the secondary maxima‟s are obtained at 3 5 7 , , ......... 2 2 2 Intensity distribution curve: The graph plotted between phase difference and intensity of the fringes is known as intensity distribution curve. The nature of the graph is as follows: Characteristics of the spectral lines or grating spectra: 1.The spectra of different order are situated on either side of central principal maximum 2.Spectral lines are straight and sharp 3.The spectra lines are more dispersed as we go to the higher orders. 4.The central maxima is the brightest and the intensity decreases with the increase of the order of spectra. Missing spectra or Absent spectra: When the conditions for minima due to diffraction at single slit and condition for central principal maxima due to diffraction at N-parallel slit is satisfied simultaneously for a particular angle of diffraction then, certain order maxima are found to be absent or missed on the resulting diffraction pattern which are known as missing spectra or absent spectra. Condition for Missing spectra: We have, The condition for central principal maxima due diffraction at N- parallel slit (e d )Sin m e sin n (e d ) Sin m m e sin n n Special case: 1. If d = e, m 2 m 2n where n 1,2,3,..... n i.e second order or multiple of 2 order spectra will found to be missed or absent on the resulting diffraction pattern. e 2. If d , m 3 m 1.5n 1 2 n 2 i.e First order spectra will found to be missed or absent on the resulting diffraction pattern. d 3. If e , m 3 m 3n 2 n i.e Third order spectra or multiple of 3 spectra will found to be missed or absent on the resulting diffraction pattern. Dispersion: The phenomenon of splitting of light wave into different order of spectra is known as dispersion. Dispersive power: The variation of angle of diffraction with the wave length d of light is known as dispersive power. It is expressed as d Where d 1 2= difference in angle of diffraction and d 1 2 =difference in wave length of light Expression for dispersive power: We have (e d )Sin m d (e d )Sin m d m d d (e d ) d Sin m d d d d (e d ) cos m d d m = d (e d ) cos d α m d 1 α (e d ) 1 α cos Determination of wave length of light using plane transmission grating: To determine the wave length of light let us consider a plane transmission grating with its rulled surface facing towards the source of light perpendicular to the axis of the spectrometer. The parallel beam of monochromatic light coming from source is allowed to incident on the transmission grating which are now defracted by different angle of diffraction.Rotating the telescope for different positions of the defracted ray the angles are measured. Using the grating equation , (e d )Sin m (e d ) Sin m We can calculate the wave length of the monochromatic light. Half period zone: The space enclosed between two consecutive circles which are differing by phase of π or by a path difference of or a time 2 T period of is known as half period zone. As it was first 2 observed by Fresnel, these are also known as Fresnel half period zone. Construction: To construct the half period zone let us consider a plane wave front of monochromatic source of light having wavelength λ coming from left to right. Let “P” be a point just ahead of the plane wave front at a perpendicular distance “b” from the plane wavefront. Taking “P” as centre and radii equal to OM1= r1,OM2= r2,OM3= r3…OMn= rn let us divide the plane wave front into large no. of concentric circles such that light coming from each consecutive half period zone will differ by a phase difference of. 2 These alternative circles which are now differing by a phase change of π are known as half period zone. These half period zones are known as Fresnel half period zone. The Fresnel‟s first half period zone is brighter than that of a second half period zone and the two half period zone are differ by a phase change of π. Properties of Half period Zone: 1. Phase of Half period Zone: Each half period zone are differ by a phase change of π 2. Area of half period zone: The space enclosed between two consecutive half period zones is called area of Half period zone. Let An-1 and An be the area of (n-1) th and nth half period zone 2 (n 1) 2 Then, An-1 = π (OMn-1) = ( PM OP ) = b 2 1 2 b 2 2 2 (n 1) 2 2 (n 1) = b 2b b2 4 2 2 = (n 1) 2 b(n 1) = π(n-1)bλ 4 (n 1) 2 2 Since λ 1 so 1 and hence neglected 4 2 n 2 and An = π (OMn) = ( PM OP ) = b b 2 2 n 2 2 2 2 n = b 2 n 2b b2 4 2 n2 2 = bn = πnbλ 4 n 2 2 Since λ 1 so 1 and hence neglected 4 Now the area of the half period zone A= An - An-1 = πnbλ- π (n-1) bλ = πbλ Thus the area of half period zone is independent of order of zone and the half period zones are equispaced 3. Radius of half period zone: We have, The area of first half period zone is πbλ i.e A1= πbλ Again, A1= r12 r12 = πbλ r12 b r1 1b Similarly, the radius of the second half period zone is r2 2b and the radius of the third half period zone is r3 3b ,……. rn nb. Thus it is found that radius of the half period zone is dependent on order of zone and the radius of the half period zone is varies directly proportional to the square root of the natural number. Factors affecting amplitude of half period zones: The factors affecting the amplitude are: a. Area of half period zone (directly) b. Average distance of half period zone (inversely) c. Obliquity factor (directly) Mathematically, If „R‟ be the radius of the half period zone, then RA (1 cos ) 1 d A(1 cos ) R d Expression for amplitude of half period zone: Let R1,R2,R3…….Rn be the amplitudes of 1st, 2nd, 3rd,……nth half period zone respectively. Then the net amplitude due to the entire half period zone is given by R R1 R2 R3 .........Rn R1 R2 R3 ........ .Rn (If n is odd) R1 R2 R3 ........ .Rn1 (If n is even) Since R 1 R2 , R 2 R3 so we have R1 R3 R R R2 and R4 3 5 and so on 2 2 R R R R R R 1 R2 3 1 R4 5 ......... n if n is even 2 2 2 2 2 R R R R R 1 R2 3 1 R4 5 ......... n1 if n is odd 2 2 2 2 2 R1 Rn R if n is odd 2 2 R1 Rn 1 = if n is even 2 2 Rn1 Rn R1 As n 1 and or 1 so, R 2 2 2 Thus the net amplitude due to entire half period zone is equal to half of the amplitude due to first half period zone. Zone plate: A special diffracting screen which obstructs the light from alternate half period zone is known as zone plate. Construction: It can be constructed by drawing a series of concentric circles on a white sheet of paper with radii proportional to square root of natural number. The alternate half period zones are painted black. A reduced photograph of this drawing is taken on a plane glass plate. The negative thus obtained act as zone plate. Depending on the initial blackening the zone plate is of two types 1. Positive zone plate: 2. Negative zone plate: the the center is bright center is dark Working: When a beam of monochromatic light is allowed to fall on a zone plate, the light is obstructed from the alternate half period zone through the alternate transparent zones. So,the rays of light differ by a phase difference of π. Hence, the resultant amplitude is sum of the individual amplitude due to light coming from alternate half period zones. Thus for any point object situated at infinite produces a bright image at a particular distance which is same as that of image produced by a convex lens. Thus a zone plate is equivalent to that of a convex lens. Theory of zone plate: Let us consider a transverse section of a zone plate placed perpendicular to the plane of the paper. Let „O‟ be a point object placed at a distance „ OP u ‟ forms a real image „I‟ at a distance „ PI v ‟ from the zone plate. Taking „P‟ as center and radii equal to PM1= r1,PM2= r2,PM3= r3…PMn= rn, , the entire plane of the paper is divided into large no. of concentric circles such that the light coming from alternate half period zone will differ by a path difference of in 2 such a way that OM1 I OPI 2 2 OM 2 I OPI 2 3 OM 3 I OPI 2 ………………….. ………………….. n OM n I OPI 2 n OM n I PM n I (1) 2 Now, in right angled ΔOPMn 1 OM n OP 2 PM 2 n 2 1 r 2n 2 r 2n 1 u2 r 2 2 n u 1 2 = u 1 2 ..... , as rn u u 2u r 2n r 2n u 1 2 u (2) 2u 2u Similarly, in right angled ΔPMnI 1 M n I M n I 2 PM 2 n 2 1 r 2n 2 r 2n 1 v r2 2 n 2 v 1 2 = v 1 2 ..... , as rn v v 2v r 2n r 2n v 1 2 v (3) 2v 2v Using eqn (2) and eqn (3) in eqn (1) we get r 2n r 2n n u v (u v) 2u 2v 2 r 2 n 1 1 n (u v) (u v) 2 u v 2 r 2n 1 1 n = 2 u v 2 1 1 r 2 n n (4) u v u v r 2n = n uv uv rn n uv rn cons. n rn n Thus the radius of the zone plate is proportional to square to natural number. Expression for primary focal length: From eqn. (4) we have, 1 1 r 2 n n u v According to sign convention, 1 1 r 2 n n u v 1 rn2 r 2 n n f (5) f n This is the required expression for primary focal length 1 Again, f fx = constant Area of zone plate: The space enclosed between two consecutive zones is known as area of zone plate. Let An-1 and An be the area of (n-1) th and nth zone Then A= A - A = r 2 r 2 = uvn uvn 1 = uv = constat n n-1 n 1 uv uv uv n Thus, the area of zone plate is independent of order of zone i.e the zones are equispaced. Multiple foci of zone plate: Now from the expression we have, 1 1 r 2 n n u v If the object is situated at infinity (∞), then the first image at distance , rn2 v1 f1 = n If we divide the half period zones into half period elements having equal area, then the 1st half period zone will divided into three half period zones,2nd half period zone will divided into five half period elements and so on The second brightest image will at 1 1 rn2 v3 f 3 = f 1 3 3 n 2 The third brightest image will at v5 f 5 = 1 f1 1 rn and so on…. 5 5 n Thus it is conclude that the zone plate has multiple foci. Comparison between the zone plate and the convex lens: Similarities 1. Both form the real image. 2. The relation between consecutive distances is same for both. 3. In both the cases focal length depends on wave length of the light. Dissimilarities Convex Lens Zone Plate a) Image is formed by a) Image is formed by refraction diffraction b) It has a single focus. b) It has multiple foci c) The focal length increases c) The focal length with increase of wave length. decreases with increase of d) Image is more intense wavelength e) The optical path is constant d) Image is less intense for all the rays of light. e) The optical path is different for different rays of light Phase reversal Zone Plate: The zone plate which is constructed in such a way that the light coming from two successive zones differ by an additional path difference of λ/2, such zone plate is known as phase reverse zone plate. Huygens’s Principle: About the propagation of the wave, Huygens suggested a theory which is based on a principle known as Huygens‟s principle. It states that:- 1) Each point on a given wave front will act as centre of disturbances and emits small wavelets called secondary wave front in all the possible direction. 2) The forward tangent envelope to these wave lets gives the direction of new wave front. Explanation/construction of secondary wave front: To explain Huygens‟s principles let us consider a source of light emits waves in all directions. Let AB be the wave front at t=0. As the time advances each point on the given wave front AB will act as centre of disturbance and emit wave lets in all possible directions. Taking a, b, c, d, e as centre and radii equal to „ct‟ (c- velocity of light &„t‟ time), we can construct a large number of spheres which represents a centre of disturbance for the new wave. The length A1B1 represents the direction of new wave front. N.B. The backward front is not visible as the intensity of the backward wave front is very small since for the backward wave front, I = k (1+cosθ) since for backward wave 0 front (θ=180 ) I=k (1-1) =0 Iback=0 POLARISATION The phenomenon of restricting the vibration of light in a particular direction perpendicular to the direction of wave motion is called as polarisation. To explain the phenomenon of polarisation let us consider the two tourmaline crystal with their optics axis placed parallel to each other.When an ordinary light is incident normally on the two crystal plates the emergence light shows a variation in intensity as T2 is rotated. The intensity is maximum when the axis of T2 is parallel to that of T1 and minimum when they are at right angle. This shows that the light emerging from T1 is not symmetrical about the direction of propagation of light but its vibration are confined only to a single line in a plane perpendicular to the direction of propagation, such light is called as polarised light. Example: Difference between Polarised and ordinary light: Polarised light Ordinary light 1. The vibrations are confined 1. The vibrations of light in a particular direction. particle are not confined in a 2. The probability of particular direction. occurrence of vibration 2. The probability of along the axis of crystal is occurrence of vibration not same in all position of along the axis of the crystal crystal is not symmetries for all 3. The intensity of light plate position of the crystal. is not same in all position of 3. The intensity of light plate the crystal plate. is same in all position of the plate. Polarised light: The resultant light wave in which the vibrations are confined in a particular direction of propagation of light wave, such light waves are called Polarised light. Depending on the mode of vibration in a particular direction, the polarised light is three types Linearly Polarised /Plane polarised: When the vibrations are confined to a single linear direction at right angles to the direction of propagation, such light is called Plane polarised light. Circularly polarised light: When the two plane polarised wave superpose under certain condition such that the resultant light vector rotate with a constant magnitude in a plane perpendicular to the direction of propagation and tip of light vector traces a circle around a fixed point such light is called circularly polarised light. Elliptically polarised light: When two plane polarised light are superpose in such a way that the magnitude of the resultant light vector varies periodically during its rotation then the tip of the vector traces an ellipse such light is called elliptically polarised light. Pictorial representation of polarised light: Since in unpolarised light all the direction of vibration at right angles to that of propagation of light. Hence it is represented by star symbol. In a plane polarised beam of light, the polarisation is along straight line, the vibration are parallel to the plane and can be represented by If the light particles vibrate along the straight line perpendicular to the plane of paper, then they can be represented by a dot. Plane of vibration: The plane containing the direction of vibration and direction of propagation of light is called as plane of vibration. Plane of polarisation: The plane passing through the direction of propagation and containing no vibration is called as plane of polarisation. Since a vibration has no component of right angle, to its own direction, so the plane of polarisation is always perpendicular to the plane of vibrations. Angle between plane of vibration and plane of polarisation is 90˚. Light waves are transverse in nature: If the light waves are longitudinal in nature, they will show no variation of intensity during the rotation of the crystal. Since during the rotation of the crystal, the variation in intensity takes place, this suggests that light waves are transverse in nature rather longitudinal. Production of polarised light: The polarised light can be produced in four different ways such as 1. Polarisation by Reflection 2. Polarisation by Refraction 3. Polarisation by Scattering 4. Polarisation by Double refraction 1. Polarisation by reflection: The production of the polarised light by the method of reflection from reflecting interface is called polarisation by reflection. When the unpolarised light incident on a surface, the reflected light may be completely polarised, partially polarised or unpolarised depending upon the angle of incidence. If the angle of incidence is 0° or 90° the light is not polarised. If the angle of incidence lies in between 0° and 90°, the light is completely plane polarised. The angle of incidence for which the reflected component of light is completely plane polarised, such angle of incidence is called polarising angle or angle of polarisation or Brewster‟s angle.It is denoted by ip. At ip the angle between reflected ray and refracted or transmitted ray is π/2. Explanation: To explain the polarisation by reflection, let us consider an interface XY on which a ray AB which is unpolarised is incident at an angle equal to polarising angle and get reflected along BC which is completely plane polarised and the ray BD which is refracted or transmitted is continues to be unpolarised. The incident ulpolarised light contain both perpendicular and parallel component of light. The parallel component of light is converted into perpendicular component and gets reflected from the interface. The parallel component of light is continues to vibrate and get refracted or transmitted. As a result of which the reflected component is polarised. Conclusion: Hence, the reflected ray of light contains the vibrations of electric vector perpendicular to the plane incidence. Thus the reflected light is completely plane polarised perpendicular to plane of incidence. Brewster’s Law: This law states that when an unpolarised light is incident at polarizing angle „ip‟ on an interface separating air from a medium of refractive index “µ” then the reflected light is fully polarized. i.e. tan i p To explain Brewster‟s law, let XY be a reflecting surface on which; AB = unpolarised incident light BC= completely polarized BD = partially polarized i p =angle of incidence, angle of polarization From fig. CBY+ DBY=90˚ 90 r 90 r 90 0 ' 0 0 90 i 90 r 90 0 p 0 0 i p r 90 0 r ' 90 0 r From Snell‟s law sin i p sin i p sin i p = = = tan i p sin r sin(90 0 i p ) cos i p Thus the tangent of the angle of polarization is numerically equal to the refractive index of the medium. NOTE: We can also prove in case of reflection at Brewster‟s angle reflected and refracted ray are mutually perpendicular to each other. From Brewster‟s law; sin i p We have tan i p cos i p According to Snell‟s law; sin i p sin r From above equations sin r cos i p sin r sin(90 0 i p ) r 900 i p r i p 90 0 90 0 CBY 90 0 DBY 90 0 CBY+ DBY=90˚ CB BD CBD 90 0 Thus, it is concluded that at polarizing angle or at Brewster‟s angle, the reflected light and the refracted light are mutually perpendicular to each other. 2. Polarisation by Scattering: When a beam of ordinary light is passed through a medium containing particles, whose size is of order of wavelength of the incident light, then the beam of light get scattered in which the light particles are found to vibrate in one particular direction. This phenomenon is called “Polarisation by scattering”. Explanation: To explain the phenomenon of scattering, let us consider a beam of unpolarised light along z-axis on a scatter at origin. As light waves are transverse in nature in all possible direction of vibration of unpolarised light is confined to X-Y plane. When we look along X- axis we can see the vibrations which are parallel to Y-axis. Similarly, when we look along Y-axis the vibration along X-axis can be seen. Hence, the light can be scattered perpendicular to incident light is always plane polarized. Polarisation by refraction: The phenomenon of production of polarised light by the method of refraction is known as polarisation by refraction. To explain the polarization by refraction, let us consider an ordinary light which is incident upon the upper surface of the glass slab at an polarizing angle i p or Brewster‟s angle B , so that the reflected light is completely polarized while the rest is refracted and partially polarized. The refracted light is incident at the lower face at an angle „r‟. Now, sin r sin r sin r g tan r a tan r g a cos r sin(90 r ) sin i p 0 Thus according to Brewster‟s law, „ r ‟ is the polarizing angle for the reflection at the lower surface of the plate. Hence, the light reflected at the lower surface is completely plane-polarised, while that transmitted part is partially polarised. Hence, if a beam of unpolarised light be incident at the polarizing angle on a pile of plates, then some of the vibrations are perpendicular to the plane of incidence are reflected at each surface and all those parallel to it are refracted. The net result is that the refracted beams are poorer and poorer in the perpendicular component and less partially polarised component. Malus law: It states that when a beam of completely plane polarized light incident on the plane of analyser, the intensity of the transmitted light varies directly proportional to the square of the cosine of the angle between the planes of the polariser and plane of the analyser. Mathematically, I cos 2 Proof: Let us consider a beam of plane polarised light coming from the plane of the polariser is incident at an angle „ ‟on the plane of the analyser. The amplitude of the light vector „E‟ is now resolved into two mutual perpendicular component i.e. E1 E0 cos which is parallel to the plane of transmission and E2 E0 sin which is perpendicular to the plane of transmission. As we are able to see only the parallel component so the intensity of the transmitted light coming from the plane of the analyser is proportional to the parallel component only. Thus, IE1 I kE02 cos 2 I 0 cos 2 I 0 kE02 2 , where I cos 2 Which is Mauls law Double refraction: The phenomenon of splitting of ordinary light into two refracted ray namely ordinary and extra ordinary ray on passing through a double refracting crystal is known as double refraction Explanation: To explain the double refraction, let us consider an ordinary light incident upon section of a doubly refracting crystal When the light passing through the crystal along the optic axis then at the optic axis the ray splits up into two rays called as ordinary and extraordinary ray which get emerge parallel from the opposite face of the crystal through which are relatively displaced by a distance proportional to the thickness of the crystal. This phenomenon is called as double refraction. Difference between the Ordinary (O-ray)and Extra ordinary ray(E-ray) Ordinary ray Extraordinary ray 1.These ray obeys the law of 1. These ray do not obey law of refraction 2.For ordinary ray refraction plane of vibration lies 2.For extraordinary ray the perpendicular to the plane of vibration parallel to direction of the direction of propagation propagation 3. The vibration of particle is 3.The vibration of particles are parallel to perpendicular to the the direction of ray. direction of ray. 4. Plane of polarisation is 4. Plane of polarisation lies in perpendicular to the its principal axis. principal plane. 5. Refractive index varies along 5. Refractive index is constant optics along axis. optics axis. 6.It travels with different speed 6. It travels with the constant in different direction.But it speed in all direction. travel with equal speed along optics axis Double refracting crystal: The crystal which splits a ray of light incident on it into two refracted rays such crystal are called double refracting crystal.It is of two types 1. Uniaxial 2. Biaxial. Uniaxial: The double refracting crystal which have one optic axis along which the two refracted rays travel with same velocity are known as uniaxial crystal Ex: Calcite crystal, tourmaline crystal, quartz Biaxial: The double refracting crystal which have two optic axis are called as biaxial crystal Ex: Topaz, Agromite Optic axis: It is a direction inside a double refracting crystal along which both the refracted behave like in all respect. Principal section: A plane passing through the optic axis and normal to a crystal surface is called a principal section Principal plane: The plane in the crystal drawn through the optic axis and ordinary ray or drawn through the optic axis and the extraordinary ray is called as principal plane these are two principal plane corresponding to refracted ray Polarisation by double refraction: To explain polarisation by double refraction let us consider a beam of light incident normally through a pair of calcite crystal and rotating the second crystal about the incident ray as axis we have the following situations as: Case 1 When principal sections of two crystals are parallel then two images O1 and E1 are seen. The ordinary ray from the first crystal passes undeviated through the 2nd crystal and emerges as O1 ray. The extraordinary ray (E-ray) from the 1st crystal passes through the 2nd crystal along a path parallel to that inside the 1 st and emerges as E1 -ray. Hence the image O1 and E1 are seen separately. When the 2nd crystal is rotated through an angle 45˚ with respect to 1st , then the two new images O2 and E2 appear.As the rotation is continued , O1 and O2 remained fixed while E1 and E2 rotate around O1 and O2 respectively and images are found to be equal intensities. When the 2nd crystal is rotated at an angle 90˚ w.r.t 1st the original images O1 and E 1 have to vanish and all the new images O2 and E2 have acquired the maximum intensity. When the 2nd crystal is rotated at an angle 135˚ w.r.t the 1st , four images once again appear with equally intense. When the 2nd crystal is rotated at an angle 180˚ w.r.t 1 st, the O2 and E2 vanishes and O1 and E1have come together in the centre. This is how we are able to produce the plane polarised light by the method of double refraction. Nicol Prism: It is an optical device made from a calcite crystal for producing and analysing plane polarised light. Principle: It is based on the principle that it eliminates the ordinary ray by total internal so that the extraordinary ray became plane polarised emerges out from it. It is based on double refraction. Construction: A calcite crystal about the three times as long as the wide is taken.Its end faces are ground such that the angles in the principal section become 68˚ and 1120 instead of 71˚ and 109˚.The crystal is cut apart along a plane which is perpendicular to both the principal section.The two cut surfaces are ground of polished optically flat. They are then cemented together by Canada balsam whose refractive index is 1.55 for sodium light and the crystal is then enclosed in a tube blackened inside. Action: When a ray of unpolarised light is incident on the nicol prism it splits up into two refracted ray as O & E ray. Since the refractive index of the canabalsum 1.55 is less than the refractive index of calcite for the ordinary ray (O- ray), so the O- ray on reaching the Canada balsam get totally reflected and is absorbed by the tube containing the crystal while E-ray on reaching the Canada balsam is get transmitted. Since E- ray is plane polarised then the light emerging from the nicol is plane polarised in which vibration are parallel to the principal section. Uses: The nicol prism can be used both as apolariser and also an analyser. When a ray of unpolarised light is incident on a nicol prism, then the ray emerging from the nicolprism is plane polarised with vibration in principal section. As this, ray falls on a second nicol which is parallel to that of 1st, its vibration will be in the principal section of 2nd and will be completely transmitted and the intensity of emergent light is maximum, thus the nicol prism behaves as a polariser. If the second nicol is rotated such that its principal section is perpendicular to that of 1st then the vibration in the plane polarisation may incident on 2nd will be perpendicular to the principal section of 2nd. Hence the ray will behave as a ray inside the 2nd and will lost by total reflection at the balsam surface. If the second nicol is further rotated to hold its principal section again parallel to that of 1st the intensity will be again maximum then the 1st prism acts as apolariser and the 2nd prism acts as an analyser. Limitations: 1. The nicol prism works only when the incident beam is slightly convergent or slightly divergent. 2. The angle of incidence must be confined with 14⁰. Quarter wave plate: A double refracting crystal plate having a thickness such as to produce a path difference of or a phase 4 difference of between the ordinary and extra ordinary wave is called 2 as quarter wave plate or plate. 4 Construction: It can be constructed by cutting a plane from double refracting crystal such that its face parallel to the optic axis. Working: When a beam of monochromatic light incident on the plate it will be broken up into O-ray and E-ray which will perpendicular to the direction of wave propagation and vibrating in the direction of incidence respectively. Let us consider a doubly refracting crystal Let t= thickness of crystal plate O be the refractive index of the crystal for O-ray E be the refractive index of the crystal for O-ray Ot = optical path for O ray E t = optical path for E ray then the path difference between the waves is ( O E )t if the plate acts as quarter wave plate , then ( O E )t = λ/4 t 4( O E ) This is for positive crystal. The crystal in which the O-ray travels with a less velocity than E-ray called positive crystal. For positive crystal VO VE and E O Ex: calcite, tormulaline etc. The crystal in which the O-ray travels with a greater velocity than E-ray called positive crystal. For a-ve crystal and VO VE and E O Ex: quartz, ice t 4( E O ) Uses: 1. It is used for producing circularly and elliptically polarised light. 2. In addition with nicol prism it is used for analysing all kind of polarised light. Half wave plate A double refracting crystal plate having a thickness such as its produces a path difference of λ/2 between the ordinary and extraordinary wave is called half wave plate. Construction: It can be constructed by cutting a plane from double refracting crystal such that its face parallel to the optic axis. Working: When a beam of monochromatic light incident on the plate it will be broken up into O-ray and E-ray which will perpendicular to the direction of wave propagation and vibrating in the direction of incidence respectively. Let us consider a doubly refracting crystal Let t= thickness of crystal plate O be the refractive index of the crystal for O-ray E be the refractive index of the crystal for O-ray O t = optical path for O ray E t = optical path for E ray then the path difference between the waves is ( O E ) t If the plate acts as quarter plate, then ( O E ) t = λ/2 t 2( O E ) This is for positive crystal. The crystal in which the O-ray travels with a less velocity than E-ray called positive crystal. For positive crystal VO VE and E O Ex: calcite, tormulaline etc. The crystal in which the O-ray travels with a greater velocity than E-ray called positive crystal. For a-ve crystal and VO VE and E O Ex: quartz, ice t 2( E O ) Uses: 1.It is used in polarimeter as half shade devices to divide the field of view into two halves presented side by side 2.It is used to produce the plane polarised light. λ/4 plate λ/2 plate 1.It produces a path difference 1.It produces a path difference of λ /4 between O and E wave of λ /2 between O and E ray. 2. The light emerging from a λ 2. The light emerging from a λ /4 plate maybe circularly /2 plates is plane polarised for elliptically or plane polarised. all orientation of the plate. 3. In this case nicol may give a 3. In this case nicol may give a non zero minimum. zero minimum always. 4. It is used for production of all 4. It is used in polarism for type polarised light. half shade device. Production and Analysis Polarised Light 1. Production of plane polarised light: To produce plane polarised light a beam of ordinary light is sent through a Nicol prism in a direction almost parallel to the long edge of the prism. Inside the prism the beam is broken upto two components „O‟ and „E‟ ray. The „O‟ component is totally reflected at the Canada balsam and is absorbed. The „E‟ component emerges out which is plane polarised with vibration parallel to the end faces of the Nicol. 2. Production of circularly polarised light: The circularly polarised light can be produced by allowing plane-polarised light obtained from the Nicol to fall normally on a quarter wave plate such that the direction of vibration in the incident plane polarised light makes an angle of 45⁰ with the optic axis of the crystal. Inside the plate the incident waves of amplitude A is divided into E A cos 450 O A sin 450 with a phase difference between them. 2 Let A cos 450 = A sin 450 Asin 45⁰= a of the axis of x Let x a sin( wt ) ?
