Engineering Physics Past Paper PDF

DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Hooke’s Law Stress is directly proportional to the strain produced, within the elastic limit. E = Stress/Strain Nm-2 Classification of elastic modulus...

DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Hooke’s Law Stress is directly proportional to the strain produced, within the elastic limit. E = Stress/Strain Nm-2 Classification of elastic modulus There are 3 types of elastic modulus based on the 3 types of strain i) Young‟s modulus(Y) ii)Bulk modulus(K) iii)Rigidity modulus(n) Youngs modulus P The ratio between the longitudinal stress to the longitudinal strain, within the elastic limits. AP Young‟s modulus(Y)=Longitudinal stress/Longitudinal strain Nm-2 Bulk modulus(K) The ratio between the volume stress to the volume strain within the elastic limits R Bulk modulus(K)= Bulk stress/ Bulk strain Nm-2 CO RIGIDITY MODULUS (n) Definition: It is defined the ratio between the tangential stress to the shearing strain with in the elastic limits. (i.e) Rigidity modulus (n)= Tangential stress / Shearing strain Nm-2 Φ U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Explanation: Let us consider a solid cube ABCDEFGH. Whose lower face CDHG is fixed as shown in fig β.4. A tangential force „F‟ is applied over the upper face AψEF. The result is that the cube gets deformed in to a rhombus shape A‟ψ‟ωDE‟F‟GH.(i.e) The lines joining the two face are shifted to an angle ᶲ. If „L‟ is the original length and „l‟ is the relative displacement of the upper face of the cube with respect to the lower fixed face, then We can write tangential stress = F/A The shearing strain (ᶲ) can be defined as the ratio of the relative displacement between the two layers in the direction of stress, to the distance measured perpendicular to the layers. P We know, Rigidity modulus(n)= Tangential stress/ Shearing strain AP n=F/AΦ Rigidity Modulus(n) = F/AΦ Nm-2 POISSON’S RATIO(σ) DEFINITION: It is defined as the ratio between the lateral strain per unit stress ( ) to the R longitudinal strain per unit stress (α) , within the elastic limits. (i.e) Poisson‟s ratio(σ) = lateral strain/longitudinal strain (or) σ = /α CO Explanation: Let us consider a wire, fixed at one end and is stretched along the other end as shown in fig 2.5. U ST Due to the force applied the wire becomes longer but it also becomes thinner (i.e) although there is an increase in its length, there is a decrease in its diameter as shown in fig. 2.5. therefore the wire elongates freely in the direction of tensile force and contracts laterally in the direction perpendicular to the force. Let „L‟ be the original length and „D‟ be the original diameter of the diameter decreases from D to d, then Longitudinal strain =l/L and Lateral strain = (D-d)/D σ = -(D-d)/D/l/L (-Ve sign indicates the decrease in length) DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS (or) σ = - L (D-d)/lD The negative sign indicates longitudinal strain and lateral strain are opposite to each other. RELATIONSHIP BETWEEN THREE MODULII OF ELASTICITY There are many relations connecting the lateral strain( ) , strain (α), Poisson‟s ratio(σ) and the three elastic moduli. Some of the relations are given below. i. Relation between α and young‟s modulus is α=1\Y ii. Relation between α and with the bulk modulus is (α-β )=1\3K iii. Relation between α and with the Rigidity modulus is (α+ )=1\2n iv. Relation between Y, n, and K is Y=9Kn\3K+n Relation between n, K and σ is σ =3K-2n\6K+2n P v. vi. Relation between Y, n and σ is σ =Y\2n-1 AP ELASTIC LIMIT When forces are applied to bodies, each and every body has a tendency to oppose the forces and try to regain its original position after the removal of the force. When the applied force is increased beyond the maximum value, the body does not regain its original position completely, R even after the removal of the external forces. Hence the maximum stress up to which a body can recover its original shape and size, after removing the external forces is called as elastic limit. CO STRESS AND STRAIN DIAGRAM Let us consider a body which is subjected to an uniformly increasing stress. Due to application of the stress , the change in dimension of the body takes place(i.e.) the strain is developed. If we plot a graph between stress and strain we get a curve as shown in fig.2.7. and is called as STRESS- STRAIN Diagram. U i. From the fig.β.7, it is found that the body obeys Hooke‟s law up to the as region OA called as elastic range. ST ii. As soon as the maximum elastic limit (i.e) yield point „ψ‟ is crossed, the strain increases rapidly than the stress. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS iii. Now, even f a small external force is applied , the body will take a new path CD and remains as plastic called as plastic range, where D is called as ultimate strength. iv. After this, the body will not come to its original state and the body acquires a permanent residual strain and it breaks down at a point as breaking stress, indicated by dotted line EF. FACTORS AFFECTING ELASTICTIY It is found that some bodies lose their elastic that even within the elastic limit, due to elastic fatigue. Therefore the manufacture should choose the material in such a way it should regain its elastic property even when it is subjected to large number of cycles of stresses. For example substance like Quartz, phosphor, bronze etc., may be employed in the P manufacturing of galvanometers, electrometers etc., after knowing their elastic properties. Apart from elastic fatigue some materials will have change in their elastic property because of AP the following factors. 1. Effect of stress 2. Effect of annealing 3. Change in temperature 4. Presence of impurities R 5. Due to the nature of crystals. CO Effect of stress: We know that when a material is subjected to large number of cycles of stresses. It loses its elasticity property even within the elastic limit. Effect of annealing: U Annealing is a process by which the material is heated to a very high temperature and then it is slowly cooled. ST Effect of temperature: The elastic property of the materials changes with the temperature. Normally the elasticity increases with the decrease in temperature and vice versa. Effect of impurities: The addition of impurities produces variation in the elastic property of the materials. The increase and decrease of elasticity depends on the type of impurity added to it. Effect of nature of crystals: The elasticity also depends on the type of the crystals, whether it is a single crystal or poly crystals. For a single crystal the elasticity is more and for a poly-crystal the elasticity is less. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS TWISTING COUPLE ON A WIRE Consider a cylindrical wire of length l and radius r fixed at one end. (fig. 1.11a). It is twisted through an angle θ by applying couple to its lower end. Now, the wire is said to be under torsion. P AP Due to elastic property of the wire, an internal restoring couple is set up inside the wire. It is equal and opposite to the external twisting couple (applied). The cylinder is imagined to consist R of a large number of thin hollow coaxial cylinders. Consider one such cylinder of radius x and thickness dx (fig. 1.11b) CO AB is a line parallel to PQ on the surface of this cylinder. As the cylinder is twisted, the line AB is shifted to AC through an angle BAC = Φ Shearing strain or Angle of shear = Φ Angle of twist at the free end = θ U From the figure (1.11 (b)) ST BC = xθ = l xθ = l Sℎearing stress Rigidity nodulus n = Sℎearing strain ∴ Sℎearing stress = n × Sℎearing strain = n nxθ = l Sℎearing force Sℎearing stress = Area over wℎicℎ tℎe force acts DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Area over wℎicℎ tℎe force acts = n x + dx 2 − nx2 = n(x2 + 2xdx + dx2 − nx2 = nx2 + 2nxdx + ndx 2 − nx2 (dx2 term is neglected since it is very snall) = 2nxdx nxθ Hence, Sℎeraing force F = l P 2πnθ 2 = x dx l AP ∴ Moment of tℎis force about tℎe axis PQ of tℎe cylinder. = Force × perpendicular distance 2 = x2dx × x 2 = x3dx R The moment of the force acting on the entire cylinder of radius r is obtained by integrating the CO expression (3) between the limits x =0 and x =r. Hence, twisting couple 2n 3 C r x dx l U 2n 3 2n  x4  l 0 0    r r l  4 0 x dx ST Applying the limits, we have 2n  r 4  2n r 4  l  4    0 4l n r 4 C 2l In the above eqn if θ = 1 radian, then we get. n r 4 Twisting couple per unit twist C  2l DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS This twisting couple required to produce a twist of unit radian in the cylinder is called torsional rigidity for material of the cylinder. Hollow cylinder For a hollow cylinder of the same length l and of inner radius r 1 and outer radius r2 2n 2 Twisting couple of the cylinder C   r2 x dx l r  r1 n   r1 4 4 P 2 2l AP r  Twisting couple per unit twist of the cylinder n C  r1 4 4 2 R 2l TORSIONAL STRESS AND DEFORMATIONS CO The shear stress set up in the shaft when equal and opposite torques (twisting moments) are applied to the ends of a shaft about its axis, is called torsional stress. U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Note that shear strain does not only change with the amount of twist, but also varies along the radial direction such that it is zero at the center and increases linearly towards the outer periphery. TORSION PENDULUM – THEORY AND EXPERIMENT A circular metallic disc suspended using a tin wire that executes torsional oscillation is called torsional pendulum. It executes torsional oscillation whereas a simple pendulum executes linear oscillations. P AP R Description CO A torsional pendulum consists of a metal wire suspended vertically with the upper end fixed. The lower end of the wire is connected to the centre of a heavy circular disc. When the disc is rotated by applying a twist, the wire is twisted through an angle θ. Then, the restoring couple set up in the wire U = Cθ — — — — — —(1) Where C – couple per unit twist. ST If the disc is released, it oscillates with angular velocity dθ/dt in the horizontal plane about the axis of the wire. These oscillations are known as torsional oscillations. If d2θ/dt2 is the angular acceleration produced in the disc and I its moment of inertia of the disc about the axis of the wire then, d2θ Applied couple = I — — — — — (2) dt 2 In equilibrium, applied couple = restoring couple DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS d2θ I = Cθ dt2 d2θ C = θ −−−−−− dt2 I This equation represents simple harmonic motion which shows that angular acceleration 2 (d θ) is proportional to angular displacement θ and is always directed towards the mean position. dt 2 Hence, the motion of the disc being simple harmonic motion, the time period of the oscillation is given by T  2 P Displacement Accelerati on  AP  2  C I T  2 I −−−−− C R CO Determination of Rigidity Modulus of the wire The rigidity modulus of the wire is determined by the following eqn T  2 U I −−−−− C ST Experiment A circular disc id suspended by a thin wire, whose rigidity modulus is to be deterrmined. The top end of the wire is fixed firmly in a vertical support. The disc is then rotated about its centre through a small angle and set it free. It executes torsional oscillations. The time taken for 20 complete oscillations is noted. The experiment is repeated and the mean time period (T) of oscillation is determined. The length l of the wire is measured. This length is then changed by about 10 cm and then experiment is repeated. The readings for five or six different lengths of wire are measured. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS The disc is removed and its mass and diameter are measured. The time period of oscillation is T  2 I −−−−− C Squaring on both sides, we have  I  T  2    − − − − − −(3) 2    2 2 2 C P 4 2 I T2  − − − − − −(4) AP C n r 4 Substituting couple per unit twist C  in eqn(4) 2l 4 2 I 4 2 I T2    n r 4 n r 4 2l R 2l CO Rearranging the eqn (5) The rigidity modulus of the material of the wire 4 2 I 4 2 I T   2l  n r 4 n r 4 2 U 2l 8 I  l  n   r 4 T2  ST I – moment of inertia of circular disc = MR2/2 M – Mass of the circular disc R – Radius of the disc RIGIDITY MODULUS BY TORSION PENDULUM (DYNAMIC TORSION METHOD) The torsion pendulum consists of a steel or brass wire with one end fixed in an adjustable chuck and the other end to the centre of a circular disc as shown in fig. The experiment consists of three parts. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P First the disc is set into torsional oscillations without any cylindrical masses on the disc. AP The mean period of oscillation T 0 is found out. T0  2 I0 C R Were IO – moment of inertia of the disc 4 2 I 0 T0  — — — — — (1) CO 2 C Two equal cylindrical masses are placed symmetrically along a diameter of the disc at U equal distance d1 on the two sides of the centre of the disc. Mean time period of oscillation T 1 is found. ST Then, T1  2 I1 C 4 2 I1 T1  2 — — — — — —(2) C Then, by the parallel axis theorem, the moment of inertia of the whole system is given by, I1= IO + 2i + 2m d21 — — — — — —(3)   Subs I1 ion eqn (2) 4 2 T1  — — — — — —(4) 2 2 I 0 + 2i + 2m d1 C Now, two cylindrical masses are placed symmetrically at equal distances d2 from the axis of the DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS wire P AP R CO U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS T2  2 I2 C 4 2 I 2 T2  2   C 4 2 T2  2 2 I 0 + 2i + 2m d 2 — — — — — —(5) C  I 2  I 0 + 2i + 2m d 2 2 Now, P I 2  I1  I0 + 2i + 2m d 2  I 0  2i  2m d1  2m (d 2  d1 ) 2 2 2 2 AP   Subtracting (4) from (5) 4 2 4 2 T2  T1  I 0 + 2i + 2m d 2  I 0  2i  2m d1  2m (d 2  d1 ) 2 2 2 2 2 2 C C 4 2 T2  T1  ( I 2  I1 ) — — — — — —(6) 2 2 R C CO Dividing eqn (1) by (6) 2  0 2 T0 I T2  T1 I 2  I1 2 2 — — — — — — (7) Substituting the value of ( I 2  I1 ) in this equations (7) we have  2 U T0 I0 T2  T1 2m (d 2  d1 ) 2 2 2 2 2m (d 2  d1 ) T0 ST I0  2 2 2 T2  T1 2 2 Thus moment of inertia of the disc about the axis of rotation is calculated substituting the values of TO, T1,T2, d1 and d2 in the above formula. Calculation of rigidity modulus of the wire We know that restoring couple per unit twist n r 4 C 2l Subs value of C in eqn (6) DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS 4 2 T2  T1  2m (d 2  d1 ) n r 2 2 2 2 4 2l 4 2  2l T2  T1  2m (d 2  d1 ) n r 2 2 2 2 4 16 l m (d 2  d1 ) T2  T1  2 2 2 2 nr 4 16 l m (d 2  d1 ) n Nm2 2 2 (T2  T1 )r 4 Or 2 2 P Using the above relation the rigidity modulus of wire is determined. AP STRESS DUE TO BENDING IN BEAMS Fig shows a beam under the action of a bending moment M. In a particular segment of the beam, bending occurs with center of curvature at O, radius of curvature R, included angle θ and neutral surface MN. R If the longitudinal stress at a filament Aψ at distance x from the neutral surface MN‟ is σ, CO then the strain in AB is given as change in length P1Q1  PQ Strain =  original length PQ U ( R  x)  R Strain = R ST R  x  R x   R R R x =  But =  stress Y Young ' s modulus   x Thus Y R Here x/R is constant for a particular cross section of the beam. Thus, the bending stress (σ) at a particular cross section is proportional to the distance from the neutral aaxis (x). For filaments above neutral axis (i.e). negative values of x), bending stress is compressive (i.e., σ is negative). For filaments below neutral axis (for positive values of x) bending stress is tensile (i.e., σ is DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS positive) DEPRESION OF A CANTILEVER LOADED AT ITS END Theory: let I be the length of the cantilever OA fixed at O. let w be the weight suspended at the free end of the cantilever. Due to the load applied the cantilever moves to a new position OA as shown in fig. 2.16. P AP Let us consider an element PQ of the beam of length dx, at a distance OP= x from the R fixed end. Let C be the centre of curvature of the element PQ and let R be the radius of curvature. Due to the applied at the free end of the cantilever, an external couple is created between CO the load W at A and the force of reaction at Q. here, the arm of the couple is (l-x). The external bending moment = W.(l-x)............. (1) We know the Internal bending moment = YI/R.............. (2) U We know under equilibrium condition ST External bending moment = internal bending moment Therefore we can write equ (1) = equ (2) (i.e) W(l-x) = YI/R R= YI/W (l-x)..............(3) Two tangents are drawn at points P and Q, which meet the vertical line AA‟ at T and S respectively. Let the smallest depression produce from T to S = dy And let the angle between the two tangents = dθ Then we can write The angle between ωP and ωQ is also dθ (i.e) PωQ = dθ DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Therefore we can write the arc of the length PQ = R dθ = dx Or dθ = dx/R......................(4) Sub, equ (3) in equ (4) dθ = dx/[YI/W (l-x)] or dθ = W/YI(l-x)....................................(5) From the QA‟S we can write sin dθ = dy/(l-x) If dθ is vey small then we can write, dy=(l-x)dθ................................................. (6) P sub, equ (5) in equ(6), we have AP dy =W/YI(l-x)2dθ.......................................... (7) Therefore the depression at the free end of the cantilever can be derived by integrating the equ (7) within the limits 0 to „l‟ Therefore y= W/YI (l-x)2 dx R = W/YI (l2+x2- 2lx) dx = W/YI (l2+x2- 2lx2/2 + x3/3) CO = W/YI (l3- l3 + l3/3) y= W/YI( l3/3) therefore the cantilever at free end U y= W l3/3YI......................................................... (8) ST Special cases (i) Rectangular cross section If „b‟ be the breadth and „d‟ is the thickness of the beam then we know I = bd3/12 Sub, the value of Ig in equ (8), we can write The depression produced at free end for a rectangular cross section Y = Wl3/3Y(bd3/12) Y = 4Wl3/Ybd3 DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS (ii) Circular cross section If „r‟ is the radius of the circular cross section, then We know, I = πr4/4 Sub, the value of Ig in equ (8), we can write The depression produced y = Wl3/γY (πr4/4) or y = 4Wl3/γπr4Y) EXPRIMENTAL DETERMINATION OF YOUNGS MODULUS BY CANTILEVER P DPRESSION Description: it consists a beam clamped rigidly at one end on the table by the use of a G-clamp. AP The weight hanger (H) is suspended at the other end of the beam, through a small groove on the beam as shown in fig. 2.17. A pin is fixed at the free end of the beam by means of a wax. A microscope (M) is placed in front of this arrangement for measuring the variation of height of the pin. R CO Procedure: The weight hanger is kept hanged in a dead load position (W) (i.e) without any U slotted weights. The microscope is adjusted and the tip of the pin is made to coincide with the horizontal cross wire. The reading in the vertical scale of the microscope is noted. ST s.no. Load(W) Microscope reading Depression M/y Increasing Decreasing mean (y) load load Unit Kg 10-2m 10-2m 10-2m Meter Kg m-1 1. W 2. W + 50 3. W + 100 4. W + 150 5. W + 200 6. W + 250 Then the weights m, 2m, 3m etc. are increased in steps to the weight hanger. Each time the microscope is adjusted to coincide the tip of the pin to the horizontal cross wire and the readings are noted from the vertical scale of the microscope. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS The experiment is repeated by decreasing the loads and the respective readings are noted from the vertical scale of the microscope and are tabulated in the tabular column as shown. The mean depression „y‟ for a load M kg is found. Theoretically, we know the depression produced for a load of M kg is Y= Wl3/3YI.........................................................(1) Where „l‟ is the length of the beam (i.e) the distance between the clamped end and the P loaded end. If „b‟ be the breadth of the beam and „d‟ is the thickness of the beam then AP The geometrical moment of inertia for a rectangular cross sectional bar I = bd3 /12.............. (2) Also the weight W= Mg........................................................ (3) R Sub, equ (2) and (3) in equ (1), we have Y = 4Mgl3/Ybd3........................................... (4) CO Rearranging equ(4) we get Young‟s modulus Y = 4Mgl3/bd3y U Or Y = 4gl3/bd3(M/y) Nm-2.......................... (5) ST UNIFORM BENDING – ELEVATION AT THE CENTR OF THE BEAM LOADED AT BOTH ENDS Theory: Let us consider a beam of negligible mass, supported symmetrically on the two knife edges A and ψ as shown in fig.β.19. let the length between A and ψ be „l‟. let equal weights W, be added to either end of the beam C and D. Let the distance CA = BD = a Due to the load applied the beam bends from position F to E into an arc of a circle and produces as elevation „x‟ from the position F to E. let „W‟ be the reaction produced at the points A and B which acts vertically upwards as shown in fig. 2.19. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P ωonsider a point „P‟ on the cross section of the beam. Then the forces acting on the part PC of the beam are AP (i) Forces W at „ω‟ and (ii) Reaction W at A as shown in fig. 2.20 Let the distance PC = a1 and PA = a2, then The external bending moment about „P‟ is Mp = W × a 1 - W × a 2 Here, the clockwise moment is taken as negative and anticlockwise moment is taken R as positive. External bending moment about P can be written as CO U ST Mp = W. (a1- a2) Mp = Wa............................................ (1) We know the internal bending moment = YI/ R................. (2) Under the equilibrium condition External bending moment = internal bending moment We can write equ(1) = equ (2) Wa = YI/R................................................ (3) DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Since for a given load (W) Y, I g, a and R are constant. The bending is called as uniform bending. Here it is found that the elevation „x‟ forms an arc of the circle of radius „R‟ as shown in fig. 2.21. From the ∆ AFO we can write OA2 = AF2 + FO2 Since OF = FE, P Therefore we can write AP OA2 = AF2 + FE2 or AF2 = OA2 - FE2 Rearranging we can write AF2 = FE [OA2 /FE -FE]................................... (4) R Here, AF= l/2 ; FE= x= R/2; OA = R CO Therefore equ (4) can be written as (l/2)2 = x [R2 / (R/2)-x] (l2/4) = X [2R -x] U (l2/4) = 2Rx –x2 ST If the elevation „x‟ is very small, then the term x2 can be neglected. Therefore we can write l2/4 = 2xR Or x = l2/8R Radius of curvature R = l2/8x............................... (5) Sub, the value of „R‟ value in equ (γ) we have Or W.a = YI/(l2/8x) Or W.a = 8YIx/l2 Rearranging the equation (6), The elevation of point „E‟ above „A‟ is DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS X= Wal2/8YI P AP R CO U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS EXPERIMENTAL DETRMINATION OF YOUNGS MODULUS BY UNIFORM BENDING Statistical method Description: It consists of a beam, symmetrically supported on the two knife edges A and B. Two weight hangers are suspended on either side of the beam at the position C and D. the distance between AC and BD are adjusted to be equal. A pin is fixed vertically at the centre of the beam. A travelling microscope is placed in front of the whole set up for finding the position of the pin. Procedure: Taking the weight hanger as the dead load (W), the microscope is adjusted and the tip of the pin is made to coincide with the vertical cross wire. The reading is noted from the vertical P scale of the microscope. AP Now the load on each hanger is increased in equal steps of m, 2m, 3m etc, kilogram and the corresponding readings are noted from the vertical scale of the microscope. The same procedure is repeated during unloading. The readings are noted from the vertical scale of the microscope. The readings are tabulated in the tabular column as shown. The mean elevation „x‟ of the centre for M kg is found. The distance between the two knife R edges is measured as „l‟ and the distance from the point of suspension of the load to the knife edge is measured as „a‟ CO S.No. Load(W) Microscope reading Elevation M/x Increasing Decreasing Mean (x) load load Unit Kg ×10-2m ×10-2m ×10-2m Meter Kg m-1 1. W U 2. W + 50 3. W + 100 4. W + 150 ST 5. W + 200 6. W + 250 7. W + 300 8. W + 350 NON- UNIFORM BENDING If the beam is loaded at its mid-point, the depression produced does not form an arc of a circle. This type of bending is called non-uniform bending. Consider a uniform cross sectional beam (rod or bar) AB of length l arranged horizontally on two knife edges K1 and K2 near the ends A and B. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS A weight W is applied at the midpoint „O‟ of the beam. The reaction force at each knife edge is equal to W/2 in the upward direction. y is the depression at the midpoint O. The bent beam is considered to be equivalent to two inverted cantilevers, fixed at O each length (l/2) and each loaded at K1 and K2 with a weight W/2. In the case of a cantilever of length (l/2) and load (W/2), depression = Wl3/3Iy Hence, for cantilever of length (l/2) and load (W/2), depression is  W  l  W  l      3 3 3 W l3 y         28 2 2 2 2 P 3IY 3IY 3IY y W l3 AP 48 IY Y W l3 48 Iy R If M is the mass, the corresponding weight W is W = Mg If the beam is a rectangular, I = bd3/12, where b is the breadth and d is the thickness of the beam. CO Hence, Mgl3 Y= bd3 48 y 12 U Mgl3 × 12 y= 48bd3y ST Mgl3 Y = Nm–2 4bd3y The value of Y can be determined by the above equation. Experiment The given beam AB of rectangular cross section is arranged horizontally on two knife edges K1 and K2 near the ends A and B. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS A weight hanger is suspended and a pin is fixed vertically at mid-point O. A microscope is focused on the tip of the pin. P The initial reading on the vertical scale of the microscope is taken. A suitable mass M is added to the hanger. The beam is depressed. The cross wire is adjusted to coincide with the tip of AP the pin. The reading of the microscope is noted. The depression corresponding to the mass M is found. The experiment is repeated by increasing and decreasing the mass step by step. The corresponding readings are tabulated. The average value of depression y is found from the observation. R CO S. No. Load(W) Microscope reading Mean depression Increasing Decreasing Mean (y) for a load of load load M Kg.cm Unit Kg ×10-2m ×10-2m ×10-2m Meter 1. W U 2. W + 50 3. W + 100 4. W + 150 ST 5. W + 200 6. W + 250 The breadth b, thickness d and length l of the beam are measured. The value of Young‟s modulus of the beam is found by the relation. Mgl3 Y = Nm–2 4bd3y DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS I SHAPE GIRDERS Definition: The girders with upper and lower section broadened and the middle section tapered, so that it can withstand heavy loads over its called as I shape girders. Since the girders look like letter I as shown in fig 2.24 they are named as I shape girder P AP EXPLANATION: In general any girder supported at its two ends as on the opposite walls of a room, bends under its own weight and a small depression is produced at the middle portion. This may also be caused when loads are applied to the beams. R Due to depression produced, the upper parts of the girder above the neutral axis contracts, while, the lower parts below the neutral axis expands. i.e the stresses have a maximum value at the top and bottom. The stresses progressively decreases as it approaches towards the neutral CO axis. Therefore the upper and lower surfaces of the girder must be stronger than the intervening part. Thus the girders are made of I shape girders. We know the depression produced in the case of a rectangular section. Y= 4Wl3/Ybd3 U Therefore for stability, the upper part and the lower part is made broader than the centre part and hence forming an I shape called as I shape girders. The depression can also be reduced by properly ST choosing the materials of high young modulus. APPLICATIONS  They are used in the construction of bridges over the rivers.  They are very much useful in the production of iron rails which are employed in railway tracks.  They are used as supporting beams for the ceilings in the construction of buildings.  They are used in the construction of iron beams to support the bridges for the heavy vehicles and also in the construction of dams.  More stability ADVANTAGES  More stronger  High durability DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P AP R CO U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P AP R CO U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS UNIT – 2 WAVES AND OPTICS Introduction: The oscillatory motion: P AP R The simple pendulum CO The oscillatory motion is the motion of the oscillating body around its rest point, where the motion is repeated through the equal intervals of the time. Examples of the oscillatory motion:  The clock, The tuning fork, The spring, The stretched string, The motion of the swing, U The rotary bee.  The movement of the Earth‟s crust during the earthquakes.  The movement of the atoms in the molecules. ST Some concepts related to the oscillatory motion: The amplitude: You should know that the amplitude is the maximum displacement done by the oscillating body away from its original position. The measuring unit of the amplitude is meter or centimeter. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS The periodic time is taken by an oscillating body to make one complete oscillation. The complete oscillation: It is very important to know that the complete oscillation is the motion P of an oscillating body when it passes by a fixed point on its path two successive times in the same direction. AP The periodic time : You should know that the periodic time is the time taken by an oscillating body to make one complete oscillation , The measuring unit of the periodic time is the second. The frequency: It is very important to know that the frequency is the number of the complete oscillations made by an oscillating body in one second. The relation between the periodic time and the frequency is an inverse relation. R CO Equation of Damped harmonic oscillation: Damped harmonic oscillators are vibrating systems for which the amplitude of vibration decreases over time. Since nearly all physical systems involve considerations such as air resistance, friction, and intermolecular forces where energy in the system is lost to heat or sound, accounting for damping is important in realistic oscillatory systems. Examples of damped harmonic oscillators U include any real oscillatory system like clock pendulum, or guitar string: after starting the clock, or guitar string vibrating, the vibration slows down and stops over time, corresponding to the decay of sound volume or amplitude in general. ST Mathematically, damped systems are typically modeled by simple harmonic oscillators with viscous damping forces, which are proportional to the velocity of the system and permit easy solution of Newton's second law in closed form. These are second-order ordinary differential equations which include a term proportional to the first derivative of the amplitude. As described below, the magnitude of the proportionality describes how quickly the vibrations of the damped oscillator damp down to nothing. Damping forces are often due to motion of an oscillatory system through a fluid like air or water, where interactions between the molecules of the fluid (e.g. air resistance) become important. At low velocities in non-turbulent fluid, the damping of a harmonic oscillator is well-modeled by a viscous damping force Fd = —bx˙. Adding this term to the simple harmonic oscillator equation DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS given by Hooke's law gives the equation of motion for a viscously damped simple harmonic oscillator. ax2 + bx + c = 0 (1) where is a constant sometimes called the damping constant. Solutions should be oscillations within some form of damping envelope: x(t) = Ae–gteiat = Aet(ai–g) = Aert (2) where A is some constant and r = ai-g will be found. Plugging this into the equation of motion yields: (nr 2 + br + k)Aert = 0 (3) P r2 + b r + k =0 (4) n n AP which is a quadratic equation in r with solutions: r = –b ± iJ( k ) — (b 2 /4n 2 ) (5) 2n n Note that there are two roots r to the quadratic as long as the imaginary part is nonzero, corresponding to the two general solutions: R k —b k b (– 2n)t iJ( n)–(b2 /4n2 )t –iJ( )–(b 2 /4n2 )t x(t) = Ae e + Be2n te n (6) CO These solutions in general describe oscillation at frequency m = J( k ) − (b 2 /2n 2 ) within a n b –( )t decay envelope of time-dependent amplitude e 2n. U Depending on the values of m, k, y and the solution exhibits different types of behavior (i) b2 < 4kn: under damping ST Under damped solutions oscillate rapidly with the frequency and decay envelope described above. For objects with very small damping constant (such as a well-made tuning fork), the frequency of oscillation is very close to the undamped natural frequency m 0 = J k n DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Underdamped oscillations within an exponential decay envelope P (ii)b2 = 4kn: Critical damping AP This case corresponds to the vanishing of the frequency J( k ) − (b 2 /4n 2 ) described previously. n At this level of damping, the solution x(t) most rapidly approaches the steady-state amplitude of zero. Larger amounts of damping (see overdamping) cause the solution to more slowly approach zero as it moves slowly through the damping fluid, whereas smaller amounts of damping cause the solution to oscillate more rapidly around zero. Notably, solutions at critical damping do not oscillate. R If the frequencyJ( k ) − (b 2 /4n 2 ) vanishes, the two linearly independent solutions can be CO n written: b bt x(t) = Ae (– 2n)t + Bte –( 2n) (7) U ST Time evolution of the amplitude of a critically damped harmonic oscillator (iii)b2 Σ kn: Over damping DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS In the overdamped case, the frequency J( k ) — (b 2 /4n 2 ) becomes imaginary. As a result, the n oscillatory terms eimtand e–imt become growing and decaying exponentials and e|m|tand e|–m|t. Overdamped solutions do not oscillate and instead slowly decay towards equilibrium. P AP An overdamped harmonic oscillator approaching equilibrium slowly Differential equation of Forced Harmonic Oscillation: R Consider a particle of mass m connected to a spring. The particle is driven by a periodic force. CO U ST The oscillations are started and the forces acting on the particles are, (i) a restoring force proportional to the displacement acting in the opposite direction. It is given by –ky where k is known as the restoring force constant. (ii) a frictional force proportional to velocity but acting on the opposite direction. It is given —r dr where r is the frictional force constant. dt (iii) The external periodic force, F sin Pt where F is the maximum value of the force and P is its angular frequency. This force opposes the restoring force as well as the frictional force and helps in motion. Therefore, net force F‟ acting on the particle F = —ky — r dy + FsinPt (1) dt DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS ψy Newton‟s second law of motion, the resultant force acting on the particle F = mass × accelaration = ma 2y F =md (2) dt2 From the equ. 1 & 2 we have, d2y dy m 2 = —ky — r + FsinPt dt dt 2y Or m d + r dy + ky = F sin pt (3) dt2 P dt d2y r ky F + + = sin pt AP dt2 m m m d2y + 2b dy + ω2y = f sin pt (4) dt2 dt Where r/m = βb, k/m= 2 and F/m=f R The equ (4) is differential equation of motion of the forced oscillation of the particle The solution of differential equation (4) CO y=A sin (Pt-θ) (5) Where A is the steady amplitude of vibration differentiating equ (5) U dy = A p cos(pt — θ) dt ST d2y and = —Ap2 sin(pt — θ) dt 2 dy d2y substituting y, and in equ (4), we have dt dt2 -Ap2 sin (pt-θ) +βbA p cos (pt-θ) + 2 A sin (pt-θ) = f sin pt = f sin {(pt-θ) +θ} (6) Or A( 2 -p2) sin (pt-θ) + βbA p cos (pt- θ) DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS = f sin (pt- θ) cos θ+ f cos (pt- θ)sin θ (7) This relation holds well for all values of t. the coefficients of sin (pt-θ) and cos (pt- θ) terms on both sides of this equation must be equal. A( 2 -p2) = f cos θ (8) 2 bAp = f sin θ (9) Squaring the equ. (8) and (9) and then adding, we have A2(m2 — p2)2 + 4b2A2p2 = f2cos28 + f2sin28 P A2[(m2 — p2)2 + 4b2A2p2] = f2 AP f2 A2 = [(m2 — p2)2 + 4b2A2p2] Taking square root, and on dividing equ (9) by (8), R 2bAp 2bp tanθ = = A(m2 — p2) (m2 — p2) CO 2bp θ = tan–1 (10) (m2 –p2 ) The equation (9) gives the amplitude of forced vibration and equ. (10) gives the U phase. ST Wave Equation for a Plane Progressive Wave: The simplest type of wave is the one in which the particles of the medium are set into simple harmonic vibrations as the wave passes through it. The wave is then called a simple harmonic wave. Consider a particle O in the medium. The displacement at any instant of time is given by Where A is the amplitude, w is the angular frequency of the wave. Consider a particle P at a distance x from the particle O on its right. Let the wave travel with a velocity v from left to right. Since it takes some time for the disturbance to reach P, its displacement can be written as DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Where f is the phase difference between the particles O and P. We know that a path difference of l corresponds to a phase difference of βπ radians. Hence a path difference of x corresponds to a phase difference of Substituting equation (1.5) in equation (1.4) P We get, AP R CO U ST Differential equation for wave motion: Differentiating equation (1.7) with respect to x, We get, DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS The velocity of the particle whose displacement y is represented by equation (1.7), is obtained by differentiating it with respect to t, since velocity is the rate of change of displacement with respect to time. Comparing equations (1.8) and (1.9) we get, P Particle velocity = wave length x slope of the displacement curve or strain Differentiating equation (1.8), Differentiating equation (1.9) AP R CO Comparing equation (1.11) and (1.12) we get, U This is the differential equation of wave motion. ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Laser Introduction: The light emitted from an ordinary light source is incoherent, because the radiation emitted from different atoms has no definite phase relationship with each other. For interference of light, coherent sources are essential. Two independent sources cannot act as coherent sources. For experimental purposes, from a single source, two coherent sources are obtained. In recent years, some sources have been developed, which are highly coherent known as LASER. The word 'Laser' is an acronym for Light Amplification by Stimulated Emission of Radiation. The P difference between ordinary light and laser beam is pictorially depicted in Fig. AP R CO U ST Characteristics of laser: The laser beam (i) is monochromatic. (ii) is coherent, with the waves, all exactly in phase with one another, (iii) does not diverge at all and (iv) is extremely intense Population of Energy levels: Spontaneous and stimulated emission: An atom may undergo transition between two energy states E 1 and E2, if it emits or absorbs a photon of the appropriate energy E2 - E1 = hν. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS In a system of thermal equilibrium, the number of atoms in the ground state (N 1) is greater than the number of atoms in the excited state (N2). This is called normal population (Fig). Consider a sample of free atoms, some of which are in the ground state with energy E 1 and some in the excited energy state with energy E2. If photons of energy hν = E2-E1 are incident on the sample, the photons can interact with the atoms in the ground state and are taken to excited state. This is called stimulated or induced absorption (Fig.) The process by which the atoms in the ground state is taken to the excited state is known as pumping. P AP If the atoms are taken to the higher energy levels with the help of light, it is called optical pumping. If the atoms in the ground state are pumped to the excited state by means of external agency, the number of atoms in the excited state (N 2) becomes greater than the number of atoms R in the ground state (N1). This is E2 called population inversion (Fig ). CO U ST The life time of atoms in the excited E1 state is normally 10-8 second. Some of the excited energy levels have greater life times for atoms (10-3s). Such energy levels are called as the metastable states. If the excited energy level is an ordinary level, the excited atoms return to the lower (or) ground energy state immediately without the help of any external agency. During this transition (Fig), a photon of energy E2-E1 = hν is emitted. This is called spontaneous emission. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P If the excited state is a metastable state, the atoms stay for some time in these levels. The atoms in such metastable state can be brought to the lower energy levels with the help of photons AP of energy hν = E2 - E1. During this process, a photon of energy E2 - E1 = hν is emitted. This is known as stimulated emission (or) induced emission (Fig.). A photon produced by stimulated emission is called secondary photon (or) stimulated photon. The secondary photon is always in phase with the stimulating photon. These photons in turn stimulate the emission further and the R process continues to give a chain - reaction. This is called laser action and by this action all the emitted photons having same energy and same frequency are in phase with each other. Hence, a CO highly monochromatic, perfectly coherent, intense radiation is obtained in laser. Conditions to achieve laser action: (i) There must be an inverted population i.e. more atoms in the excited state than in the ground U state. ST (ii) The excited state must be a metastable state. The emitted photons must stimulate further emission. This is achieved by the use of the reflecting mirrors at the ends of the system. Resonant cavity & Optical amplification: An optical cavity, resonating cavity or optical resonator is an arrangement of mirrors that forms a standing wave cavity resonator for light waves. Optical cavities are a major component of lasers, surrounding the gain medium and providing feedback of the laser light. They are also used in optical parametric oscillators and some interferometers. Light confined in the cavity reflects DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS multiple times producing standing waves for certain resonance frequencies. The standing wave patterns produced are called modes; longitudinal modes differ only in frequency while transverse modes differ for different frequencies and have different intensity patterns across the cross section of the beam Einstein’s A and B Coefficient derivation and discussion: Einstein showed the interaction of radiation with matter with the help of three processes called stimulated absorption, spontaneous emission and stimulated emission. He showed in 1917 that for proper description of radiation with matter,the process of stimulated emission is essential.Let us first derive the Einstein coefficient relation on the basis of above theory: P Let N1 be the number of atoms per unit volume in the ground state E 1 and these atoms exist in the radiation field of photons of energy E2-E1 =h v such that energy density of the field is E. AP Let R1 be the rate of absorption of light by E1 -> E2 transitions by the process called stimulated absorption This rate of absorption R1 is proportional to the number of atoms N1 per unit volume in the ground state and proportional to the energy density E of radiations. R That is R1∞ N1 E R1 = B12N1 E (1) CO Where B12 is known as the Einstein‟s coefficient of stimulated absorption and it represents the probability of absorption of radiation. Energy density e is defined as the incident energy on an atom as per unit volume in a state. Now atoms in the higher energy level E2 can fall to the ground state E1 automatically after 10-8 U sec by the process called spontaneous emission. The rate R2 of spontaneous emission E2 -> E1 is independent of energy density E of the radiation ST field. R2 is proportional to number of atoms N2 in the excited state E2 thus R2∞ N2 R2=A21 N2 (2) Where A21 is known as Einstein‟s coefficient for spontaneous emission and it represents the probability of spontaneous emission. Atoms can also fall back to the ground state E1 under the influence of electromagnetic field of incident photon of energy E2-E1 =hv by the process called stimulated emission DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Rate R3 for stimulated emission E2-> E1 is proportional to energy density E of the radiation field and proportional to the number of atoms N2 in the excited state, thus R3α N2 E Or R3=B21N2 E (3) Where B21 is known as the Einstein coefficient for stimulated emission and it represents the probability of stimulated emission. In steady state (at thermal equilibrium), the two emission rates (spontaneous and stimulated) must balance the rate of absorption. Thus R1=R2+R3 P Using equations (1,2, and 3) ,we get AP N1B12E=N2A21+N2B21E Or N1B12E –N2B21E=N2A21 Or (N1B12-N2B21) E =N2A21 Or E= N2A21/N1B12-N2B21 R = N2A21/N2B21[N1B12/N2B21 -1] CO [by taking out common N2B21 from the denominator] Or E=A21/B21 {1/N1/N2(B12/B21-1) (4) Einstein proved thermodynamically,that the probability of stimulated absorption is equal to the probability of stimulated emission.thus U B12=B21 ST Then equation(4) becomes E=A21/B21(1/N1/N2-1) (5) From ψoltzman‟s distribution law, the ratio of populations of two levels at temperature T is expressed as N1/N2=e(E2 –E1)/KT N1/N2=ehv/KT Where K is the ψoltzman‟s constant and h is the Planck‟s constant. Substituting value of N1/N2 in equation (5) we get DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS E= A21/B21(1/ehv/KT-1) (6) Now according to Planck‟s radiation law, the energy density of the black body radiation of frequency v at temperature T is given as E= 8πhv3/c3(1/ehv/KT) (7) By comparing equations (6 and 7), we get A21/B21=8πhv3/c3 This is the relation between Einstein‟s coefficients in laser. Significance of Einstein coefficient relation: This shows that the ratio of Einstein‟s coefficient P of spontaneous emission to the Einstein‟s coefficient of stimulated absorption is proportional to cube of frequency v. It means that at thermal equilibrium, the probability of spontaneous AP emission increases rapidly with the energy difference between two states. Semiconductor Laser: Homojunction semiconductor laser: Specifications × Type: Homojunction semiconductor R × Active medium: p-n junction diode × Active centre: recombination of electrons and holes × Optical resonator: junction of diodes polished CO × Nature of output: pulsed or continuous × Power output:1mw × Wavelength of output:8400 – 8600 Ǻ × Band gap: 1.44 ev U Principle × The electrons in the conduction band (CB) combines with a hole in the valence band (VB) ST × Recombination of hole and electron produces energy in the form of light × This photon induces another electron in CB to VB stimulating another photon Construction: × The active medium is made of pn junction diode made of GaAs × The P region is doped with Germanium × The N region is doped with tellurium × Thickness of the light is very narrow so that emitted laser has large divergence × The refractive index ( µ) of GaAs is high, so it acts as optical resonator(no need of mirrors) × The upper and lower electrodes fixed on p and n region helps for the flow of current while biasing × Experimental set up of homojunction semiconductor laser : DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS P Working: × Population inversion is achieved by heavily doping p and n materials. Fermi level lies AP within CB of N type and VB of P type × Junction is forward biased with an applied voltage nearly equal to band gap voltage direct conduction takes place × Active region is generated near the depletion layer × Radiation with frequency ν is incident on the pn junction × Emitted photons increase the rate of recombination of electrons and holes R × Frequency of the incident radiation should be in the range ( EG / h) < ν < (Efc – Efv) / h CO × Emitted photons are in the same phase as that of the incident photon × Energy level diagram of homo. Jn. Laser : U ST Calculation of wavelength × Bandgap of gaas = 1.44 eV × Eg = hc / λ = 1.44 × Sub. The values of c & h in above equation × The wavelength can be obtained × λ= 86β6Ǻ DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS × Wavelength is in IR region Heterojunction semiconductor laser Specifications: Type: Heterojunction semiconductor laser Active medium: p-n junction Active center: recombination of electrons and holes Pumping: direct pumping Optical resonator: polished junctions of diode Power output: 10mw Nature of output: continuous wave form P Wavelength of output: 8000 Ǻ Band gap: 1.55eV AP Principle: × Same as that of homojunction semiconductor laser × Electrons in the conduction band (CB) combines with a hole in the valence band(VB) × Recombination of hole and electron produces energy in the form of light × This photon induces another electron in CB to VB stimulating another photon Construction: R × There are five layers in this type. They are: 1) first layer : a contact layer of GaAs (p type) 2) Second layer : GaAlAs (p type) with wider band gap CO 3) Third layer : a layer of GaAlAs (n type) with narrow band gap 4) Fourth layer: GaAs (n type) with wider band gap 5) Fifth layer: GaAlAs (n type) 6) Sixth layer: GaAs (n type) × The first and fifth layers are polished well and they act as optical resonator U × Upper and lower electrodes are connected to top and bottom. They help in forward biasing ST Working: × Similar to homojunction semiconductor laser × Diode is forward biased with the help of upper and lower electrodes DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS × Charge carriers are produced in wider band gap layers(2 & 4) × These charge carriers are injected to active region(3) × population inversion is achieved × Some of the injected carriers recombine and produce spontaneously photons × These photons stimulate more ejections × Photons are reflected back and forth by the polished layers × Intense and coherent beam emerges from layer 3 and 4 Calculation of wavelength: × Band gap of GaAlAs = 1.55eV × Eg = hc / λ = 1.55 eV P × Sub. The values of c & h in above equation, × The wavelength can be obtained AP × λ= 8014 Ǻ × Wavelength is in IR region Advantages: × Power output is high × Produces continuous wave R × High directionality and coherence × Diodes are highly stable and has longer life time CO Disadvantages: × Cost is higher than the other type × Practical difficulty in growing different layers of p – n junction U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS FIBRE OPTICS Introduction: The development of lasers and optical fiber has brought about a revolution in the field of communication systems. Experiments on the propagation of information – carrying light waves through an open atmosphere were conducted. The atmospheric conditions like rain, fog etc affected the efficiency of communication through light waves. To have efficient communication systems, the information carried by light waves should need a guiding medium through which it can be transmitted safely. P Optical fiber: The optical fiber is a wave guide. It is made up of transparent dielectrics (SiO), (glass or plastics). AP Fiber Construction: It consists of an inner cylinder made of glass or plastic called core. The core has high refractive index n1. This core is surrounded by cylindrical shell of glass or plastic called cladding. The cladding has low refractive index n. This cladding is covered by a jacket which is R made of polyurethane. It protects the layer from moisture and abrasion. The light is transmitted through this fiber by total internal reflection. The fiber guides light waves to travel over longer distance without much loss of energy. Core diameters range from 5 to 600µ m while cladding CO diameters vary from 125 to 750µ m. Core transmits the light waves. The cladding keeps the light waves within the core by total internal reflection. U ST DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Numerical aperture Acceptance angle and fractional index change Acceptance angle: P Consider a cylindrical fiber as shown AP It consists of core and cladding Let n0 be the refractive index of the medium It can be noted from the figure below Incident ray travels through AO Enters at an angle “i” to fiber axis Ray is refracted along Oψ at an angle θ in the core R Further it also falls at a critical angle φc ,the reflected ray graces the surface For any angle of incidence more than critical angle, ray is total internally reflected (φ> φc) CO It is the acceptance angle Those rays which passes within the acceptance angle will be total internally reflected Light incident on core within this maximum external incident angle i m can be propogated through the fiber This angle im is called waveguide acceptance angle Mathematical relation U ψy applying snell‟s law to AO; n0 sin i = n1 sin θ ST Sin i = (n1/n0) sin θ = (n1 / n0) 1 – cos2 θ (since sin2 θ +cos2 θ=1) (1) Mathematical relation Applying the same to „Oψ‟ interface n1 sin φ = n2 sin 90 0 Sin φ = (n2/n1) {since φ = 90 – θ } Sin(90 – θ ) = (n2/n1) ωos θ = (n2/n1) (2) DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Sub the value of cos θ from (β) in (1) the value of angle of incidence is given by, 2 2 Im = sin -1 ( n1 - n2 ) Definition: The maximum angle at which or below which the light suffers total internal reflection is called acceptance angle The cone which contains this angle is called acceptance cone P Numerical aperture: AP Sine of acceptance angle is called numerical aperture Na = sinim 2 2 = n1 - n2 R CO U Fractional index change: It is the ratio of refractive index difference in core and cladding to the refractive index of ST the core ∆ = (n1- n2) / n1 Na= n12 - n22 = (n1- n2) (n1+ n2) = (n1+ n2) (n1 δ) = 2n1 2δ DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Na= n1 2δ Types of Optical fibers: MODES REFRACTIVE TYPES INDEX P GLASS PLASTIC SINGLE MULTI STEP GRADED AP INDEX MODE MODE INDEX SINGLE MULTI MODE MODE R Glass : made of mixtures of metal oxides and silica. eg: core : SiO2, CO Cladding : P2O3 – SiO2 Plastics: made of plastics, can be handled tough due to its durability. Core: polystyrene ,cladding: methyl methacrylate Classification based on modes: U Mode describes the propagation of electromagnetic waves in the ave guides. Single mode : ST 1. Core diameter is very small so that it allows only one mode of propagation 2. Cladding diameter is very large 3. Optical loss is reduced. DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Multimode fiber: P 1. Core diameter is very large compared to single mode 2. Allows many modes to propagate. AP 3. It is made up of glass-clad glass, silica –clad silica R CO U Classification based on refractive index: Further classified into step index and graded index fiber ST Step index fiber: refractive index of air cladding and core varies step by step. This fiber is sub classified into single mode and multimode. 2. graded index fiber:  Refractive index of the core varies radially from the axis of the fiber  Refractive index of the core is maximum along fiber axis and gradually decreases DOWNLOADED FROM STUCOR APP DOWNLOADED FROM STUCOR APP PH8151/ENGINEERING PHYSICS Refractive index is minimum at interface. Losses in optical fiber: P Attenuation is defined as the ratio of optical output power from fiber of length „l‟ to input power AP Attenuation α = -(10 / l)log(pin / p out ) Three types of losses. They are a) absorption b)scattering and c)radiative losses Absorption: R Occurs due to imperfections of atomic structures(eg : missing molecules) There are three bands of wavelength where absorption increases drastically is 950 nm , CO 1250 nm , 1380 nm. Scattering: Loss is dependant on wavelength Glass is used in manufacture of optical fiber Due to disordered structure of glass, there is variation in refractive index U Some portion of light is scattered where as some travels through the fiber This is called Rayleigh scattering ST Rayleigh scattering α (1 / λ 4) Radiative loss: Due to bending of finite radius of curvature in optical fiber There are two types of radiative losses They are due to macroscopic bends and microscopic bends in the fiber Macroscopic Bends: R core > R cladding This may cause large cur