2 Thus, the total energy decreases more rapidly than the amplitude with time. 9.2. Thus, all the points on this wavefront become the origin of ether disturbance which is exactly the same to the ether disturbance originated at O at t time earlier. Putting this value of D into Eq. We will discuss a few of them like (i) determination of wavelength of monochromatic light, (ii) resolution of spectral lines, (iii) determination of refractive index of materials, and (iv) determination of thickness of a thin sheet. 3.8.1 Theory The directions or positions of the principal maxima of the nth order when a parallel beam of light is incident normally on a diffraction grating is obtained from Eq. Solution The data given are n = 2, l1 = 5048 Å = 5048 × 10–8 cm, l2 = 5016 Å = 5016 × 10–8 cm and on the grating there are 6000 lines/cm. In the propagation of sound waves, air particles oscillate (but do not move) about their mean position in the direction in which sound travels. Calculate the wavelength of the light. For example, a system of particles held together by springs turns out to be a useful model of the behavior of atoms mutually bound in a crystalline solid. 4.11 What is the difference between linearly polarized light and plane polarized light? (6.55), we have H OI = EOI (C ) η1 Putting the values of h1 and EOI into Eq. Calculate the change in diameter of the 5th bright ring when the air medium is replaced by water of refractive index 1.33 and is observed by the reflected light. No remarkable changes in the transmitted light are observed when tourmaline plate N1 is rotated. This expression gives the amount of radiant energy per unit volume having a frequency interval of dn. 9.2.2 Non-inertial frame of reference Opposite of the inertial frame of reference is the non-inertial frame of reference or accelerated frame of reference. The limits on the measurement imposed by Eq. Also calculate the propagation constant and wave speed in the given aluminum. As we know, the refractive index m of the prism material in terms of the minimum angle of deviation d is given by µ= sin[(α + δ ) / 2] sin(α / 2) For a small refracting angle a, this equation gives µ ≈1+ δ /α or δ ≈ (µ − 1)α 162 Principles of Engineering Physics 1 Here d and a have been depicted in Fig. Thus, no light emerges from the analyzer N2 in case of crossed Nicols. (iii) 6. Thus, we conclude that energy flows in the electromagnetic field with the same speed as the wave itself. (1.17) is found out to be 1 EK max = mω02 r 2 2 (1.18) The total energy of a simple harmonic oscillator at any instant or at any position is given by E = EP + EK 1 1 = mω02 r 2 cos2 (ω0t + θ ) + mr 2ω02 sin2 (ω0t + θ ) 2 2 = 1 2 2 mω = E= EP max K max 0r 2 (1.19) 6 Principles of Engineering Physics 1 Equation (1.19) shows that the total energy of a simple harmonic oscillator at any instant or at any position is constant. Thus, we have cos θT EOI cos θ I − EOR cos θ= I n1 (EOI + EOR ) n2 n1 n = − EOI cos θT 1 + EI cos θOI n2 n2 or EOR cos θ I + EOR cos θT or n cos θ I − 1 cos θT  EOR  n2   =  EOI P cos θ + n1 cos θ I T n2 (6.127) The subscript P in Eq. The position of secondary spherical wavelets and secondary ellipsoidal wavelets originating from all the points between B and F are obtained in the same manner as described earlier. (6.125) is always positive thereby implying that  EOR     EOI P is to be positive. The diameter of the 8th bright ring as seen by the reflected light is 0.72 cm. The resultants of secondary wavelets originating from these Fresnel’s half period zones will have equal amplitudes and alternately, opposite phases. The points at which particles are oscillating with maximum displacements are called anti nodes iv. Calculate the angular separation between two wavelengths 5890 Å and 5896 Å of a sodium vapour light for a second order spectrum. 2 (1.108) (Heisenberg's uncertainty principle for energy and time) iii. The Electromagnetism 355 corresponding function is called a point function. Half angular width of the principal maximum The half angular width of the central maximum in a single slit Fraunhofer diffraction pattern is defined as the angle between two points on the central maximum where intensity is one half of the intensity at the centre of the central maximum, i.e., mathematically I q = half width of the central maximum in a single slit diffraction pattern when I = 0 . Polarization 2 Intensity of the ordinary image = O0 kr= cos2 θ I 0 cos2 θ 293 (A) PE0 = r sin θ = Amplitude of O0 along the principal section of C2. Polarization 343 4.61 How many times is the length of a half-wave plate more than that of a quarter-wave plate for a particular monochromatic light. (i) spherical (ii) cylindrical (iii) plane 21. S. Panigrahi is Senior Professor at the Department of Physics and Astronomy, National Institute of Technology (NIT), Rourkela. (2.28) as is defined by e = 2dD e= 4d 2 − ∆ 2 ∆D 4d 2 − ∆ 2 or e= 2d ∆ (2.32) Thus, eccentricity of the hyperbola described by Eqs (2.27) or (2.28) is obtained 2d approximately as e = . Solution Data given are Dt0 = 30 yrs – 20 yrs = 10 yrs and v = 0.985 c 714 Principles of Engineering Physics 1 ∆t = γ∆t 0 = ∆t 0 2 1− v c 2 = 10yrs 1 − (0.985c)2 c 2 = 57.95yrs Example 9.4 The period of a seconds’ pendulum is 2.0 s in the reference frame of the pendulum. The superposition principle holds good for matter waves. That is, as ionization density N increases continuously, the medium becomes more and more rarer for the electromagnetic wave. There are certain points in the medium at which particles of the medium are oscillating with maximum amplitude. Example 5.4 Find the gradient of the scalar function ϕ (x , y , z ) = xy + yz + zx at the point (1, 2, 3). 4.16. Also calculate the wavelength. [Ans 2.7 × 108 Hz] 6.28 In a plasma medium, there are 1012 electrons per unit volume. diffracting through a zone plate travel different optical paths. The energy carried by one quantum or photon hn is equal to the energy difference of the two stationary orbits, i.e., hn = E2 – E1. The fundamental concepts of physics have paved the way for the development of technologies. 2.94 Explain the formation of Newton’s rings. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. iii. (A), we get sin2 θT = 1 2n2 or cos2 θT = 1 − 1 2n2 Electromagnetic Waves 537 Putting this equation into Eq. The resultant amplitudes in Eq. A progressive wave is the disturbance produced in the medium by the repeated periodic motion of the particles of the medium, motion being handed over from particle to particle. [Ans 0 872 c] 9.14 An electron in a typical television picture tube moves with a kinetic energy 0.017 MeV Find its speed. For different wavelengths, the diffraction angle q will be different and the position of diffraction maxima of each wavelength will be different. In dextro-rotary substances, the velocity of the right-handed circularly polarized vibration is greater than the velocity of the left-handed circularly polarized vibration while in levo-rotary substances, the velocity of the left-handed circularly polarized vibration is greater than the velocity of the right-handed circularly polarized vibration. π d sin θ into Eq. (3.104) as Hence, the grating element = e + b =  nλ    e +b  θ = sin −1  Diffraction 249 The positions of the spectral lines of the nth order spectrum from the central line when a parallel beam of light is incident normally on a diffraction grating is obtained from the equation tan θ = x f or x = f tan θ  nλ  or x = f tan sin −1    e +b  The position of the spectral line from the central line (n = 0) of the sodium light 5890 Å in the first order spectrum is  1 × 5890 × 10−8  x1 = 100 × tan sin −1  72.88 cm = −4  1.00 × 10  The position of the spectral line from the central line (n = 0) of the sodium light 5896 Å in the first order spectrum is  1 × 5896 × 10−8  x2 = 100 × tan sin −1  73.00 cm = −4  1.00 × 10  The separation in cm of the two lines in the first order spectrum of the sodium light is. 6.10 Speed of Electromagnetic Waves For charge-free non-conducting medium = ( ρ 0= and σ 0), Eqs (6.28) and (6.31) become, respectively ∇2 E − µε ∂2 E 0 = ∂t 2 (6.56) ∇2 H − µε ∂2 H 0 = ∂t 2 (6.57) These equations are of the same form as that of the wave equations we are familiar with. Calculate the angular frequency of oscillation. Hence, we can express the propagation vector k in the form 514 Principles of Engineering Physics 1  kT= α + iβ (6.140) Squaring both sides of this equation, we have kT2 = α 2 + i 2 β 2 + i2αβ = α 2 − β 2 + i2αβ Comparing this equation with Eq. 2.51 Explain how the interference fringes obtained in Young’s double slit experiment appear straight although they are actually hyperbolae. 4.34). 5.69 Would Gauss’s law hold true if the surface area of a sphere is not 4π r 2 ? (6.11), we get       F M ⋅ v =− E ⋅ J =− J ⋅ E (6.13) This is the rate at which mechanical force does work on unit volume and is called power density. 5.59 Using Gauss’s law, calculate the field intensity due a cylindrically symmetric charge distribution of uniform density of infinite length when the point lies outside the charge distribution. (8.36), we have 2 2 k − k0 2k0 2 ϕR + ϕT A v0 + A v1 = k + k0 k + k0 2 A2 v 0 or = ( k + k0 ) A2 v 0 = ( k + k0 ) = A2 v 0 ( k + k0 ) 2 2 2  v  2 2  (k − k0 ) + 4k0 ×  v0    k 2 2  ( k − k0 ) + 4k0 ×  k0   ( k + k0 ) Thus, jR + jT = jI 2 = A2 v 0 = ϕ I 658 Principles of Engineering Physics 1 Refractive index m In the usual manner, the refractive index of the second medium (x > 0) with respect to the first medium (x < 0) is defined as µ= speed in second medium v1 = speed in first medium v0 Putting the value of E − V0 = E µ = v1 from Eq. (i) 1 ∂ 2ψ ∇ψ = v 2 ∂x 2 2 ∂ 2ψ ∂x 2 (iii) ∇2ψ = v2 13. The two waves are represented by Ψ1 =r1 sin ω1t and Ψ 2 =r2 sin ω2t (1.120) The resultant wave according to the superposition principle will be given by Ψ = Ψ1 + Ψ 2 = r1 sin ω1t + r2 sin ω2t Ψ r1 sin ω1t + r2 sin {ω1t − (ω1t − ω2t )} or= Ψ r1 sin ω1t + r2 sin ω1t cos (ω1t − ω2t ) − cos ω1t sin (ω1t − ω2t )  or= = Ψ sin ω1t r1 + r2 cos (ω1t − ω2t )  − r2 cos ω1t sin (ω1t − ω2t )  or (1.121) Now let R cos θ = r1 + r2 cos (ω1t − ω2t ) (1.122) Oscillations and Waves R sin θ r2 sin (ω1t − ω2t ) and = 81 (1.123) Substituting Eqs (1.122) and (1.123) into Eq. In case of a plane wave, the wavefront is a plane perpendicular to the direction of propagation. How old will she find her daughter? The principal focal length of a zone plate is given by f1 = or f1= r12 λ λR = R= 1m λ Example 3.5 The radius of the first half period zone of a zone plate is 0.05 cm. Figure 8.4 Ideal potential energy barrier of height V0. Thus, to avoid the total reflection of E-ray and transmission of O-ray by the Canada balsam, the angle between the extreme rays of light incident on the Nicol is limited to about 28°. 6.96 Show that the refracted electromagnetic wave is in phase with the incident wave when refraction occurs from a dielectric rarer medium to a denser medium. 3.31 Light, which is a mixture of two wavelengths 5500 Å and 5555 Å, is incident normally on a plane transmission grating having 10000 lines/cm. As we know that the intensity distribution pattern follows from the equation I = I1 + I 2 + 2 I1 I 2 cos δ , it will vary with the variation of phase difference d. If the phase difference d varies with time, the interference pattern will change with time. Mathematically, tan ϕ p = µ where jp = Polarizing angle. Figure 3.4 (a) Positive zone plate where the central circular zone is transparent. 1.9.4 Characteristics of progressive waves The following are the characteristics of progressive waves. b. Polarization 325 Figure 4.31 OL is the circularly polarized vibration rotating in the anti-clockwise direction and OR is the circularly polarized vibration rotating in the clockwise direction. The strongest and next strongest images of the source are formed on the other side of the zone plate at distances of 30 cm and 6 cm respectively. There are also arrangements in the uprights so that the slit and the biprism can be rotated in their own planes. When we say that the wave packet is spread out in space, it does not mean that the particle itself is spread out in space. Moreover, interference fringes can be seen through a powerful eye-piece it its focal plane. (4.1) into the above equation we have ( = sin ϕ p µ sin 90° − ϕ p or ) tan ϕ p = µ (4.2) Example 4.1 A glass plate of refractive index 1.7 is to be used as a polarizer. A lens of focal length 100 cm is used to observe the spectrum on a screen. 4.8.1 Principle When ordinary light is transmitted through a calcite crystal it splits into an O-ray and an E-ray plane polarized with the planes of polarization perpendicular to each other. 4 As earlier, let us consider the case of a positive zone plate. 3 The magnitude of conduction current depends The magnitude of displacement current depends upon the applied potential difference and upon the rate of change of electric flux with respect resistivity of the conductors. Hence, it is again the case that increased precision in the knowledge of the position of the electron is gained only at the expense of decreased precision in knowledge of its momentum. (2.24) is β= Dλ . Thus, we have λ (µ AD + DE ) − (µt + BC ) = n 2 From Fig. (1.111) = r1 sin (ωt − kx ) + r2 sin (ωt − kx ) cos δ + cos (ωt − kx ) sin δ  = r1 sin (ωt − kx ) + r2 sin (ωt − kx ) cos δ + r2 cos (ωt − kx ) sin δ or = Ψ sin (ωt − kx )( r1 + r2 cos δ ) + r2 sin δ cos (ωt − kx ) Let r1 + r2 cos δ = R cos θ (1.112) (1.113) 78 Principles of Engineering Physics 1 and r2 sin δ = R sin θ (1.114) Putting these substitutions into Eq. Let a beam of parallel light wave be incident obliquely on the plane surface of the uniaxial crystal slab as shown in the figure. The tangential components of the electric vector are continuous along the interface. The mathematics of the problem is left as an exercise to the students. 2.17.1 Michelson interferometer In the Michelson interferometer, different types of interference patterns are produced due to the splitting and reunion of two coherent light beams by partial reflection and refraction. In other words, the Brackett series is composed of the lines which are emitted when the electron jumps from outer energy levels to the fifth energy level with principal quantum number n1 = 5. By symmetry, E has the same magnitude at any point on this spherical Gaussian surface ofradius r and its direction is always perpendicular to the surface S, i.e., the angle between E and nˆ is 0° at any point on this Gaussian surface. This was what happened to Tacoma Narrows Bridge (Washington, U.S.A.). (6.137) is given by     ET (r , t ) = E OT e i (ωT t − kT ⋅r ) . A positive crystal is defined as a doubly refracting crystal in which the speed of the ordinary ray along the optic axis is greater than or equal to the speed of the extraordinary ray. 9.2) is t1 = 1  2  1  + 1 = 1 2 2  c −v c +v c 1− v / c  (9.2) Special Theory of Relativity 705 The path of beam 2, travelling from M to M2 and back through the ether, is shown in Fig. Then according to Ampere’s circuital law, we have  ˆ = µ0 I 0 ∫ B ⋅ rd (A) C Here I0 is the current flowing through the cross-sectional area of the cylindrical conductor enclosed by the circular path C. However, in our present case, the amount of current flowing through the cross-sectional area of the cylindrical conductor is not the same as the amount of current flowing through the cross-sectional area enclosed by the circular path C. Therefore, we shall first calculate the amount of current flowing through the cross-sectional area enclosed by the circular path C. 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