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Tutorial 5: Solutions 1. The electric field of an electromagnetic wave is given as: ~ = E0 ĵ sin πz cos(kx − ωt) E z0 (1) (a) Describe the field given above. (b) Determine an expression for k in terms of other quantities. (c) Find the phase velocity of the wave. (a) The given electric field of an electromagnetic wave is ~ = E0 ĵ sin πz cos(kx − ωt) E z0 ~ = 1 E0 ĵ[sin( πz + kx − ωt) + sin( πz − kx + ωt)] E 2 z0 z0 So, the given electric field is the superposition of two fields. ~ 1 = 1 E0 ĵ sin(kx + πz − ωt) E 2 z0 (2) (3) (4) ~ 2 = − 1 E0 ĵ sin(kx − πz − ωt) (5) E 2 z0 ~ 1 is vibrating along positive y direction and E ~ 2 is vibrating along negative y direction E with amplitude 12 E0 and the propagation directions ~k1 = k î + π1 k̂ and ~k2 = k î − π1 k̂. The z0 z0 resultant field is linearly polarised along y and varies sinusoidally along z. (b) We know that electric field of the electromagnetic wave satisfies the following wave equation ~ 1 ∂2E (6) c2 ∂t2 ~ ~ ~ ∂2E 1 ∂ 2E ∂2E + = (7) =⇒ ∂x2 ∂z 2 c2 ∂t2 ~ = E0 ĵ sin πz cos(kx − ωt) twice with respect to x, z, and t Now, after differentiating E z0 ~ = ∇2 E respectively, we get from wave equation k= s ω2 π2 − 2 c2 z0 (8) (c) The phase velocity (vp ) is vp = ω ω =q 2 ω k − c2 1 π2 z02 (9) 2.(a) The electric field of a standing electromagnetic wave is given by E = 2E0 sin kx cos ωt. cos 5πt. Derive an expression for B(x, t). (b) A standing wave is given by E = 100 sin 2πx 3 Determine the two waves that can be superimposed to generate it. 2.(a) From Faraday’s law we have ∂B ∂E =− ∂x ∂t (10) Integrate to get B(x, t) = − But E0 k ω = E0 c Z Z ∂E dt = −2E0 k cos kx cos ωtdt ∂x 2E0 k cos kx sin ωt =− ω (11) = B0 ; thus B(x, t) = −2B0 cos kx sin ωt (12) (b) The given standing wave is E = 100 sin 2πx cos 5πt 3 (13) It can be rewritten as E = 100 sin 2πx 2πx + 5πt + sin − 5πt 3 3 (14) 3.(a) Describe completely the state of polarization of each of the following waves: ~ = îE0 cos(kz − ωt) − ĵE0 cos(kz − ωt) (i)E ~ = îE0 sin(ωt − kz) − ĵE0 cos(ωt − kz − π ) (ii)E 4 ~ (iii)E = îE0 cos(ωt − kz) − ĵE0 cos(ωt − kz − π ) 2 ~ = E0 [î cos ωt + ĵ cos(ωt − π )] sin kz. What kind of (b) Consider the disturbance given by E 2 wave is it? 3.(a)(i) The given wave is ~ = îE0 cos(kz − ωt) − ĵE0 cos(kz − ωt) = îE0 cos(kz − ωt) + ĵE0 cos(kz − ωt + π) (15) E 2 Thus, Ey lags ex by π. We can also write it as, ~ = (î − ĵ)E0 cos(kz − ωt) E Now, tan θ = E0 E0 (16) = 1; So, θ = 45o . Now, the direction of vibration 1 n̂ = î cos 45o − ĵ sin 45o = √ (î − ĵ) 2 (17) √ ~ = n̂ 2E0 cos(kz − ωt) E (18) So, Since, the given wave has a fixed amplitude, it represents a linearly polarised wave with its electric vector making an angle 45o with both x and y axis. (ii) The given electric field vector is ~ = îE0 sin(ωt − kz) − ĵE0 cos(ωt − kz − π ) E 4 ~ = îE0 cos(ωt − kz − π ) − ĵE0 cos(ωt − kz − π ) =⇒ E 2 4 π π ~ = îE0 cos(ωt − kz − − ) − ĵE0 cos(ωt − kz − π ) =⇒ E 4 4 4 ~ = îE0 [cos(ωt − kz − π ) √1 + sin(ωt − kz − π ) √1 ] + ĵE0 cos(ωt − kz − π ) =⇒ E 4 2 4 2 4 So, from the above equation, π π E0 Ex = √ [cos(ωt − kz − ) + sin(ωt − kz − )] 4 4 2 (19) (20) (21) (22) (23) And, Ey = E0 cos(ωt − kz − Now, √ π ) 4 2Ex − Ey = E0 sin(ωt − kz − (24) π ) 4 (25) And, Ey = E0 cos(ωt − kz − π ) 4 (26) Now squaring and adding these two equations we obtain √ ( 2Ex − Ey )2 + Ey2 = E02 3 (27) =⇒ Ex2 − √ 2Ex Ey + Ey2 = E02 2 (28) This describe an ellipse. As there is a cross term in the above equation, there is a shift of major axis by an angle θ with respect to x-axis (or, Ex axis). Let us give a rotation of coordinate system by an angle θ. The components of the electric field in the new coordinate system is, Ex = Ex′ cos θ − Ey′ sin θ (29) Ey = Ex′ sin θ + Ey′ cos θ (30) And So, in the new co-ordinate system, the equation is Ex′2 + Ey′2 − √ 2(Ex′2 cos θ sin θ − Ey′2 sin θ cos θ) − √ 2Ex′ Ey′ (cos2 θ − sin2 θ) = E02 2 (31) So, in the new co-ordinate system the cross term should be zero. Hence, √ 2Ex′ Ey′ (cos2 θ − sin2 θ) = 0 (32) So, cos 2θ = 0. =⇒ θ = π4 . Now putting θ = π 4 in the equation (38) we obtain, Ex′2 + Ey′2 − √ Ex′2 Ey′2 E2 2( − )= 0 2 2 2 (33) Ey′2 E2 E ′2 = √0 (34) =⇒ √ x + √ 2+1 2−1 2 This is an equation of ellipse whose major and minor axes are shifted by an angle θ = π4 Further, at z = 0 plane, At t = 0; Ex = 0; Ey = At some later time t = E0 √ . 2 π ; Ex 4ω = E0 √ ; 2 Ey = 0. So, the given wave is a right handed elliptical polarised wave. (iii) The given electric field is ~ = îE0 cos(ωt − kz) − ĵE0 cos(ωt − kz − π ) E 2 (35) ~ = îE0 cos(kz − ωt) + ĵE0 sin(kz − ωt) =⇒ E (36) 4 Now, from the above equation Ex = E0 cos(kz − ωt). Ey = E0 sin(kz − ωt). Now, squaring and adding the above two equation, we get Ex2 + Ey2 = E02 (37) So, the above equation is an equation of a circle. Further, at z = 0 plane Ex = E0 cos ωt, and Ey = −E0 sin ωt. Now, at t = 0; Ex = E0 and Ey = 0. At a later time t = π ; 2ω Ex = 0 and Ey = −E0 . So, the given wave equation is a right circularly polarised wave. ~ = E0 [î cos ωt + ĵ cos(ωt − π )] sin kz represents a left circularly polarised (b) The wave, E 2 standing wave. 4.(a) Suppose that the intensity of the sunlight falling on the ground on a particular day is 140W/m2 . What are the peak values of electric and magnetic fields associated with the incident radiation? (b)A linearly polarised harmonic plane wave with a scalar amplitude of 10V /m is propagating along a line in the xy plane at 45 degrees to the x axis, with the xy plane as the plane of vibration. Write down a vector expression for the wave, assuming kx and ky are both positive. Calculate the flux density taking the wave to be in vacuum. 4.(a) The intensity of radiation (I) is defined as the magnitude of the time average of the Poynting vector. I =|< ~S >|= E0 B0 |< cos2 (kx − ωt) >| µ0 Since, the average value of < cos2 (kx − ωt) > is 1 2 So, E0 B0 E2 cB02 = 0 = 2µ0 2cµ0 2µ0 p E0 = 2Icµ0 ≈ 324.9V /m I= 5 (38) (39) (40) And, B0 = r 2Iµ0 ≈ 1 × 10−6 T esla. c (41) (b) Since, the propagation is along a line in the xy plane at 45 degrees to the x-axis ~ = kx î + ky ĵ K (42) k k kx = k cos 45 = √ ; ky = k sin 45 = √ 2 2 (43) k | kx |=| ky |= √ 2 (44) So, Now the radiation is also in the xy plane.So, the direction of vibration of the harmonic wave aî + bĵ n̂ = √ a2 + b 2 (45) Where a and b are any real quantity. Now the propagation direction and the direction of vibration should be perpendicular. So, ~ =0 n̂.K akx + bky =0 =⇒ √ a2 + b 2 =⇒ a = −b Now the vibration is along n̂ = ˆ √1 (i 2 (46) (47) (48) − ĵ) So the vector expression of the wave is kx ky 10 ˆ ~ =√ (i − ĵ) cos( √ + √ − ωt) E 2 2 2 (49) Now the energy flux density is given as the magnitude of the time average of the Poynting’s vector. Energy flux density = |< ~S >| So, |< ~S >|= E0 B0 E2 E0 B0 |< cos2 (kx − ωt) >|= = 0 ≈ 63.3 mJ m−2 s−1 µ0 2µ0 2cµ0 6 (50) 5.(a) A plane, harmonic linearly polarised light light wave has an electric field Ez = E0 cos[π1015 (t − x )] 0.65c (51) Write down the frequency and wavelength of the light wave and obtain the index of refraction of the glass. (b) Consider electromagnetic wave of wavelength λ = 30cm in air. What is the frequency of such waves? If such waves waves pass from air into the block of quartz, for which the dielectric constant K = 4.3, what is their new speed, frequency, and wavelength? 5.(a)The given equation of the electric field is Ez = E0 cos[π × 1015 (t − x )] 0.65c (52) From the above equation angular frequency (ω) = π × 1015 rad/s. So, the frequency ν = ω 2π = 5 × 1014 Hz. Again from the given equation k = So, the wavelength (λ) = 2π k π×1015 0.65c m−1 . ≈ 390 nm. Hence, the refractive index of glass is (ng ) = c νλf ≈ 1.538. ; Where c is the speed of light in vacuum. (b) Given that the wavelength (λ) = 30cm = 0.3m.; and the refractive index of the quartz √ √ block nq = K = 4.3. Now the frequency of the wave in air is f = c na λ ≈ 109 Hz. ; Where c is the speed of light in vacuum. Now, the velocity of such waves in the quartz medium v = c nq ≈ 1.45 × 108 m/s. Since, the frequency of the source is constant, the wavelength will change. So, the new wavelength (λ′ ) = v nq f ≈ 7 × 10−4 m. 6. Imagine that we have a non-absorbing glass plate of index n and thickness ∆y, which stands between a source S and an observer P . 7 y (i) If the unobstructed wave (without the plate present) is Eu = E0 eiω(t− c ) show that with the plate in place the observer sees the wave given by, Ep = E0 eiω[t−(n−1) ∆y − yc ] c (53) (ii) Show that if either n ≈ 1 or ∆y is small, then, Ep = Eu + iπ ω(n − 1)∆y Eu e− 2 c (54) 6.(i) The phase difference (∆φ) due to the presence of the non-absorbing glass plate of thickness ∆y is ∆φ = 2π 2πc 2πf ∆y (n∆y − ∆y) = (n − 1)∆y = (n − 1)∆y = ω (n − 1) λ λc c c (55) Where n is the refractive index of the glass plate. Due to this extra phase (∆φ), the new wave will be y Ep = E0 eiω(t− c ) e−∆φ =⇒ Ep = E0 eiω[t−(n−1) (56) ∆y − yc ] c (57) (ii) Now, Ep = Eu + iπ ω(n − 1)∆y Eu e− 2 c =⇒ Ep = Eu e−iω(n−1) ∆y c (58) (59) Now if n ≈ 1 or ∆y is small then the exponent term is very small. Hence, expanding it in Taylor’s series and neglecting the higher order terms, we obtain Ep ≈ Eu [1 − iω(n − 1) =⇒ Ep = Eu + ∆y ] c iπ ω(n − 1)∆y Eu e− 2 c 8 (60) (61) 7. (a) Draw a graph of θt versus θi for an air-glass boundary where the index of refraction of glass is ng = 1.5. (b) A glass prism whose cross section is an isosceles triangle stands with its (horizontal) base in water; the angles which its two equal sides make with the base are each equal to θ (see Figure 1.). An incident ray of light, above and parallel to the water surface and perpendicular to to the prism’s axis, is internally reflected at the glass-water interface and subsequently re-emerges into the air. Taking the refractive indices of glass and water to be 3 2 and 4 3 show that θ must be atleast 25.9 degree. 7(a). From Snell’s law θt = sin −1 The graph: sin θi ng 40 Θt H in o L 30 20 10 0 0 50 100 Θi H in o 150 L FIG. 1: θt versus θi 7(b). From the geometry of the prism, it can easily be seen that the angle of incidence at the first surface is θi = 90o − θ. Using Snell’s law, we have ng sin θt′ = sin(90o − θ) or, ng sin θt′ = cos θ 9 glass θ water FIG. 2: Figure for Problem 7(b) where, ng (refractive index of the glass) = 3 2 and θt′ is the refraction angle at that surface. Suppose, θi′ be the angle of incidence at the glass and water interface. Then, again from geometry of the prism, we have θi′ = θt′ + θ Now, for internal reflaction at the glass-water interface, following inequality must be satisfied. sin θi′ ≥ nw ng where, nw (refractive index of water) = 43 . Using trigonometric identities and the previous two equations, we get the following inequality 2 cos θ + q n2g − cos2 θ sin θ ≥ nw Further simplifying the above inequality we get s n2g − n2w n2g + 1 − 2nw s ! n2g − n2w n2g + 1 − 2nw cos θ ≤ or, θ ≥ cos−1 Using the numerical values, finally we get θmin ≈ 25.9o . 10 ** 8. (a) Consider Snell’s law of refraction. If the medium of incidence has an index n1 > 0 and the other medium has an index n2 < 0, then draw the refracted ray by assuming some angle of incidence θi . What difference do you notice in comparison to the case when n1 , n2 > 0? (b) Recall that the refractive index can be written as n = √ ǫr µr where ǫr = ǫ ǫ0 and µr = µ . µ0 It is known that ǫ and µ are complex quantities and also functions of ω, with their real and imaginary parts related to physical quantities. Show, with an example, that using ǫ < 0 and µ < 0 it is possible to have the refractive index n < 0. Note: Materials exhibiting negative index of refraction have actually been made in recent times though the original proposal was by the Russian physicist V. Veselago, in 1967. Such materials are known as metamaterials. 8.(a) Snell’s law n1 sin θi = n2 sin θt n1 sin θ1 or, sin θt = n2 If n1 > 0 and n2 < 0, then sin θt < 0. Let us assume n1 = 1, n2 = −1.5, θi = 30 degrees. Then we get sin θt = − 31 . Thus, θt = −19.27 degrees. Note the minus sign. The refracted ray is now bent in a different way. Draw the incident and refracted rays, with the above values and you can see how different it is. (b)The definition of refractive index is n= √ √ ǫµ ǫr µ r = √ ǫ0 µ 0 It is known that ǫ and µ are complex quantities. So, written in the polar form of a complex number, ǫ = |ǫ|eiφ1 and µ = |µ|eiφ2 . So p |ǫ||µ| i(φ1 +φ2 )/2 n= √ e ǫ0 µ 0 Now, if ǫ < 0 and µ < 0 then φ1 = φ2 = π. So, p or, |ǫ||µ| iπ n = √ e ǫ0 µ 0 p |ǫ||µ| n = −√ <0 ǫ0 µ 0 11