Title 10.WAVE OPTICS - Explanations.pdf ✅

2 (c) e sin θ = λ ⇒ sin θ ≈ θ = λ e x = f, θ = fλ e ∴ 2x = separation between minima on either side of central maximum’ = 2fλ e Hence, e = 2fλ 2x = 2 × 1 × 600 × 10−9 4 × 10−3 = 0.3 mm 3 (b) β = λD d 4 (a) β = λD d ⇒ β ∝ λ 5 (d) Ultrasonic waves are longitudinal waves 6 (c) Imax = (√I1 + √I2) 2 = (√I + √4I) 2 = 9I Imin = (√I1 − √I2) 2 = (√I − √4I) 2 = I 7 (d) λRed > λBlue > λX−ray > λγ 8 (b) Distance of n th minima from central bright fringe xn = (2n − 1)λD 2d For n = 3 i. e. 3 rd minima x3 = (2 × 3 − 1) × 500 × 10−9 × 1 2 × 1 × 10−3 = 5 × 500 × 10−6 2 = 1.25 × 10−3m = 1.25 mm 9 (b) Infrasonic waves are mechanical waves 10 (b) EM waves transport energy, momentum and information but not charge. EM waves are unchanged 11 (c) The angular wave number k = 2π λ ; where λ is the wave length. The angular frequency is w = 2πv The ratio k ω = 2π/λ 2πv = 1 vλ = 1 c = constant 12 (a) I = 1 2 ε0CE0 2 ⇒ E0 = √ 2I ε0c = √ 2 × 5 × 10−16 8.85 × 10−12 × 3 × 108 = 0.61 × 10−6 V m Also E0 = V0 d ⇒ V0 = E0d = 0.61 × 10−6 × 2 = 1.23μV 13 (a) The diffraction pattern of light waves of wavelength (λ) diffracted by a single, long narrow slit of width is shown. For first minimum. e sin θ = λ sin θ = λ e When e is decreased for same wavelength, sin θ increases, hence θ increases. Thus, width of central maxima will increase. 14 (d) Intensity of EM wave is given by I = P 4πR2 = vav. c = 1 2 ε0E0 2 × c ⇒ E0 = √ P 2πR2ε0c = √ 800 2 × 3.14 × (4) 2 × 8.85 × 10−12 × 3 × 108 = 54.77 V m 15 (a) For secondary maxima d sin θ = 5λ 2 ⇒ dθ = d. x D(≈ f) = 5λ 2 ⇒ 2x = 5λf d = 5 × 0.8 × 6 × 10−7 4 × 10−4 = 6 × 10−3m = 6mm 17 (d) Distance of nth dark fringe from central fringe xn = (2n − 1)λD 2d
∴ x2 = (2 × 2 − 1)λD 2d = 3λD 2d ⇒ 1 × 10−3 = 3λ × 1 2 × 0.9 × 10−3 ⇒ λ = 6 × 10−5 cm 18 (a) For second dark fringe d sin θ = 2λ ⇒ 24 × 10−5 × 10−2 × sin 30 = 2λ ⇒ λ = 6 × 10−7m = 6000 Å 19 (d) Momentum transferred in one second p = 2U c = 2SavA c = 2 × 6 × 40 × 10−4 3 × 108 = 1.6 × 10−10kg − m/s 2 20 (a) Oil floating on water looks coloured only when thickness of oil layer=wavelength of light=10000Å 21 (d) β = λD d = 600 × 10−9 × 2 1 × 10−3 = 12 × 10−4 m So, distance between the first dark fringes on either side of the central bright fringe = 2β = 2 × 12 × 10−4 m = 24 × 10−4 m = 2.4 mm 22 (d) Sound waves cannot be polarised as they are longitudinal. Light waves can be polarised as they are transverse. 23 (c) Doppler shift (Source moving towards observer) λ ′ = λ (1 − V C ) 5400Å = 6200Å (1 − V C ) V = [1 − 54 62] C = 3.9 × 107 approx 24 (b) β = (a + b)λ 2a(μ − 1)α , i. e. , β ∝ λ (μ − 1) When placed in water β′ ∝ λ μ′ ( μ μ′−1) i. e. , β′ ∝ λ (μ−μ′) but < μ ∴ β ′ β = (μ − 1) (μ − μ′) ∵ μ ′ > 1λ ∴ β ′ > β i. e., the fringe width increases 25 (a) β ∝ 1 d ⇒ If d becomes thrice, then β becomes 1 3 times 26 (b) nλr = (n + 1)λb n + 1 n = λr λb = 600 480 = 4 5 1 n = 4 5 − 1 = 1 4 n = 4 27 (b) β = (a + b)λ 2a(μ − 1)α Where a = distance between source and biprism = 0.3 m b = distance between biprism and screen = 0.7 m α = Angle of prism = 1°, μ = 1.5, λ = 6000 × 10−10m Hence, β = (0.3+0.7)×6×10−7 2×0.3(1.5−1)×(1°× π 180) = 1.14 × 10−4m = 0.0114 cm 28 (d) The phase difference (φ) between the wavelets from the top edge and the bottom edge of the slit is φ = 2π λ (d sin θ) where d is the slit width. The first minima of the diffraction pattern occurs at sin θ = λ d , so φ = 2π λ (d × λ d ) = 2π 29 (b) Δλ λ = v c = 6 × 107 3 × 108 = 0.2 Δλ = λ ′ − λ = 0.2λ ⇒ λ ′ = 1.2λ = 1.2 × 4600 = 5520Å 30 (a) Position of n th minima xn = nλD d ⇒ 5 × 10−3 = 1 × 5000 × 10−10 × 1 d ⇒ d = 10−4m = 0.1 mm 31 (a) Shift = β λ (μ − 1)t = β (5000×10−10) × (1.5 − 1) × 2 × 10−6 = 2β i. e. ,2 fringes upwards 32 (c) Position of nth bright fringe x1 = nλD d For first bright fringe n = 1 ∴ x1 = λD d Position of nth dark fringe x2 = (2n−1)λD 2d For first dark fringe n = 1
∴ x2 = λD 2d Now, x1 − x2 = λD 2d If B is the band width, then x1 − x2 = B 2 33 (a) Here, λ = 6250 Å = 6250 × 10−10m a = 2 × 10−2 cm = 2 × 10−4m D = 50 cm = 0.5 m Width of central maximum=2λD a = 2 × 6250 × 10−10 × 0.5 2 × 10−4 312.5 × 10−3 cm 34 (d) Greater is the wavelength of wave higher will be its degree of diffraction. 35 (b) Fringe width (β) ∝ 1 prism Angle (α) 36 (a) Angular spread on either side is θ = λ a = 1 5 rad 37 (a) Photoelectric effect explain the quantum nature of light while interference, diffraction and polarization explain the wave nature of light 38 (c) Path difference, x = (SS1 + S1O) − (SS2 + S2O) Ifx = nλ, the central fringe at O will be bright. Ifx = (2n − 1)λ/2, the central fringe at O will be dark. 39 (d) E = hc λ = 6.6 × 10−34 × 3 × 108 21 × 10−2 = 0.94 × 10−24 = 10−24J 41 (b) Average energy density of electric field is given by ue = 1 2 ε0E 2 = 1 2 ε0 ( E0 √2 ) 2 = 1 4 ε0E0 2 = 1 4 × 8.85 × 10−12(1) 2 = 2.2 × 10−12J/m3 44 (a) Distance between the slits, d = √d1d2 = √4.05 × 10−3 × 2.90 × 10−3 = 3.427 × 10−3m 45 (d) If shift is equivalent to n fringes then n = (μ − 1)t λ ⇒ n ∝ t ⇒ t2 t1 = n2 n1 ⇒ t2 = n2 n1 × t t2 = 20 30 × 4.8 = 3.2mm 47 (d) Given single slit of width d = 0.1 mm d = 0.1 × 10−3 m Or d = 1 × 10−4 m Light of wavelength a = 600 Å Or α = 6 × 10−7 m The angle of diffraction θ = nλ d θ = 2 × 6 × 10−7 1 × 10−4 θ = 12 × 10−3 θ = 0.012 rad 48 (a) As velocity of light is perpendicular to the wavefront, and light is travelling in vacuum along they − axis, therefore, the wavefront is represented byy = constant. 49 (a) d1 = 7λ1 D d And d2 = 7λ2 D d ∴ d1 d2 = λ1 λ2 50 (b) λBlue < λRed. Therefore fringe pattern will contract because fringe width ∝ λ 51 (d) Interference is shown by electromagnetic as well as mechanical waves 52 (a) As, a sin θ = nλ λ = a sinθ n = 5 sin 30° 1 = 2.5 cm 53 (a) Distance covered by T.V. signals = √2hR ⇒ maximum distance ∝ h 1/2 54 (d) In the given options none of sources generates plane wavefront, it can be artificially produced by
reflection from a mirror or by refraction through a lens. 56 (c) The number of fringes on either side of centre C of screen is n1 = [ AC β ] = [ 0.5 0.021] = [23.8] = 23 ∴ Total number of fringes = 2n1 + fringe at centre = 2n1 + 1 = 2 × 23 + 1 = 46 + 1 = 47 In Young’s experiment, the number of fringes should be odd. 57 (c) Wave theory of light is given by Huygen 59 (c) In interference, we use two sources while in diffraction, we use light from two points of the same wavefront. 60 (c) Speed of light of vacuum c = 1 √μ0ε0 and in another medium v = 1 √με ∴ c v = √ με μ0ε0 = √μrK ⇒ v = c √μrK 61 (c) For a slit of width a, light of wavelength λ, when light falls on the slit, the diffraction patterns so obtained as The first diffraction minimum occurs at the angles given by sin θ = λ a From the equation, it is clear that width of the central diffraction maximum is inversely proportional to the width of the slit. On increasing the width sizea, the angle θ at which the intensity first becomes zero decreases, resulting in a narrower central band and if the slit width is made smaller, the angle θ increases, giving a wider central band. 64 (b) When a Young’s double slit set up for interference is shifted from air to within water then the fringe width decreases. 65 (d) v = c √μrεr = 3 × 108 √1.3 × 2.14 = 1.8 × 108m/s 66 (d) Let λ be wavelength of monochromatic light, used to illuminate the slit S, and d be the distance between coherent sources, then width of slits is given by W = Dλ d When D is distance between screen and source. Given, d = 3 mm, λ = 5000 Å = 5 × 10−7 m = 5 × 10−4 mm D = 90 cm = 900 mm ∴ W = 5 × 10−4 × 900 3 = 15 × 10−2 mm = 0.15 mm 67 (d) By using fp = r 2 (2p−1)λ For first HPZ r = √fpλ = √0.6 × 6000 × 10−10 = 6 × 10−4m 68 (d) As reflected and refracted rays are perpendicular to each other, therefore, ip = i = 60°