Unit 6ab kho tài liệu bách khoa

VERY SHORT AND SHORT-ANSWERS QUESTIONS 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 What are red and blue shifts ? What is Fresnel distance ? Write an expression for it Write an expression for the resolving power of a telescope What is the limit of resolution of a microscope ? Write an expression for it What is optical activity ? Give an expreiment to show that light waves are transverse in nature (AISSCE 1994) Write some uses of polaroids Write the conditions for maxima and minima of diffraction pattern The refractive index of glass w.r.t air is 5/3, and that of water is 4/3 Find the refractive index of glass w.r.t water The refractive indices of two media X and Y are 5/3 and 5/4 respectively Which medium is denser ? What happens to the path when a ray passes from a rarer to a denser medium ? The wavelength of light is 0.00006 cm Express this wavelength in micron and in angstrom In Young’s double slit experiment, violet, yellow and red lights are successively used For which colour will the fringe width be maximum ? What is the phase difference corresponding to a path difference of λ ? What is the shape of the wave front for a point source ? Two slits are separated by a distance less than the wavelength of light used Can you observe interference pattern on the screen ? A Young’s double slit arrangement is shifted from air to inside water What happens to the fringe width ? S Chand & Company Limited 68 In Young’s double slit experiment, one of the two slits is covered with a cellophane sheet which absorbs half the light intensity What will happen to the fringes ? 69 What is the ratio of the fringe widths for bright and dark fringes in Young’s double slit experiment 70 Can we obtain a interference pattern on a screen by using two identical sodium lamps 71 Do we get any information about the longitudinal and transverse nature of light from the phenomenon of diffraction 72 How will the width of the central maximum change if a single slit diffraction set up is immersed completely in water without changing any other factor ? 73 When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle Explain Why ? 74 Name one phenomenon which is exhibited by light waves but not by sound waves 75 Can electromagnetic waves be polarised ? 76 Does the value of Brewster’s angle depend on the colour of light ? 77 What is Doppler shift 78 A star is accelerating towards the earth It is emitting orange colour Whether its colour, as seen from the earth, will turn gradually yellow or red ? 79 State Malus law 80 Light of wavelength 5000 Å in air enters a medium of refractive index 1.5 What will be its frequency in the medium ? 81 A and B are two points on a water surface where waves are generated What is the phase difference if (a) A and B are on the same wavefront but separated by one wavelength, (b) A and B are on successive crests 82 In Young’s double slit experiment, if the distance between the two slits is halved and the distance between the slits and the screen is made four times, then by what factor will the fringe width change ? S Chand & Company Limited 83 Widths of the two slits in Young’s experiment are in the ratio : What is the ratio of the amplitudes of light waves coming through them 84 The phase difference between two waves reaching a point is π/2 What is the resultant amplitude if the amplitudes of the two waves are mm and mm 85 A ray of light strikes a transparent medium of refractive index If the reflected and refracted rays are multually perpendicular, what is the angle of incidence ? 86 Two coherent sources have intensity ratio 64 :1 Calculate their amplitude ratio 87 Refractive index of glass for lights of yellow, green and red colours are µy, µg, and µr respectively Rearrange these symbols in an increasing order of values (AISSCE 1997) 88 Deduce the laws of reflection on the basis of Huygens’ principle (AISSCE 1992) 89 Using Huygens’ principle, derive Snell’s law (AISSCE 1993) 90 Derive an expression for the angular width of the central maximum of the diffraction pattern produced by a single slit, illuminated with monochromatic light (AISSCE Delhi (comp) 1991) 91 In Young’s double slit experiment derive an expression for fringe width 92 Explain polarisation by reflection and obtain Brewster’s law 93 State and explain Huygens’ principle 94 Explain the terms Fresnel distance and Fresnel zone 95 The polarizing angle of a medium is 60º What is the refractive index of the medium ? (AISSCE Delhi 1999) 96 What is the polarizing angle of a medium of refractive index ? (AISSCE 1999) 97 What changes in the interference pattern in Young’s double slit experiment will be observed when (i) the distance between the slits is reduced, and (ii) the apparatus is immersed in water ? (AISSCE 1999) S Chand & Company Limited 98 State Huygens’ postulates of wave theory Sketch the wavefront emerging from (i) a point source of light and (ii) a linear source of light like slit (AISSCE Delhi 2000) 99 In a single slit diffraction experiment, if the width of the slit is doubled, how does the (i) intensity of light (ii) width of the central maximum change? Give reasons for your answer (AISSCE Delhi 2000) 100 What is the effect on the interference pattern observed in a Young’s double slit experiment in the following cases: (i) Screen is moved away from the plane of the slits (ii) Seperation between the slits is reduced (iii) Widths of the slits are doubled Giver reasons for your answers (AISSCE 2000) ANSWERS 50 The phenomenon of apparent change in the frequency of light due to relative motion of source and observer is called Doppler effect 51 Due to Doppler effect, a wavelength in the middle of the visible spectrum will be shifted towards red if the source and observer move away from each other, and towards blue if they approach each other The terms red shift and blue shift are used for wavelength increase and wavelength decrease, respectively, even when the wavelength is not in the visible region 52 Fresnel distance (DF) is the distance of the screen from the slit at which the spreading of light due to diffraction becomes equal to the size of the slit It is given by DF = a λ a , 53 Resolving power of a telescope = 1.22 λ where a is the diameter of the aperture of the objective S Chand & Company Limited 54 The limit of resolution of a microscope is defined as the smallest distance between two objects so that they can be seen distincity It is given by d = λ 2µ sin θ where µ is the refractive index of the medium between the object and the objective of the microscope and θ is the semi vertical angle 55 Certain crystals or compounds in solution have the ability to rotate the plane of polarization of plane polarized light This property is called optical activity 56 A tourmaline crystal cut parallal to its axis behaves like a very fine slit When unpolarised light is incident normally on a pair of such tourmaline crystal, placed with their axes parallal, the intensity of the emergent light is maximum Now we place the two crystals with their axes perpendicular to each other The intensity of outcoming light in this case is minimum This experiment shows that light waves are transverse in nature 57 Polaroids are used in (a) sun glasses, (b) window panes, (c) photography as filters λ n = 1, 2,3, For minima a sin θ = nλ, n =1, 2,3, Symbols have their usual meanings 59 a µ g = , a µ w = 3 53 aµg ∴ = = = 1.25 wµg = 43 aµw 60 X is denser than Y 58 For maxima a sin θ = (2n + 1) S Chand & Company Limited 61 When a ray of light passes from a rarer to a denser medium it bends towards the normal 62 0.00006 cm = 0.6 micron = 6000 Å 63 Fringe width β ∝ λ The wavelength is maximum for red, so the fringe with will be maximum for red 64 Phase difference = 2π (path difference) λ 2π = × 2λ λ = 4π radians 65 Spherical 66 No In this case the fringe width will become very large λD  67 In water, wavelength λ will decrease and so the fringe width will also decrease  β = d   68 Bright fringes will be less bright and dark fringes will be less dark Thus the contrast between the fringes will decrease 69 : 70 No 71 Diffraction does not give any information about the transverse or longitudinal nature of light 72 The width of the central maximum in the single - slit diffraction pattern is 2λD/a Since the wavelength λ decreases in water, the width of the central maximum will decrease on immersing the set-up in water S Chand & Company Limited 73 At the centre, the light waves diffracted from the edge of the obstacle interfere constructively So a bright spot is seen at the centre of the shadow 74 Polarisation is exhibited by light waves but not by sound waves 75 Electromagnetic waves are transverse in nature So they can be polarised 76 Yes Brewster’s angle is tan–1(µ), and µ is different for different colours 77 The apparent change in the frequency of waves due to relative motion of observer and source is called Doppler shift 78 Since the star is accelerating towards the earth, the wavelength of the light emitted by it will go on decreasing Therefore its colour will gradually turn yellow 79 When plane polarised light is passed through an analyser, the intensity of the outcoming light changes as the analyser is rotated in its own plane It is given by I = I cos θ, where I0 is the maximum intensity of the transmitted light and θ is the angle between the axes of the polariser and the analyser This equation is called Malus law 80 × 108 c = = × 1014 Hz λ 5000 × 10 –10 Frequency does not change in going from one medium to another Frequency ν = 81 (a) If points A and B are on the same wavefront, the phase difference between them is zero (b) If A and B are on successive crest, then the path difference between them is λ and so the phase difference is 2π S Chand & Company Limited 82 Dλ d β= 4D λ 8D λ = d /2 d β′ 8D λ / d = =8 β Dλ / d β ′ = 8β β′ = ⇒ or 83 Let d1 and d2be the width of two slits, then A1 d = A2 d2 =3 = 84 We have A = A12 + A22 + A1 A2 cos θ Here A1= mm, ∴ A A2 = 4mm, φ = π/2 = (3)2 + (4)2 + 2(3) (4)cos(π / 2) = 5mm 85 When the refracted and reflected rays are mutually perpendicular, the angle of incidence is given by i = tan–1 (µ) = tan–1 ( 3) = 60º 86 Amplitude ratio a1 = a2 I1 I2 S Chand & Company Limited Here I1 64 = I2 ∴ a1 = a2 64 = 1 87 µr < µy< µg 88 Let XY be a reflecting surface and AB be a plane wavefront just incident on the surface at A The normals LA and MB to the wavefront represent incident rays If AN is the normal to the surface at A, then ∠ LAN = i, M B N N' R P L i r X A Q r Y C Fig A.2 the angle of incidence CB′ is the reflected wavefront and r is the angle of reflection To prove the laws of reflection, consider any point P on the incident wave front Starting from P we consider a ray PQR such that PQ is perpendicular to AB and QR is perpendicular to B′ C Let c be the velocity of light The distance travelled by the ray from P to R = PQ + QR S Chand & Company Limited Total time taken PQ + QR AQ sin i + QC sin r = c c AQ sin i + ( AC – AQ) sin r = c AC sin r + AQ (sin i – sin r ) = c This time is to be the same for all wavelets on the incident wavefront., So it must be independent of AQ ⇒ sin i – sin r = or sin i = sin r or i = r , i.e., angle of incidence = angle of reflection This proves the first law of reflection Also, the incident wavefront, the reflected wavefront and the normal to the surface, all lie in the same plane i.e., the plane of the paper This is the second law 89 Let XY be a plane surface separating two media and let PA be a plane wavefront just incident on it Normals LA and MP to the incident wavefront represent incident rays AN is normal to the surface at the point A ∠ NAL = i, the angle of incidence The wavefront first strikes at A, so secondary wavelets will start first from A, which travel with velocity c2 in the second medium During time t the disturbance reaches from P to P ′ In this time the secondary wavelet from A will travel a distance c2 t in the second medium With A as centre and c2t as radius, we draw an arc and from P ′ we draw a tangent P ′ A ′ P ′ A ′ is the refracted wavefront t = S Chand & Company Limited M P N Q L i i K X A P' Y r r Q' A' Fig A.3 Consider a ray QKQ′ such that QK is perpendicular to AP and KQ′ is perpendicular to A ′P ′ The time taken by the ray to travel a distance QK + KQ ′ is S Chand & Company Limited t= or QK KQ ′ + c1 c2 AK sin i KP′ sin r + c1 c2 AK sin i ( AP′ – AK ) sin r = + c1 c2 t= =  sin i sin r  AP′ sin r – + AK   c2 c2   c1 The rays from different points on the incident wavefront will take the same time to reach the corresponding points on the refracted wavefront It is possible only if t is independent of AK sin i sin r – ⇒ =0 c1 c2 c sin i = = 1µ sin r c2 µ is called the refractive index of the second medium with respect to the first medium and the above equation is called Snell’s law 90 Fig A.4 shows the diffraction of a monochromatic wavefront from a single slit (AB) The secondary wavelets that go straight across the slit arrive at the lens in same phase and are brought to focus on the screen at O Thus the intensity at O is maximum or S Chand & Company Limited θ A a P C O θ B N P' a sin θ Fig A.4(a) ↑ I Central Maximum –λ —— a O λ –— a Fig A.4(b) Consider now the rays making an angle θ with the direction of normal to the slit, such that a sin θ = λ S Chand & Company Limited In this case, if we imagine the slit to be divided into two equal parts AC and CB, then for every point in AC, there is a point in CB such that the path difference between the rays from these points is λ/2 Thus the points P and P ′ at which these rays converge will be minima Since θ is very small, this gives θ = λ/a 2λ Angular width of central maximum = 2θ = a 91 S is a source of monochromatic light illuminating two slits S1 and S2 S1and S2 are very close to each other and equidistant from source S S1and S2 will act as coherent sources of light and wavefronts from them will spread in all directions Due to their superposition, alternate bright and dark bands are obtained on the screen O is equidistant from S1 and S2 , so maximum is obtained at O Let y be the distance of a point P on the screen from the centre O P S1 A d C O S2 B y D Fig A.5 S Chand & Company Limited The path difference between the rays reaching point P from S1and S2 is given by P = S2 P – S1 P d  Now, S1 P = S1 A2 + AP = D +  y –  2  and d  S2 P = D +  y +  2  2 So, or d  d  S2 P – S1 P =  y +  –  y –  2  2  (S2 P – S1 P ) ( S2 P + S1 P ) = yd We can take S2 P + S1P = 2D = S P – S1 P = Path difference For maxima or yd = nλ , D n λD y= d 2 yd yd = 2D D n = 0,1, 2, This is the position of the nth maximum on the screen Fringe width β = Distance between two successive maxima = yn+1 – yn = (n + 1) λD nλD – d d or β = λD d S Chand & Company Limited 92 Unpolarized light, on reflection from a transparent surface, gets polarised The degree of polarization depends on the angle of incidence At a particular angle of incidence the reflected beam becomes completely polarised This angle is called polarising angle or Brewster’s angle (θp) The plane polarised light has vibrations perpendicular to the plane of the paper The refracted light is partially polarised It is found that when light is incident at polarising angle, the reflected and the refracted rays are perpendicular to each other Brewster’s law In the figure θp + r = 90º N θp or r = 90 – θp According to Snell’s law sin i =µ sin r sin θp or =µ sin (90 – θp ) or or sin θp cos θp =µ tan θp = µ This is Brewster’s law S Chand & Company Limited θp Air B 90° r N Fig A.6 Medium 93 Huygens’ principle is a geometrical construction which enables us to find the new position of the wavefront at a later time from its given position at any instant It is based on the following assumptions: (a) Each point on a given wavefront acts as a source of disturbance called secondary wavelets, which travel in all the directions with the speed of light in that medium A" B′ A A′ A a a b A" b c c d d e e f B" B f g A′ B" Fig A.7(a) B B′ Fig A.7(b) S Chand & Company Limited (b) The tangential surface or the envelope of these secondary wavelets at any instant gives the new position of the wavefront at that instant Consider a point source of light In figure A (a), AB is a section of a spherical wavefront at time t To find the new position of the wavefront after time ∆t we consider points a, b, c, on the wavefront AB These points will act as sources of secondary wavelets The distances travelled by the secondary wavelets in time ∆t is equal to c ∆t Now taking a, b, c, as centres we draw spheres of radius c ∆ t Draw the forward envelop (A ′B′) of these secondary wavelets A′B′ gives the new position of the wavefront after time ∆t Fig A (b) shows a plane wavefront λ 94 An aperture or slit of size a, illuminated by a parallal beam sends diffracted light into an angle θ ! a This is the angular size of the bright central maximum In travelling a distance D, the difDλ due to diffraction fracted beam acquires a width a The distance at which the diffraction spreading of a beam equals the size of the slit or aperture is called Fresnel distance (DF) Thus, DF λ a = a or DF = a λ If we want a beam to travel a distance D without too much broadening by diffraction, we must have DF > D or a2 > D or a > λ λD S Chand & Company Limited for a given value of D, quantity λ D is called the size of the Fresnel zone, aF: aF = λ D 95 Polarizing angle ip = 60º Refraction index µ = tan ip = tan 60º = = tan–1(µ) 96 Polarizing angle ip = tan–1 ∴ ( 3) = 60º 97 Fringe width, β = λD d (i) If d is reduced, β will increase (ii) Wavelength (λ) of light decreases in water Therefore, if the apparatus is immersed in water, β will decrease 98 (i) See Q 93 (ii) (a) Point source - Spherical wavefront (Fig a) (b) Linear source - cylindrical wavefront (Fig b) S Chand & Company Limited (a) (b) Fig A.8 99 (i) The intensity of light increases (ii) Width of the central maximum, ω = If a increases, ω decreases 100 (i) Dλ a Dλ d If D is increased, β will increase Fringe width, β = (ii) If d is reduced, βwill increase (iii) The intensity of light from each slit will be doubled Therefore the intensity of bright fringes will be doubled However the fringes may be blurred due to overlapping S Chand & Company Limited