Unit 6bc tủ tài liệu training

VERY SHORT AND SHORT-ANSWERS QUESTIONS 81 A lens placed in a medium behaves as an ordinary glass plate What is the refractive index of the medium relative to that of the lens ? 82 A ray of light, after refraction through a concave lens, emerges parallal to the principal axis Under what conditions does this happen ? 83 The focal length of a biconvex lens is half the radius of curvature of either face Calculate the refractive index of the material of the lens 84 What is the inclination of the refracted ray inside the prism with the base of the prism, if the prism is set in the position of minimum deviation ? 85 When a beam of white light is passed through a prism, the deviation suffered by violet colour is more than that suffered by red colour Why ? 86 Write the names of the three main components of a spectrometer 87 What is the function of the collimator in a spectrometer ? 88 Why is red coloured light used in danger signals or traffic stops ? 89 Why is the sequence of colours in the secondary rainbow reverse of that in the primary rainbow ? 90 Why we prefer a magnifying glass of short focal length ? 91 An object is seen through a microscope first in red light and then in violet light In which case is the magnification larger ? 92 How can you increase the resolving power of a telescope ? 93 The focal length of a convex lens is 25 cm Where should the object be placed in front of the lens so as to get a real image equal to the size of the object ? 94 The power of a lens is + 4D What is the nature and focal length of the lens ? 95 Two lenses of powers + 8D and + 4D are kept in contact, what is the focal length of the combination ? S Chand & Company Limited 96 Prove the relation aµ b = bµ a 97 What is meant by apparent depth ? Derive a relation between apparent depth, real depth and refractive index for near normal incidence 98 How optical fibres transmit light without significant absorption ? Mention one practical application of optical fibres (AISSCE 1991) µ1 µ µ – µ1 + = 99 Derive the relation – u v R for refraction at a spherical surface The symbols have their usual meaning 100 Derive the thin lens formula (equation) 101 Derive the lens maker’s formula in the case of a double convex lens 102 Define linear magnification produced by a lens Derive an expression for it 103 Show that the refractive index of the material of a prism is given by µ= 104 105 106 107 sin [( A + δm ) / 2] , where the symbols have their usual meanings sin ( A / 2) Derive the expression for the angle of deviation for a ray of light passing through an equilateral prism of refracting angle A (AISSCE 1993) What is angular dispersion ? Derive an expression for the angular dispersion produced by a small angled prism for small angles of incidence What is emission spectrum ? Distinguish between the three types of emission spectra Explain the working of a convex lens of short focal length as a simple microscope OR Explain the working of a simple microscope and show that its magnification is given by S Chand & Company Limited m = 1+ 108 109 110 111 112 113 114 115 116 117 118 D f where f is the focal length and D is the least distance of distinct vision (AISSCE 1992) With the help of a ray diagram explain the working principle of a compound microscope (AISSCE 1993) Draw a labelled diagram of an astronomical telescope forming the final image at the near point Write down the formula for its magnifying power (AISSCE 1993) Draw a labelled diagram of a telescope and explain its working when the final image is formed at infinity Give an expression for its magnifying power (AISSCE 1992, AISSCE Delhi 1999) Write two common applications of simple microscope (magnifying glass) Which one has higher refractive index - crown glass or flint glass ? What would be the colour of the sky if viewed from the moon ? (a) Draw a labelled ray diagram for a refracting type astronomical telescope (b) How will its magnifying power be affected on increasing for its eye piece (i) the focal length (ii) the aperture ? (AISSCE 1997) (a) Draw a labelled ray diagram for a compound microscope (b) how will its resolving power be affected when (i) the frequency of light used to illuminate the object is increased, and (ii) the focal length of the objective is increased (AISSCE 1997) Draw a labelled ray diagram for a reflecting type telescope Write four advantages of a reflecting type telescope over a refracting type telescope? (AISSCE 1999) How does (i) the magnifying power (ii) the resolving power of a telescope change on increasing the diameter of its objective ? Give reasons for your answer (AISSCE 1998) Violet light is incident on a converging lens of focal length f State, with reason, how the focal length of the lens will change if the violet light is replaced by red light (AISSCE - 1999) S Chand & Company Limited 119 A convex lens, made of a material of refractive index µ1,is kept in a medium of refractive index µ2 Parallal rays of light are incident on the lens Complete the path of the rays of light emerging from the convex lens if (i) µ1 > µ2 (ii) µ1= µ2 (iii) µ1 < µ2 (AISSCE - 2000) 120 A concave lens, made of a material of refractive index µ1 , is kept in a medium of refractive index µ2 A parallal beam of light is incident on the lens Complete the path of the rays of light emerging from the concave lens if (i) µ1 > µ2 (ii) µ1 = µ2 (iii) µ1 < µ2 (AISSCE - 2000) ANSWERS 81 One 82 When the incident ray is directed towards the principal focus of the lens, then after refraction it emerges out parallal to the principal axis (Fig B.6) 83 F  1  = µ –  –  f  R1 R2  ( Here Therefore ) f = R/2, R1 = R, R2 = – R R/2 R or 1 1 +  R R = (µ – 1)  = (µ – 1) R µ – = or µ = S Chand & Company Limited Fig B.6 84 The refracted ray inside the prism is parallal to the base of the prism 85 We know that more the refractive index of the prism for a colour, more is the deviation Since µ V > µR, violet colour suffers larger deviation than red 86 The three main components of a spectrometer are (i) collimator, (ii) prism table, and (iii) telescope 87 The function of the collimator is to produce a beam of parallal rays 88 Red colour, being of the largest wavelength in the visible spectrum, scatters least Therefore, it can easily be seen from large distances, even in foggy conditions 89 The sequence of colours in the secondary rainbow is reverse of that in the primary rainbow because it is formed by two internal reflections of light in water droplets, whereas primary rainbow is formed by one internal reflection  D  f 90 The magnifying power of magnifying glass is given by m =  +  This shows that magni fying power is large when focal length is small 91 We have m = + (D/f) Focal length of red colour is greater than that of violet colour, so magnification is larger for violet light 92 The resolving power of a telescope can be increased by increasing the aperture of the objective of the telescope 93 To have a real image equal to the size of object, the object must be placed at a distance 2f from the lens, i.e., at × 25 = 50 cm from the lens 1 = m = 25 cm 94 f = P Since focal length is + ve, the lens is convex S Chand & Company Limited 95 Total power P = P1 + P2 = + = 12 D Focal length of the com bination, f = = P = 12 m 100 cm = 8.33 cm 12 96 Let a ray of light travel from a medium a to a medium b Refractive index of medium b w.r.t medium a is given by sin i (1) a µb = sin r On reversing the path, light travels from medium b to medium a Now the angle of incidence is r and the angle of refraction is i Therefore the refractive index of the medium a w.r.t medium b is sin r (2) b µa = sin i Medium a i Medium b r Fig B.7 S Chand & Company Limited Eqs (1) and (2) give a µb = b µa 97 The depth of an object lying in or below an optically denser medium appears to be less than its real depth This is called the apparent depth P C Medium a A B r I i Q Medium b O Fig B.8 In the figure (O) is a point object below a transparent slab Ray OA strikes the surface normally and moves undeviated Another ray OB strikes at B at an angle of incidence i and is refracted along BC The two refracted rays appear to diverge from I So I is the virtual image of O AI is the apparent depth and AO is the real depth S Chand & Company Limited Now or b µa = a µb = sin i sin r sin r sin i ∠ AOB= ∠ OBQ = i (alternate angles) ∠ AIB = ∠ PBC = r (corresponding angles) In the right triangles AOB and AIB AB AB sin i = and sin r = OB BI AB BI BO = So, a µb = AB BO BI For near normal incidence B is very close to A We have BO ! AO and BI ! AI Therefore, a µ b = AO AI = Realdepth Apparent depth 98 Optical fibres consist of thousands of long fine quality glass or quartz fibres of high refractive index of about 1.7 They are coated with a material of refractive index 1.5 The thickness of each fibre (strand) is about 10–4 cm When light is incident on one end of the fibre at a small angle, it passes through the fibre such that it suffers total internal reflection along the fibre and finally comes out at the other end S Chand & Company Limited 99 In Fig B.9 XY is a convex refracting surface separating media and O is a point object on the principal axis OA is an incident ray which is refracted along AI i is the angle of incidence and r is the anlge of refraction Another ray OP strikes normally and goes without any deviation The two rays meet at I, so I is the image of O Let ∠AOP = α, ∠ AIP = β and ∠ ACP = γ From Snell’s law N X A i β ϒ O P M R C I Medium '1' Medium '2' Y sin i sin r or = 1µ2 = µ2 Fig B.9 µ1 µ1 sin i = µ sin r As angles i and r are small, we have sin i ! i, sin r ! r So µ1 i = µ2 r (1) Now in ∆ OAC , ∠ OAN is external angle of ∠OAC S Chand & Company Limited So i=α+γ (2) Similarly in ∆ ACI γ = r + β or r=γ–β (3) Putting the value of i and r from equation (2) and (3) in equatiion (1), we have µ1 (α + γ) = µ2 (γ – β) (4) For small aperture of the refracting surface, we have AM α ≈ tan α = OM β ≈ tan β = AM MI γ ≈ tan γ = AM MC Equation (4) becomes  AM AM   AM AM  µ1 + = µ2  OM MC   MC − MI  or µ1    OM + MC  = µ    MC − MI  As the aperture is small, M is very close to P so that OM = OP, MC = PC and MI = PI Therefore, µ1 or      OP + PC  = µ  PC − PI  µ1 + µ2 = µ – µ1 OP PI PC Now, OP = – u, PI = v and PC = R Therefore, S Chand & Company Limited B P F A' O A B' u v Fig B.10 But PO = AB Therefore, AB = FO A′B′ FA′ From equations (1) and (2) FO = (2) AO FA′ A′ O Using the new Cartesian sign convention AO = – u (3) A′ O = v FA′ = OA′ – OF = v – f and FO = f Equation (3) becomes S Chand & Company Limited f v– f = –u v vf = – uv + uf or uf – vf = uv Dividing by uvf 1 – = v u f This is the lens formula (equation) 101 Consider a convex lens of refractive index µ2 placed in a medium of refractive index µ1 (< µ2 ) [Fig B.11] Let O be a point object, lying on the principal axis of the lens I1 is the image of O formed by the spherical surface XP1 Y Using the relation for refraction at a spherical surface, we have – µ1 µ2 µ2 – µ1 + = u v1 R2 .(1) where u is the distance of the object from C, v1 is the distance of the image I1 from C and R1 is radius of curvature of the face XP1 Y Now we consider the refraction at face XP2 Y For this face, AB is the incident ray, which is refracted along BI Here I1 acts as a virtual object whose real image is formed at I In this case the ray is passing from a denser to a rarer medium So we have µ – µ1 µ1 µ = − v v1 R2 .(2) S Chand & Company Limited where R2 is the radius of curvature of the face XP2 Y and v is the distance of the final image from C Adding Eq (1) and (2) µ – µ1 µ1 – µ µ1 µ – = + v u R1 R2 or or or  1  1 µ  –  = (µ – µ )  –  u R2 v  R1   1  µ2  – = –1  –  v u µ R R 2     1 1  – = (µ – )  – , v u R2  R1 where µ = µ2 µ1 This is the lens maker’s Formula S Chand & Company Limited X A O B µ1 µ2 µ3 P1 C P2 Y I I1 Fig B.11 102 Linear magnification is defined as the ratio of the size ( y ′ ) of the image to the size (y) of the object : m= y′ y In the figure, AB is a linear object placed on the principal axis of a lens A′ B ′ is the image of AB formed by refraction through the lens S Chand & Company Limited B P y A' C A F y' B' v u Fig B.12 In similar ∆s ACB and A′CB ′ A′ B′ CA′ = AB CA Now, A′ B ′ = – y ′ AB = y CA′ = v CA = – u y′ v The above equation becomes – y = – u y′ v = or y u or m= v u S Chand & Company Limited 103 In the figure, a ray of light PQ strikes the refracting face AB of the prism at Q It is refracted along QR QR is incident on the face AC and is refracted along RS PQ is the incident ray, and RS is the emergent ray ∠ RTK is called the angle of deviation, δ ∠ NQP = ∠ TQO = i and ∠ RQO = r1 ∠ TQR = ∠ TQO – ∠ RQO = i – r1 Similarly ∠ TRQ = e – r2 δ = (i – r1) + (e – r2) = (i – e) – (r1 + r2) Now ∠ A + ∠ QOR = 180º In ∆ QRO r1 + r2 + ∠ QOR = 180º (1) (2) (3) From Eqs (2) and (3) ∠ A = r1 + r2 (4) Let the prism be set in the position of minimum deviation, so that δ = δm Then r1 = r2 = r (say) and i = e From Eq (1) δm = 2i – 2r From Eq (2) A = 2r ⇒ δm = i – A A + δm or i= sin i Now, µ= sin r S Chand & Company Limited A + δm µ= sin A/2 sin or 104 See Q 103 δ=i+e–A 105 Angular dispersion for a pair of colours is the difference in the deviations suffered by those two colours on passing through the prism If δV and δR are the deviations suffered by violet and red colours, then the angular dispersion for red and violet colours is ∆δ = δV – δR Let µR and µV be the refractive indices of the material of the prism for red and violet colours respectively Then for a small angled prism and for small angles of incidence, we have δV = (µV – 1) A and δR = (µR – 1) A, where A is the prism angle ⇒ ∆δ = δV – δR = A (µV – µR) 106 Emission spectrum Spectrum of light emitted by luminous bodies is called emission spectrum Such spectra are of the following three types : (i) Line spectrum A line spectrum consists of narrow bright lines separated by dark intervals Line spectra are emitted by substances in atomic state and are characteristic of the substances emitting them (ii) Band Spectrum A band spectrum consists of a number of bright bands separated by dark intervals Band spectra are emitted by substances in molecular state A band consists of a large number of close lines (iii) Continuous spectrum A continuous spectrum consists of an unbroken sequence of wavelengths (or colours) over a wide range Continuous spectra are emitted by solids, liquids and highly compressed gases heated to high temperature A continuous spectrum is not characteristic of the nature of the source but depends only on the temperature S Chand & Company Limited 107 A convex lens of short focal length is used as a simple microscope When a small object is placed between the optical centre and the focus of a convex lens, its virtual and enlarged image is formed The position of the object is so adjusted that the final image is formed at the least distance of distinct vision D [See Fig B 14] v f –v v = = 1– u f f Magnification m= Here v =–D So, m = 1+ D f B' B B" A' ' F α β F C A u Fig 6B.14 S Chand & Company Limited 108 The compound microscope consists of two lenses of short focal lengths The lens facing the object is called the objective and the other one, which is towards the eye, is called the eye piece The objective forms the image of an object placed just outside the focus This image is real, inverted and magnified It acts as an object for the eye - piece The adjustment is so done that the final image formed by the eyepiece is at the least distance of distinct vision EYE PIECE OBJECTIVE B Q α A" A ' FO ' FO C uo ' Fe f0 A′ β C' B' B" vo ƒe D Fig B.15 Let AB be an object placed just outside the focus Fo of objective A′ B ′ is the real image formed on the other side of the objective which lies in between the focus and the optical centre of the eyepiece Now eyepiece acts as a simple magnifier The distance between AB and the objective is so adjusted that a virtual and magnified image is formed at the least distance of distinct vision S Chand & Company Limited Magnifying power of a compound microscope is m= vo  D L  D 1 + ! 1 +  uo  fe  f o  fe  109 The astronomical telescope consists of two lenses The lens facing the eye is of short focal length and is called the eyepiece The other lens, which is near the object, is of large focal length and is called the objective The objective forms a real and inverted image of a distant object at its focus Now the position of the eye piece is so adjusted that the final image is formed at the least distance of distinct vision Fig B.16 S Chand & Company Limited magnifying power is given by m= fo fe  D 1 +  fe   fo and fe are the focal lengths of the objective and the eyepiece, respectively , and D is the least distance of distinct vision 110 The objective forms a real, inverted and diminished image of a distant object at its focus on the other side of the objective This image A′B′ acts as an object for the eyepiece The position of the eyepiece is so adjusted that A′B′ lies at the focus of the eyepiece (foci of objective and eyepiece coincide in this case) This image acts as an object for the eyepiece Fig B.17 S Chand & Company Limited The object being at the focus, the final, highly magnified image will be formed at infinity In this case the magnifying power is given by m= 111 112 113 114 fo fe fo and fe are the focal lengths of the objective and the eyepiece respectively (i) In laboratories magnifying glass is used to read vernier scales etc (ii) It is widely used by watch makers and jewellers Flint glass Due to absence of atmosphere on the moon, no scattering of light will take place and the sky will appear dark (a) For ray diagram of astronomical telescope see answer to Q 110 (b) (i) m = fo , so the magnifying power decreases when the focal length of the eyepiece is fe increased (ii) The magnifying power is not affected by increasing the aperture of the eyepiece 115 (a) For diagram of compound microscope see answer to Q 108 (b) The resolving power of a microscope is given by But λ= 2µ sin θ λ c ν S Chand & Company Limited 2µ ν sin θ c (i) The resolving power of a compound microscope will increase if the frequency of light used is increased (ii) The resolving power of a compound microscope will not be affected by increasing the focal length of the objective So, R.P = 116 Fig B.18 The main advantages of a reflecting type telescope over a refracting type telescope are as follows: (1) It is free of chromatic aberration (2) Spherical aberration gets removed by using parabolic mirror (3) The image is brighter (4) Since a mirror weighs much less than a lens of equivalent optical quality, mechanical support is much less of a problem S Chand & Company Limited 117 (i) Magnifying power m = tive fo Obviously, m does not depend on the diameter of the objecfe d 1.22 λ Where d is the diameter of objective Thus resolving power will increase with increase in the diameter of the objective (ii) Resolving power = 118  1  = (µ – 1)  –  f R2   R1 As µR < µV , so fR > fV Therefore, the focal length will increase 119 µ2 (i) µ1>µ2 µ1 (ii) µ1=µ2 µ1 µ2 (Fig B 19) S Chand & Company Limited (iii) µ1µ2 (ii) µ1=µ2 (iii) µ1