allen physics part 2

Chaoptnertents PHYSICS CBSE–XII www.notesdrive.com PART – II Syllabus Unit-VI 1-58 9. Ray Optics and Optical Instruments 59-76 10. Wave - Optics 77-104 Unit-VII 105-118 119-154 11. Dual Nature of Radiation & Matter Unit-VIII 12. Atoms 13. Nuclei Unit-IX 14. Semiconductor Electronics CBSE Sample Paper & Question Paper 15. CBSE Sample Question Paper (Term-I & II) 16. CBSE Paper 2021-2022 (Term-I & II)

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ALLENÒ CBSE 1 UNIT-VI : OPTICS CHAPTER- 9 : RAY OPTICS & OPTICAL INSTRUMENTS 1. REFLECTION OF LIGHT Reflection : When light travelling in a medium strikes a reflecting surface, it goes back into the same medium obeying certain laws. This phenomenon is known as reflection of light. Laws of reflection of light (i) The angle of incidence (i.e.Ði ) is equal to the angle of reflection (i.e. Ðr). node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx(ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray lie in the same plane. www.notesdrive.com Spherical Mirror : (A) Concave Mirror (Converging Mirror) (B) Convex Mirror (Diverging Mirror) Sign Convention : Relation between f and R : f = R/2 Mirror Formula : 1 = 1 + 1 f u v (Where, u ® Object Distance; v ® Image Distance; f® Focal Length) Linear Magnification : m = I = - v = f = f - v O u - f f u E

2 Physics ALLENÒ 2. REFRACTION OF LIGHT The phenomenon of bending of light, as it goes from one medium to another, is called refraction of light. 2.1 Laws of Refraction (at any refracting surface) : (i) Incident ray, refracted ray and normal always lie in the same plane. n n1 i n2 r www.notesdrive.com(ii) The product of refractive index and sine of angle of incidence at a point in a medium is constant. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx n1 sin i = n2 sin r (Snell's law) Snell's law : Sin i = 1 n2 = n2 = v1 = l1 Sin r n1 v2 l2 1n2 = Refractive index of medium 2 with respect to 1. Note : Frequency of light does not change during refraction . 2.2 Refraction through a parallel slab : Emergent ray is parallel to the incident ray, if medium is same on both sides. A AIR N GLASS i B r N' t 90° Cx i D Lateral shift x= t sin(i - r) ; t = thickness of slab cos r Note : Emergent ray will not be parallel to the incident ray if the medium on both the sides are different. E

ALLENÒ CBSE 3 2.3 Apparent depth of submerged object: (h¢ < h) µ1 > µ2 m2 m1 h h' x O For near normal incidence h¢ = m2 h m1 Note : h and h' are always measured from surface. 3. TOTAL INTERNAL REFLECTION (TIR) & CRITICAL ANGLE : TIR : When a ray of light enters from denser to rarer medium at an incident angle more than critical angle then the ray of light totally reflected back in the same medium. This phenomenon is known as TIR. NN N node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx I RARER I' www.notesdrive.com i C i>C DENSER O N' N' N' Conditions of TIR : (i) Ray is going from denser to rarer medium (ii) Angle of incidence should be greater than the critical angle (i > C) . Critical angle : It is the angle of incident at which ray of light totally reflected back into the same medium. Critical angle : (C ) = sin-1 mR = sin -1 vD = sin -1 l D mD vR lR Some illustrations of total internal Reflection Sparkling of diamond : The sparkling of diamond is due to total internal reflection inside it. As refractive index for diamond is 2.42 so C = 24.41°. Now the cutting of diamond are such that i > C. So TIR will take place again and again inside it. The light which beams out from a few places in some specific directions makes it sparkle. Optical Fibre : In it light through multiple total internal reflections is propagated along the axis of a glass fibre of radius of few microns in which index of refraction of core is greater than that of surroundings (cladding) m1 m1 > m2 light pipe E

4 Physics ALLENÒ • Mirage and looming : Mirage is caused by total internal reflection in deserts where due to heating of the earth, refractive index of air near the surface of earth becomes rarer than above it. Light from distant objects reaches the surface of earth with i > q C so that TIR will take place and we see the image of an object along with the object as shown in figure. cold air hot air hot surface www.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Similar to 'mirage' in deserts, in polar regions 'looming' takes place due to TIR. Here n decreases with height and so the image of an object is formed in air if (i > qC) as shown in figure. 4. REFRACTION AT SPHERICAL SURFACE : • n2 - n1 =n2 - n1 v u R n1 n2 +ve OP CI v, u & R are to be kept with sign as v = PI u = –PO R = PC (Note : Radius is with sign) 5. THIN LENS FORMULA : 1m 1 11 1 v-u= f • +ve Lens maker formula • 1 = (n - 1) æ 1 - 1 ö f ç R1 R2 ÷ è ø • magnification (m) = v u E

ALLENÒ CBSE 5 • Power of Lenses Reciprocal of focal length in meter is known as power of lens. • SI unit : dioptre (D) • Power of lens : P = 1 = 100 dioptre f(m) f(cm) Combination of Lenses (Two thin lens are placed in contact to each other) Power of combination. P = P1 + P2 Þ 1 = 1 + 1 f f1 f2 f1 f2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Use sign convention while solving numericals www.notesdrive.com 6. REFRACTION THROUGH PRISM P angle of deviation A dmax i r d i' r' dmin QR i=ig i=e e=ig angle of e=90° i=90° incidence • d = (i + i¢) – (r + r¢) • r + r¢ = A • There is one and only one angle of incidence for which the angle of deviation is minimum. When d = dm then i = i¢ & r = r¢ , the ray passes symetrically about the prism, & then n= sin é A + dm ù , where n = R.I. of glass w.r.t. surroundings. ë2û sin ëé A ùû 2 • For a thin prism ( A £10o) ; d=(n–1)A E

6 Physics ALLENÒ 7. DISPERSION OF LIGHT When white light is incident on a prism then it split into seven colours. This phenomenon is known as dispersion. speWchtirtue lmightwww.notesdrive.comWhite light beam node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Glass Prism Angle of Dispersion : Angle between the rays of the extreme colours in the refracted (dispersed) light is called Angle of Dispersion. q = dv – dr 8. OPTICAL INSTRUMENTS 8.1 Simple microscope : A simple microscope is a converging lens of small focal length. Eye focussed on near point u D (a) h q0 D (b) Magnifying power when image is formed at D : MP = 1+ D f When image is formed at infinity : MP = D f E

ALLENÒ CBSE 7 8.2 Compound microscope : MP = - v0 æ D ö u0 ç ue ÷ è ø u f fe A FBe 'B' Eyepiece h b hh' ' B\" B b E O A' Objective ue D node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxvDe = –D www.notesdrive.com A\" Ray diagram for the formation of image by a compound microscope (Final image is at D) Magnifying power when final image is formed at D, MP = - v0 æç1 + D ö u0 è fe ÷ ø Tube length (L ) = v 0 + u e When final image is formed at infinity MP = - v0 ´ D u0 fe Tube length (L) = v0 + fe 8.3 Astronomical Telescope (Refracting telescope) : MP = - f0 ue (i) When final image is formed at D Lo(Objective lens) f0 =V0 ue Le(eyepiece) fe MP = - f0 èæç1+ fe ö Fo Fe F0 fe D ø÷ B\" o1 B' o2 Fe A' Tube length (L) = f0 + |ue| A\" Ve=D E

8 Physics ALLENÒ (ii) When final image is at infinity fe Eyepiece f0 Objective MP = - f0 fe a a B' F0,Fe A' b Tube length ( L) = f0 + fe (Normal adjustment) Image at ¥ 8.4 Cassegrain Telescope (Reflecting telescope): www.notesdrive.com Telescope with mirror objectives are called reflecting telescope. Spherical aberration and and node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx chromatic aberration are removed by using reflecting telescope. Secondary Objective mirror mirror Eyepiece Schematic diagram of a reflecting telescope (Cassegrain). 9. IMPORTANT POINTS · A fish in water at depth h sees the whole outside world in horizontal circle of radius. r = h tanqC= h m2 -1 · A convex lens behaves as a concave lens when placed in a medium of refractive index greater than the refractive index of its material. Similarly a concave lens behaves as a convex lens when placed in a medium of refractive index greater than the refractive index of concave lens. · If lower half of a lens is covered with a black paper, the full image of the object is formed because every portion of lens forms the full image of the object but sharpness of image decrease. E

ALLENÒ CBSE 9 CHAPTER-10 : WAVE OPTICS 1. HUYGEN'S WAVE THEORY According to Huygens principle \"each point on a wave front is a source of new disturbance, called secondary wavelets. These wavelets are spherical and travel with speed of light in that medium\". Wavefront : The locus of all the particles of the medium vibrating in the same phase at a given instant is called a wavefront. • The forward envelope of the secondary wavelets at any instant gives the new wavefront. • In homogeneous medium, the wave front is always perpendicular to the direction of wave propagation. Plane wavefront Spherical wavefront AB secondary wave node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Primary www.notesdrive.com source Secondary source A' B' Types of wavefront: The type (shape) of wavefront depends upon the shape of the light source from the wavefront originates. On this basis there are three types of wavefronts. (i) Spherical (ii) Cylindrical (iii) Plane Comparative study of three types of wavefront S.No. Wavefront Shape Diagram of Variation of Variation of Intensity (I) of light shape of amplitude (A) with distance source wavefront with distance 1. Spherical Point source A µ 1 or A µ 1 I µ 1 d r r2 2. Cylindrical Linear source or slit O Aµ 1 or A µ 1 I µ 1 3. Plane O' d r r Extended large source or point A = constant I = constant source situated at very large distance E

10 Physics ALLENÒ • Refraction of a plane wave using Huygens principle : Incident wavefront A' B v1t v1 i r CC Medium1 i Refracted A v2t wavefront Medium2 r v2 < v1 E • Reflection of a plane wave using Huygens principle : www.notesdrive.com E Incident i node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxwavefront A B Reflected M wavefront rC N 2. COHERENT SOURCES Two sources will be coherent if and only if they produce waves of same frequency (and hence wavelength) and have a constant phase difference w.r.t. time. (example laser light) Incoherent sources : Two sources are said to be incoherent if they have different frequency and initial phase difference is not constant w.r.t. time. 3. INTERFERENCE When two light waves of same frequency with constant phase difference superimpose over each other, then the resultant amplitude (or intensity) in the region of superimposition is different from the amplitude (or intensity) of individual waves. This modification in intensity in the region of superposition is called interference. Redistribution of energy take place during interference. The law of conservation of energy holds good in the phenomenon of interference. • Resultant intensity (coherent sources) I = I1 + I2 + 2 I1I2 co s f 0 • Intensity µ width of slit µ (amplitude)2 ((ÞI1= W1 = a12 Þ Imax = I1 + )I22 = æ a1 + a2 2 I2 W2 a22 I m in I- 2 ç a1 - a2 )I2 è ö 1 ÷ ø E

ALLENÒ CBSE 11 Types of Interference : (i) Constructive Interference When both waves are in same phase then phase difference is an even multiple of p Þ f = 2 np; n = 0,1,2 ... l • Path difference is an even multiple of 2 Q f = d Þ 2np =d Þ d = 2n ælö Þ d = nl (where n = 0,1,2...) 2p l 2p èç 2÷ø l • When time difference is an even multiple of T \\ Dt = 2n æTö 2 èç 2 ø÷ • In this condition the resultant amplitude and intensity will be maximum. Amax = (a1 + a2) Þ Imax = I1 + I2 + 2 I1 I2 = ( I1 + I2 )2 (ii) Destructive Interference When both the waves are in opposite phase. So phase difference is an odd multiple of p . f = (2n–1) p ; n = 1, 2 ... node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx • When path difference is an odd multiple of l , d = ( 2n –1) l , n = 1, 2 ... www.notesdrive.com 2 2 • When time difference is an odd multiple of T , D t = (2n–1) T , ( n=1,2...) 2 2 In this condition the resultant amplitude and intensity of wave will be minimum. Amin = (a1 – a2) Þ Imin = ( I1 - I2 )2 3.1. Young's Double Slit Experiment (YDSE) : P G ®nth fringe O x P D S1 xn S2 d OZy z S S2 D (a) G' Young’s arrangement to produce interference pattern. During expriment Young observed alternative dark and bright fringes (bands) on the screen. S1 dq S2 dsinq • Distance of nth bright fringe from centre of screen xn = nlD d Path difference = nl where n =0, 1, 2, 3, ..... E

12 Physics ALLENÒ • Distance of nth dark fringe xn = (2n - 1)lD 2d Path difference=(2n–1) l where n = 1,2, 3,..... 2 • Fringe width b = lD d • Angular fringe width b = l D d 4. DIFFRACTION Bending of light rays from sharp edges of an opaque obstacle or aperture and its spreading in the www.notesdrive.com geometrical shadow region is defined node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx as diffraction of light or deviation of light from its rectilinear propogation tendency is defined as diffraction of diffraction from obstacle diffraction from aperture light. Fraunhofer's diffraction : The source and screen are placed at large distances from the aperture or the obstacle and converging lens is used to observed the diffraction pattern. The incident wavefront is planar one. To P Lq From S M1 q To C M Q M2 M2 N q The geometry of path differences for diffraction by a single slit. Difraction phenomenon define the limit of ray optics • For minima : a sinqn = nl • For maxima : a sinqn = (2n + 1) l 2 • Linear width of central maxima : Wx = 2lD a • Angular width of central maxima Wq = 2l a E

ALLENÒ CBSE 13 NCERT IMPORTANT QUESTIONS CHAPTER-9 RAY OPTICS & OPTICAL INSTRUMENTS 1. The focal lengths of an objective lens and eyepiece are 192 cm and 8 cm respectively in a small telescope. Calculate it’s magnifying power and the separation between the two lenses. Sol. fo = 192 cm and fe = 8 cm Magnifying power M= f0 = 192 Þ M = 24 fe 8 Distance between two lenses - L = fo + fe node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx L = 192 + 8 www.notesdrive.com L = 200 cm 2. A double convex lens is made of a glass of refractive index 1·55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm. Sol From 1 = (µ21 - 1) é 1 - 1ù f ê R1 ú ë R 2 û 1 = (1.55 -1) é 1 - æ - 1 öù QR1 = R2 = R f ê R çè R ÷øûú ë 1 = (.55)ëéêR2 ùúû = 1.1 f R 1 = 1.1 Þ R = 1.1 ´ 20 = 22cm 20 R 3. A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid ? Could the liquid be water ? Sol. The refractive index of the liquid must be equal to refractive index of the lens for lens disappear in a liquid. i.e. n1 = n2 = 1.47 so refractive index of the liquid is 1.47. Then n = n2 =1 n1 Now from lens maker’s formula 1 = (n - 1) æ 1 1 ö f ç R1 - R2 ÷ è ø 1 = (1 - 1) æ 1 - 1 ö Þ 1 =0 then f = ¥ f ç R1 R2 ÷ f è ø So the lens in the liquid will act like a plane sheet of glass. Since refractive index of water is 1.33. So given liquid is not water. E

14 Physics ALLENÒ 4. Find the position of the image formed by the lens combination given in the Fig. Sol. Lens formula for first lens 111 1 1 1 v1 - u1 = f1 Þ v1 - (-30) = 10 www.notesdrive.com 1 = 1 - 1 = 3-1 = 2 = 1 v1 10 30 30 30 15 v1 = 15cm. The image formed by the first lens serves as the object for the second lens at distance 15 - 5 = 10 cm to the right of the second lens. lens formula for second lens. 1 1 1 11 1 Þ 1 = -110 + 1 = 0 v2 - u2 = f2 Þ v2 -10 = (-10) v2 10 v2 = ¥ cm This virtual image is formed at an ¥ distance to the left of the second lens. This acts as an object for the third lens. at distance ¥ + 10 = ¥ cm to the left of the third lens. 1 1 1 11 1 1 -0 = 1 v3 - u3 = f3 Þ v3 - ¥ = (30) Þ v3 30 v3 = 30 cm Thus the final image is formed 30 cm to right of the third lens. 5. A ray PQ incident on the refracting face BA is refracted in the prism BAC as shown in the figure and emerges from the refracting face AC as RS such that AQ = AR. If the angle of prism A = 60° node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx and refractive index of material of prism is 3 , calculate angle q. A || 60° || Q Rq P S BC E

ALLENÒ CBSE 15 Sol. As AQ = AR, ray QR | | BC This is the condition of min deviation. From- sin æ A + dm ö sin æ 60o + dm ö èç 2 ÷ø ç 2 ÷ m = Þ 3 = è ø 12 sin æ A ö çè 2 ÷ø 3 æ 60 o + d ö 2 ç 2 ÷ = sin è m ø or sin 60o = sin æ 60o + dm ö ç 2 ÷ è ø 60o = 60o + dm Þ dm = 60o 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx 6. Draw a ray diagram to show the image formation by a combination of two thin convex lenses in www.notesdrive.comcontact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses. f1 f2 O I I1 v AB Sol. u u v1 Object is placed at point O, whose image is formed at I1 by first lens. As image I1 is real, it works as a virtual object for second lens B, producing the final image at I. Since the lenses are thin, assume the optical centres to be coincident. Let this central point is P. For the image formed by the lens A, 111 ....(1) v1 - u = f1 Similarly, for the lens B, 1 - 1 = 1 ....(2) v v1 f2 Add (1) & (2) we get, 11 1 1 v - u = f1 + f2 If two lens system is taken as equivalent to a single lens of focal length f, we get, P = 1 = 1 + 1 ....(3) f f1 f2 In term of power, equation (3) can be written as, P = P1 + P2 where P is the net power of this combination. E

16 Physics ALLENÒ 7. Draw a labelled ray diagram of a refractive telescope. Deduce an expression of magnifying power of it.Write two main limitations of a refracting type telescope over a reflecting type telescope. Sol. Refracting telescope : Lo(Objective lens) f0 =V0 ue Le(eyepiece) fe Fo Fe F0 B\" o1 B' o2 Fe A' A\" Ve=D Refracting type telescope consists of an objective lens of large aperture and large focal length whereas eyepiece is of small aperture and small focal length. Magnifying Power : It is the ratio of visual angle subtended by final image at eye to the visual angle subtended by an object. www.notesdrive.com M= b ìif a and b are verysmall a node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxîía » tanaand b » tanb æ A'B' ö =èç O2B' ÷ M =ttaann b Þ M æ A'B' ø Þ M =OO12BB'' Þ M = f0 ...(i) a ö ue ç ÷ è O1B ' ø (i) When final image is formed at least distance of disinct vision (Ve = D) - Applying lens formula for eyepiece : 1 = 1 - 1 ìapplying sign convention fe ve ue îíve = -D, ue = -ue 1 11 1 11 fe = - D + ue Þ ue = fe + D Hence magnification power from eq. (i) M = f0 é1 + 1ù Þ fo èæç1 + fe ö ê ú fe D ø÷ ë fe D û (ii) When final image is formed at infinity (ue = fe) From eq. (i) & M = f0 fe Drawbacks of refracting telescope : (1) Defect of chromatic aberration occurs in refracting type telescope. (2) It has small resolving power. E

ALLENÒ CBSE 17 8. Describe the construction of a compound microscope. Derive an expression for its total magnification.Draw a ray diagram for the formation of image by a compound microscope. Sol. Compound microscope : Figure shows ray diagram of a compound microscope. It consists of two convex lenses one nearer to object is known as objective and other close the eye is eyepiece lens. Here objective lens is of small focal length (f0) and small aperture whereas eyepiece is also a small focal length but larger than objective lens and relatively large aperture. L Le A Lo vo B' fe Fe Object Fe ue o1 B F0uBfoo\" Fo O2 A' node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx A\" Ve=D www.notesdrive.com A BD The image A' B' formed by the objective lens L0 of object AB and Final image is A\"B\". Magnification power (M) = Angle subtended by Angle subtended by final image at eye distinct vision the object when it is placed at the least distanceof M= b a ìif a and b are very small, then íîa»tana and b » tanb M = tan b tan a æ A\"B\" ö èç ÷ø M = æ D Þ M = A\"B\" AB ö AB èç D ÷ø M = A\"B\" ´ A'B' Þ M = m e ´ m o ìïïm0 = v0 A'B' AB í = u0 îïïme ve ue M = vo ´ ve ...(1) uo ue E

18 Physics ALLENÒ Applying lens formula for eyepiece - 1 = 1 - 1 îíìvuee Þ -D fe ve ue Þ -ue 1 = - 1 + 1 fe D ue D = -1+ D fe ue D =1+ D ....(2) ue fe (i) When final image is formed at least distance of distinct vision (ve=D) Mwww.notesdrive.com=vo´D uo ue node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx M = vo æ 1 + D ö ....(3) uo ç fe ÷ è ø (ii) When final image is formed at infinity (ue=fe) M = vo ´ D ....(4) uo fe 9. Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point source lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence derive lens maker’s formula. Sol. Lens maker’s formula, 1 = (n - 1) é 1 - 1 ù f ê R1 R2 ú ë û Assumptions – (a) Lens is considered as a thin lens. (b) Object is a point object which is situated on the principal axis. (c) Aperture of the lens is small. (d) Incident & refracted ray makes small angle with pricipal axis. Consider a thin convex lens of absolute refractive index n2 placed in rarer medium of absolute refractive index n1. Also R1 & R2 are the radii of curvature of surfaces XP1Y & XP2Y resp. X O P1 C P2 I I¢ v u Y v¢ E

ALLENÒ CBSE 19 For refraction at surface XP1Y : ‘O’ is the object & I’ is its real image. using formula, n2 - n1 = n2 - n1 we get, v u R n2 - n1 = n2 - n1 ....(1) v' u R1 For refraction at surface XP2Y : I’ is the virtual object & I is its real image (final image). Using formula n1 - n2 = n1 - n2 v u R n1 - n2 = n1 - n2 ....(2) v v' R2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Adding equations (1) & (2), we get www.notesdrive.com n1 æ 1 - 1 ö = (n2 - n1) é1 - 1 ù èç v u ÷ø ê R2 ú ë R1 û 1 - 1 = n2 - n1 é1 - 1 ù v u n1 ê R2 ú ë R 1 û or 1 = 1 - 1 = æ n2 - 1 ö é 1 - 1ù f v u ç n1 ÷ ê R1 ú è ø ë R2 û or 1 = (n - 1) é 1 - 1 ù Q n2 =1n 2 =n f ê R1 R2 ú n1 ë û 10. Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism. Sol. angle of AT deviation dmax i1 Q M R i2 dmin d d1 dr22 r1 angle of PN S incidence i=ig i=e e=ig BC e=90° i=90° Let PQ & RS are incident & emergent rays. Let incident ray gets deviated by (d) in prism, i.e. ÐTMS = d d1 & d2 are deviation produced at surfaces AB & AC respectively. E

20 Physics ALLENÒ \\ d = d1 + d2 .....(1) d = (i1 – r1) + (i2 – r2) d = (i1 + i2) – (r1 + r2) QN, RN are normals In quadrilateral AQNR, A + ÐQNR = 180o Also, in DQNR, ÐQNR + r1 + r2 = 180o Þ A = r1 + r2 .....(2) www.notesdrive.com From eq (1) & (2) , we get Þ d = (i1 + i2) – A .....(3) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Angle of deviation produced by prism varies with angle of incidence. When prism is adjusted at angle of minimum deviation, then i1 = i2 = i (suppose) at d =dm Þ r1 = r2 = r (suppose) From (1) & (2) we have, dm = 2i – 2r & 2r = A Þ i= A+dm 2 r = A/2 \\ Refractive index of material of prism is. sin æ A + dm ö çè 2 ø÷ m = sin i = sin r æ A ö sin çè 2 ÷ø E

ALLENÒ CBSE 21 NCERT IMPORTANT QUESTIONS CHAPTER-10 WAVE OPTICS 1. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. Sol. Distance of fourth bright fringe from centre, Xn = nlD Þ X4 = 4lD d d 1.2 ´ 10-2 = 4´l´1.4 0.28´10-3 0.28´10-3 5.6 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx1.2´10-2´=lÞl=600nm www.notesdrive.com2. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. Sol. Angular fringe width q= l d. In water angular fringe width q' = l' \\ l' = l d m q' = l = q Þ q' = 0.2 = 0.15º md m 43 3. A slit 4.0 cm wide is irradiated with microwave of wavelength 2.0 cm. Find the angular spread of central maximum, assuming incidence normal to the plane of the slit. Sol. Here a = 4.0 cm, = 4.0 x 10-2 m. l = 2.0 cm = 2.0 x 10-2m. Angular speed of central maximum (2q) is 2q = 2l = 2 ´ 2.0 ´10-2 = 1 rad. a 4.0´10-2 4. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10–2m towards the slits, the change in fringewidth is 3 × 10–5m. If the distance between the slits is 10–3 m, calculate the wave length of the lightused. Sol. d= 10–3 m; DD = 5 × 10–2m; Db = 3 × 10–5m. If D is decreased by DD, the fringe width will also decrease by Db, such that Db = DDl d or l= Dbd = 3´10–5 ´10–3 = 6´10-7m DD 5´10–2 E

22 Physics ALLENÒ 5. In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength l’. Find the value of l’ . Sol. The angular position of first minimum for red light of wavelength lr is given by sin q1 = lr ………………..(i) a The angular position of first maximum for light of wavelength l’ is given by sin q1' = 3l ' ……………..(ii) 2a As the first minimum for red light coincides with the first maximum for wavelength l’, sinq1 = sinq'1 Therefore, from (i) and (ii), we have 3l ' = lr orl ' = 2lr 2a a 3 2 ´ 660 3 l’ = www.notesdrive.com = 440 nm 6. Define wavefront. Use Huygens’ principle to verify the laws of refraction. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Sol. Wavefront : It is defined as the locus of all the points vibrating with zero or constant phase difference. S' S N Rarer X i B Denser X i v1t C A r v2t r D T' T Proof of SNELL'S Law : Consider a plane wavefront (AB) incident on the surface XY, separating two media. Let the secondary wavelets from point (B) reach upto point (C) in time (t). So draw an arc of length (v2t) from point A to locate the position of refracted WF. Now we draw a tangent (CD) on this arc where CD represents refracted WF. Clearly incident ray, refracted ray & the normal are respectively ^ to incident WF, refracted WF & surface XY. Also, sin i = BC = v1t ....(i) AC AC & sin r =AACD =Av2Ct ....(ii) From (1) & (2), v1 = sin i v2 sin r or m2 = sin i m1 sin r SNELL 'S Law E

ALLENÒ CBSE 23 7. Two harmonic waves of monochromatic light y1 = a coswt and y2 = acos(wt + f) are superimposed on each other. Show that maximum intensity in interference pattern is four times the intensity due to each slit. Hence write the conditions for constructive and destructive interference in terms of the phase angle f. Sol. Given, y1 = a coswt, y2 = acos(wt + f) (a) Resultant displacement is given as : y = y1 + y2 = acoswt + acos(wt + f) = acoswt + acoswtcosf – asinwtsinf = acoswt(1 + cosf) – a sinwt sinf Put Rcosq = a(1 + cosf) .....(i) Rsinq = asinf ....(ii) By squaring & adding Eqs. (i) & (ii), we get node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxR2 = a2(1+cos2f + 2cosf) + a2sin2f = 2a2(1 + cosf) = 4a2cos2f/2 www.notesdrive.com ff \\ I = R2 = 4a2cos2 2 = 4I0cos2 2 (b) For constructive interference, ff cos 2 = ±1 or 2 = np or f = 2np For destructive interference, ff p cos 2 = 0 or 2 = (2n + 1) 2 or f = (2n + 1)p 8. Write any two necessary conditions for interference of light. Obtain and expression for fringe width in Young’s double slit experiment. Draw curve for intensity distribution in Young’s doble slit experiment. Sol. Conditions for interference of light : (1) Both sources of light must be coherent. (2) Both sources of light should be monochromatic. Expression for fringe width in young's double slit experiment : Path difference between the light waves reaching at point P from S1 and S2 – Dx = S2P – S1P from DS2PN, S2P2 = D2 + æ yn + d ö2 ....(1) P (yn– d ) yn d èç 2 ÷ø 2 2 S1 M (yn+ ) d O Similarly in DS1PM d/2 N S2 d/2 d ö2 S1P2 = D2 + æ y n - 2 ø÷ ....(2) èç D subtract eq. (2) from (1) Screen S2 P2 - S1P2 = æ yn + d ö2 - æ yn - d ö2 çè 2 ø÷ çè 2 ø÷ (S2P + S1P) (S2P – S1P) = y2n d2 + ynd - yn2 d2 + ynd +4 -4 E

24 Physics ALLENÒ (S2P + S1P) (S2P – S1P) = 2 ynd If point P and O are very close to each other, then (S1P » S2P » D) 2D(S2P – S1P) = 2 ynd S2P – S1P = ynd D Path difference ( Dx) = ynd ....(3) D Case (i) Position of bright fringes : If nth bright fringe occurs at point P, then path difference - Dx = nl yn d = www.notesdrive.comnlÞyn=nlDwhere n = 0, 1, 2, 3, ..... D d node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx yn Þ position of nth bright fringe on screen from central bright fringe. Case (ii) Position of dark fringes : Path difference for nth dark fringe - Dx = (2n – 1)l/2 y 'n d = (2n -1)l / 2 D y 'n (2n -1)lD where n = 1, 2, 3, .... = 2d y'n Þ position of nth dark fringe. Fringe width : The distance between two consecutive bright or dark fringes is called fringe width. fringe width (b) = yn+1 – yn [Taking condition for brigt fringe yn = nlD ] d b = (n + 1)lD - nlD d d b= lD d Intensity distribution curve : Intensity (I) -5p -4p -3p -2p -p O p 2p 3p 4p 5p Phase difference E

ALLENÒ CBSE 25 9. What is meant by diffraction of light ? Draw a graph to show the relative intensity distribution for a single slit diffraction pattern. Obtain an expression for the first minima of diffraction. Sol. Diffraction of light– The phenomenon of bending of light waves around corners or edges of aperture or obstacle is called diffraction of light. For diffraction of light waves, the size of aperture or obstacle is comparable to wavelength of light waves. a»l a Þ size of aperture or obstacle Single slit diffraction : L1 L2 P1 S Aq P qN y1 q node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxaCM O www.notesdrive.com q B P f D If lens L2 is very close to slit AB then (Df) Light rays from source S incident on lens L1 it becomes parallel and plane wavefront is obtained. When plane wavefront is incident on single slit AB, after diffraction from AB, diffraction pattern will be obtained on screen. Position of Ist minima– If the path difference between extreme rays from AB is l at point P on the screen, then destructive interference will take place & first minima will be obtained. Path difference, BM = a sin q a sin q = l If q is very small, then (sin q » q) aq = l Þ q = l a Where, q Þ angular position of first minima Intensity distribution graph :- In Fraunhofer diffraction intensity is given by I = é (2n 2 1) p ù2 I0 ê + ú ëê úû (I) I0 (Ist bright) I0/22 (2nd bright) I0/22 I0/61 I0/61 –3l –5l –2al –3l –la l 3l 2l 5l 3l a 2a 2a a 2a a 2a a E

26 Physics ALLENÒ 10. Define a wavefront. Using Huygen’s principle, verify the laws of reflection at a plane surface. Sol. The wave front is defined as a surface of constant phase. (Alternatively : The wavefront is the locus of all points that are in the same phase). Let speed of the wave in the medium be 'n' Let the time taken by the wave front, to advance from point B to point C is 't' Hence BC = nt Let CE represent the reflected wave front, Distance AE = nt = BC D AEC and D ABC are congruent \\ ÐABC = ÐAEC(90º ) AC = AC AE = BC, So, ÐBAC = ÐECA (by C.P.C.T.) Þ Ði =Ðr (proved) www.notesdrive.com CHAPTER-9 : RAY OPTICS & OPTICAL INSTRUMENTS node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx PREVIOUS YEARS QUESTION EXERCISE-I ONE MARK QUESTIONS : 1. How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason. Sol. d = A (µ – 1) Q µ =c , i.e. refractive index (µ) is less for red light as compare to violet. So (dm) will be l decreased with red light. 2. A concave mirror and convex lens are held in water. What change, if any, do you expect to find in the focal length of the either ? Ans. No change in focal length of mirror, but focal length of convex lens will increase. 3. A bi-convex lens with both faces of same radius of curvature, is to be manufactured from a glass of refractive index 1.55. What should be the radius of curvature for the focal length of the lens to be 20cm ? Ans. using lens makers formula, it is 22 cm. E

ALLENÒ CBSE 27 4. The length of an astronomical telescope is 16cm and its magnifying power is 3. The focal length of the lenses will be- (A) 4cm, 12cm (B) 4cm,8cm (C) 4cm,2cm (D) 8cm,4cm Ans,. (A) 5. Calculate the refractive index of the material of an equilateral prism for which the angle of min. deviation is 60°. Ans. 3 = 1.732 EXERCISE-II TWO MARK QUESTIONS : 1. State, with the help of a ray diagram, the working principle of optical fibres. Write one important use of optical fibres. Sol. Working principle of optical fibers: Optical fibres are extensively used for transmitting audio and video signals in the form of light from one end to other on the basis of multiple total internal reflection at the interface of inner and outer layer which are known as core and cladding. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Light Cladding (low µ) www.notesdrive.com Core(high µ) Use: Used in visual examination of internal organs like stomach and intestine by the method of endoscopy in medical science. 2. Under what conditions does the phenomenon of total internal reflection take place ? Draw a ray diagram showing how a ray of light deviates by 90° after passing through a right-angled isosceles prism. Sol. Conditions - (a) Light must pass from denser to rarer medium. (b) Incidence angle must be greater then critical angle. 45° 45° 45° E

28 Physics ALLENÒ 3. A beam of light converges at a point P. Draw ray diagrams to show where the beam will converge if (i) a convex lens, and (ii) a concave lens is kept in the path of the beam. Sol. (i) P (ii) P 4. Under which conditions can a rainbow be observed? Distinguish between a primary and a secondary rainbow. Sol. An observer can observe rainbow only when his back is towards the sun. Primary Secondary www.notesdrive.com More brighter Less brighter node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx More intensity Less intensity Formed due to two Formed due to two refractions & one TIR refractions & two TIR 5. A ray PQ incident normally on the refracting face BA is refracted in the prism BCA made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge ? Justify your answer. A P Q B 60° C Sol. Refractive index (m) = 1.5 (given) Ray PQ will go straight in DBAC as it falls normally on the surface BA. From condition of TIR siniC = 1 = 1 = 2 = 0.66 A m 1.5 3 P 30o or iC = 42o (Q sin 42° ; 0.66) Q sin i = m2 = 2 30o N sin r m1 3 48.5° or sin30° = 2 60° C sin r 3 B or sin r = 0.75 or r = 48.5° (Qsin 48.5° ; 0.75 ) So ray PQ will emerge from surface AC at an angle = 48.5° E

ALLENÒ CBSE 29 EXERCISE-III THREE MARK QUESTIONS : 1. (a) A convex lens of focal length 30 cm is in contact with concave lens of focal length 20 cm. Find out if the system is converging or diverging. (b) Obtain the expression for the angle of incidence of a ray of light, which is incident on the face of a prism of refracting angle A so that it suffers total internal reflection at the other face. (Given, the refractive index of the glass of the prism is µ.). Sol. (a) Focal length of convex lens f1 = 30 cm focal length of concave lens f2 = –20 cm Equivalent focal length of combination node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx 11 1 www.notesdrive.comf = f1 + f2 1 =310 - 1 f 20 1 =- 1 Þ f = -60 cm f 60 The system is diverging. (b) For total internal reflection, sin ic =1 A m For prism we know that QR r1 + r2 = A and r2 = ic i r1 r2 90° r1 =A - ic P Applying snell's law at point 'Q' BC sini =m sin r1 sin i = µ sin r1 sin i = µ sin (A – ic) i =sin -1[µ sin(A - ic )] E

30 Physics ALLENÒ 2. A triangular prism of refracting angle 60° is made of a transparent material of refractive index 2 / 3 . A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation. PK 60° Q L M Sol. sin(60°) 1 K sin(e) www.notesdrive.com=2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx360° 30° 33 60° N 2sine = 2 e = 90° e60° Emergent angle, e = 90° d LM Deviation, d = 30° 3. (a)Draw a schematic diagram of a reflecting telescope (Refer to theory topic 8.4) (b) State the advantages of reflecting telescope over refracting telescope. Sol. (b) ¨ Reflecting telescope does not suffer from chromatic aberration. ¨ Intensity of light is more in case of reflecting telescope. ¨ Reflecting telescope’s mirrors are easier to mount. ¨ Manufacturing of mirror is cheaper compared to lens. 4 . (a) Draw a ray diagram depicting the formation of the image by an astronomical telescope in normal adjustment. (b) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope ? Give reason. Lenses Power (D) Aperture (cm) L1 38 L2 61 L3 10 1 E

ALLENÒ CBSE 31 f0 fe Eyepiece Objective Sol. (a) a a B' F0,Fe A' b (Normal adjustment) Image at ¥ (b)node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx'L1' used for an objective as it has largest focal length which is needed, Mq =f0for better fe www.notesdrive.com magnifying power. 'L3' used for an eyepiece to get better angular magnification, as from Mq = f0 , least fe is fe needed. 5. A symmetric biconvex lens of radius of curvature R and made of glass of refractive index 1.5, is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with it's tip on the principal axis of the lens is moved along the axis until it's real, inverted image coincides with the needle itself. The distance of the needle itself. The distance of the needle from the lens is measured to be x. On removing the liquid layer and repeating the experiment, the distance is found to be y. Obtain the expression for the refractive index of the liquid in terms of x and y. ///////////////////////////// Sol. Let m1 denote the refractive index of the liquid. When the image of the needle coincides with the needle itself ; its distance from the lens, equals the relevant focal length. With liquid layer present, the given set up, is equivalent to a combination of the given (convex) lens and a concavo plane / plano concave ' liquid 'lens'. E

32 Physics ALLENÒ We have 1 = (m -1)çæ 1 - 1 ö f R1 R2 ÷ è ø 11 1 and f = f1 + f2 Þ 111 x = fl + y as per the given data, we then have 1 1 é 1 æ 1 öù 1 ì1 = 1 = 1 ü f2 y êë R èç R ø÷úû R ï R f2 ï = = (1.5 -1) - - = Þ í y ý îï f = x þï \\ 1 = www.notesdrive.com(m1-1) æèç -1ö+1= -m1 + 2 x R ÷ø y y y node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx \\ m1 = 2 - 1 = æ 2x - y ö or m1 = æ 2x - yö y y x ç xy ÷ èç x ÷ø è ø 6. (i) A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used. (ii) A converging lens is kept coaxially in contact with a diverging lens both the lenses being of equal focal length. What is the focal length of the combination ? Sol. (i) For first position of the lens, we have 1 = 1 - 1 ...(i) f y (-x) screen object L1 20 cm L1 x(cm) 100 cm y(cm) ...(ii) For second position of the lens, we have 1 = y 1 - [-(x 1 20)] f - 20 + E

ALLENÒ CBSE 33 From (i) and (ii) 1 + 1 = (y 1 + (x 1 y x - 20) + 20) x+y = (x + 20) + (y - 20) xy (y - 20)(x + 20) \\ xy = (y – 20) (x + 20) = xy – 20 x + 20 y – 400 \\ x – y = –20 Also, x + y = 100 \\ x = 40 cm and y = 60 cm node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx \\ 1 11 2+3 = 5 www.notesdrive.comf=60 - -40 =120 120 \\ f = 24 cm (OR) (i) Distance between the image (screen) & the object, D = 100 cm. Distance between two location of the convex lens, d = 20cm Focal length (f) = D2 - d2 = (100)2 -(20)2 4D 4´100 f = 24 cm (ii) 11 1 11 f = f1 + (-f2) = f1 - f2 1 =f2f2-f1f1 f1 f2 f f =f f2 f1 2 - f1 7. Three rays (1, 2, 3) of different colours fall normally on one of the sides of an isosceles right angled prism as shown. The refractive index of prism for these rays is 1.39, 1.47 and 1.52 respectively. Find which of these rays get internally reflected and which get only refracted from AC. Trace the paths of rays. Justify your answer with the help of necessary calculations. A Red (1) 45° Green(2) 45° Blue (3) BC E

34 Physics ALLENÒ Sol. sin i = gµa A sin r sin 45° = 1 Red 45° sin r a µg Green B 45° C 1.39 Blue sin r = 2 = 1.39 × 0.707 = 0.98273 By geometry, angle of incidence (i) of all three rays is 45o. For total internal reflection angle of incidence > critical angle. i > iC sini > siniC www.notesdrive.com or sin45o > siniC or 11 or 2 <m node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxsin 45o < sin iC Hence, TIR (total internal reflection) takes place on AC for green & blue light whereas red undergoes refraction. ( )8. (i) For a glass prism µ = 3 the angle of minimum deviation is equal to the angle of the prism. Calculate the angle of the prism. (ii) Draw ray diagram when incident ray falls normally on one of the two equal sides of a right angled isosceles prism having refractive index µ = 3 . Sol. (i) Given, dm = A sin æ A + dm ö sin æ A+A ö èç 2 ø÷ çè 2 ÷ø from m = Þ 3 = = sin A sin (A 2) 1 sin (A 2) sin A 2 Þ 3 =2sinsAin2Aco2sA 2 Þ 3 = cosA/2 2 Þ cos30o = cosA/2 Þ 30o = A/2 A =60 o (ii) Given m = 3 (prism) For total internal reflection, angle of incidence (i) > critical angle (ic) i.e. i > iC A N 45o or sin i > sin iC P Q 45o T 1 or sin i > m 1 B 45o C or sin 45o > m or m > 2 S Hence, ray PQ suffers TIR & goes along path TS. E

ALLENÒ CBSE 35 9. (i) What is total internal reflection ? Under what conditions does it occur ? (ii) Find a relation between critical angle and refractive index. (iii) Name one phenomenon which is based on total internal reflection. Sol. (i) The complete reflection of a light ray at the boundary of two media, when the ray is going from denser to rarer medium is called TIR. Condition : (a) Light ray must pass from a denser to a rarer medium. (b) Angle of incidence must be greater than critical angle, (i > iC). Normal m2(rarer) (ii) O B m1 & m2 = refractive indices of different medium Iicc m1(denser) A In the above fig. ray AO goes from denser (say water) to rarer (say air) medium, node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx So, using snell’s law, sin i = m2 www.notesdrive.com sin r m1 or sin ic = m2 [Here, angle of incidence = ic sin 90° m1 and angle of refraction = 90°] If refractive index of rarer medium = 1 (i.e. m2 = 1) or sin ic = 1 1 æçQ m2 = m21 ö = m è m1 ÷ ø m1 (iii) Mirage or Brilliance of diamond. 10. The figure shows a ray of light falling normally on the face AB of an A equilateral glass prism having refractive index 3 , placed in water of 2 refractive index 4 . Will this ray suffer total internal reflection on B C 3 striking the face AC ? Justify your answer. Sol. The angle of incidence, of the ray, on striking the face AC is i = 60º (as from figure) Also , relative refractive index of glass, with respect to the surrounding water, is mr = 32 = 9 43 8 Also sin i = sin 60º = 3 = 1.732 = 0.866 A 2 2 N For total internal reflection, the required critical angle, in this case, is given by. sin iC = 1 = 8 ; 0.89 BC m 9 \\ i < iC Hence the ray would not suffer total internal reflection on striking the face AC. [The student may just write the two conditions needed for total internal reflection without analysis of the given case. The student may be awarded ( )+12 12 mark in such a case.] E

36 Physics ALLENÒ 11. (i) Name the phenomenon on which the working of an optical fibre is based (ii) What are the necessary conditions for this phenomenon to occur ? (iii) Draw a labelled diagram of an optical fibre and show how light propagates through the optical fibre using this phenomenon. Sol. (i) Total internal reflection (TIR) (ii) (a) Ray should go from denser to rarer medium. (b) Angle of incidence should be greater than critical angle (i > iC) Cladding (µ = 1.4) (iii) Light which enters from one end suffers multiple TIR and then Core (µ = 1.7) emerges out from other end only 12. In the following diagram, an object ‘O’ is placed 15 cm in front of a convex lens L1 of focal www.notesdrive.com length 20 cm and the final image is formed at ‘I’ at a distance of 80 cm from the second lens L2. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Find the focal length of the lens L2. 15cm 80cm 15cm I' O I L1 L2 80cm 15cm 15cm Sol. I¢ I O L1 L2 For Lens L1 : 111 111 1 1 1 v1 - u1 = f1 Þ v1 = f1 + u1 Þ v1 = 20 + -15 v1 = – 60 Now image formed (v1) acts as an objects for L2. For Lens L2 : 1 1 1 111 1 = 31 Þ f2 = 1200 Þ f2 = 39 cm v2 - u2 = f2 Þ 80 + 75 = f2 Þ f2 1200 31 13. (a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. If m for water is 1·33, find the wavelength, frequency and speed of the refracted light. (b) A double convex lens is made of a glass of refractive index 1·55, with both faces of the same radius of curvature. Find the radius of curvature required, if the focal length is 20 cm. E

ALLENÒ CBSE 37 Sol. (a) lair = 589 nm, µwater = 1.33 (Given) (i) For lwater, µw = la Þ lw = 589 µa lw 1.33 \\ lw = 442.8nm (ii) Frequency of light remains constant. (iii) velocity, v= c = 3 ´108 = 2.25´108 m / s µ 1.33 (b) From 1 = (µ21 - 1) é 1 - 1ù f ê R1 ú ë R2 û 1 =(1.55 - 1) é 1 - æ - 1 öù QR1 = R2 = R node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxfëêRèçRø÷ûú www.notesdrive.com1=(.55)éëêR2 ùûú=1R.1Þ 1 = 1.1 Þ R = 1.1´ 20 = 22cm f 20 R EXERCISE-IV FIVE MARK QUESTIONS : 1. (a) A point object O on the principal axis of a spherical surface of radius of curvature R separating two media of refractive indices n1 and n2 forms an image ‘I’ as shown in the figure. Prove that : n2 - n1 = n2 -n1 v u R N n1 •M n2 O R CI v u (b) Use this expression to derive lens maker’s formula. Draw the necessary diagram. (Refer to chapter-9, NCERT- Q. No. 9) (c) A convex lens is placed over a plane mirror. A pin is now positioned so that there is no parallax between the pin and its image formed by this lens-mirror combination. How will you use this observation to find focal length of the lens ? Explain briefly. Sol. (a) Assumptions :- · Aperature of the spherical refracting surface is small. · Object is a point object & lies on the pricipal axis. · Incident ray, refracted ray & normal to the spherical surface makes small angles with PA. E

38 Physics ALLENÒ Let, XPY = Convex spherical refracting surface. O = Point object in rarer medium I = Real image in denser medium X n2 n1 i A r a gb I O uPM C R v Y From DAOC, i = a + g From DAIC, g = r + b Þ r = g – b www.notesdrive.com From snell’s law, sin i = n2 Þ n1sini = n2sinr node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxsinrn1 Since the angles are small, \\ n1i = n2r Substituting for i & r, in the above eqn, we get n1(a + g) = n2 (g - b) or n1 íì AM + AM ýü = n2 ìíîAMMC - AM üý î PO MC þ MI þ Since the aperature is small, \\ MC = PC, MI = PI \\ íì n1 + n1 ýü = ìí n2 - n2 üý î PO PC þ î PC PI þ Acc. to sign convention, PO = –u, PC = R, PI = v \\ ì n1 + n1 ü = ì n2 - n2 ü í -u R ý í R v ý î þ î þ or n2 - n1 = n2 - n1 v u R (c) The rays must fall normally on the plane mirror so that the image of the pin coincides with itself. P O \\ P is the position of the focus of the lens. i.e. Distance OP = focal length E

ALLENÒ CBSE 39 2. (i) Draw a labelled schematic ray diagram of astronomical telescope in normal adjustment. (ii) Which two aberrations do objectives of refracting telescope suffer from ? How are these overcome in reflecting telescope ? Sol. (i) Refer to Three Marks Q.4 (a) (ii) Two aberrations :– spherical aberration & chromatic aberration. Spherical aberration in mirrors is corrected by using parabolic mirror & in lenses it is corrected by using small parabolic aperture lens. Chromatic abberation is reduced by increasing the focal length of the lens or by using an achromatic lens in which materials with differing dispersive prop. are assembeld to form a compound lens. 3. (i) Draw a ray diagram showing the formation of image of a point object situated on the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx principal axis and on the convex side of a spherical surface of radius of curvature R. www.notesdrive.com Taking the rays as incident from a rarer medium of refractive index n1 to a denser medium of refractive index n2, derive the relation n2 - n1 = n2 -n1 , where symbols have their v u R usual meaning. (ii) Explain briefly how the focal length of a convex lens changes with increase in wave length of incident light. (iii) Explain briefly how the focal length of a convex lens changes when it is immersed in water ? Refractive index of the material of lens is greater than that of water. Sol. (i) Refer to Five Marks Q.1 (ii) From m = c = c v nl Here, refractive index m µ1 l So, m of a lens increases with the decreases in wavelength of the incident light. From. 1 = (m -1) æ 1 - 1 ö f ç R1 R2 ÷ è ø We can say focal length will increases with increase in wavelength and vice versa. (iii) From, 1 = æ m2 - 1 ö é1 - 1 ù m2 (lens) = 1.5 f ç m1 ÷ ê R2 ú m1 (Water) = 1.33 è ø ë R û 1 1 = æ 1.5 - 1÷øö é1 - 1 ù When lens placed f çè 1.33 ê R2 ú in water ë R1 û Cleary, æ m2 ö gets decreased which means focal length gets increased. ç m1 ÷ è ø E

40 Physics ALLENÒ 4. (i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism. (Refer to chapter-9, NCERT Q. No. 10) (ii) What is dispersion of light ? What is its cause ? (iii) A ray of light incident normally on one face of a right angled isosceles prism is totally reflected as shown in figure. What must be the minimum value of refractive index of glass? Give relevant calculations. Sol. (ii) Splitting of white light into band of seven different colours (V,I,B,G,Y,O,R) is calledwww.notesdrive.com dispersion of white light. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Cause – Different colours travel with different velocity while passing through prism so they get separated. BC (iii) M 45o N 45o Normal A From geometry, angle of incidence at surface (AC) is 45o. Now as shown ray MN gets totally reflected at angle 45o, i.e. i > iC From m= 1 = 1 = 2 sin iC sin 45o 5. (i) Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point source lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence derive lens maker’s (Refer to NCERT Q. No. 9) formula. (ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ? Sol. (ii) Given : m1 = 1, m2 = 1.5, R = 20cm, u = 100cm Þ from m2 - m1 = m2 -m1 v u R Þ 1.5 - 1 = 0.5 v -100 20 Þ 1.5 = 5 -1010 = 3 or v = 100 cm v 200 200 E

ALLENÒ CBSE 41 6. (a) Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses. (b) A ray of light passing from air through an equilateral glass prism undergoes minimum 3 deviation when the angle of incidence is 4 th of the angle of prism. Calculate the speed of light in the prism. f1 f2 Sol. (a) O I I1 v AB u v1 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Object is placed at point O, whose image is formed at I1 by first lens. As image I1 is real, it works www.notesdrive.comas a virtual object for second lens B, producing the final image at I. Since the lenses are thin, assume the optical centres to be coincident. Let this central point is P. For the image formed by the lens A, 1 - 1 = 1 ....(1) v1 u f1 Similarly, for the lens B, 11 1 ....(2) v - v1 = f2 Add (1) & (2) we get, 1 - 1 = 1 + 1 v u f1 f2 If two lens system is taken as equivalent to a single lens of focal length f, we get, P = 1 = 1 + 1 ....(3) f f1 f2 In term of power, equation (3) can be written as, P = P1 + P2 where P is the net power of this combination. (b) Given : ABC is an (1) Equilateral triangle. (2) Ði = 3 A [A = 60°] 4 A NM i rr e N' BC E

42 Physics ALLENÒ (A = angle of prism) we know that, d=i+e–A Now, in the condition of min. deviation, i = e, therefore we can write, dm = i + i – A or dm = 2i – A or dm = 2 æ 3A ö - A èç 4 ø÷ dm =www.notesdrive.com3A-A=A = 30° 2 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Þ dm = 30° sin æ A + dm ö sin æ 60° + 30° ö çè 2 ÷ø èç 2 ø÷ Now, m = = æ A ö æ 60° ö sin çè 2 ø÷ sin èç 2 ÷ø m = sin 45° = 2 sin 30° So, v = c = 3´108 = 2.1´108 m/s m 2 7. Optical fibres :- Now-a-days optical fibres are extensively used for transmitting audio and video signals through long distances. Optical fibres too make use of the phenomenon of total internal reflection. Optical fibres are fabricated with high quality composite glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding. When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end. Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal. Optical fibres are fabricated such that light reflected at one side of inner surface strikes the other at an angle larger than the critical angle. Even if the fibre is bent, light can easily travel along its length. Thus, an optical fibre can be used to act as an optical pipe. E

ALLENÒ CBSE 43 Refractive m2 < m1 index, m2 Core Cladding Refractive Buffer coating index, m1 (i) Which of the following statement is not true. (A) Optical fibres is based on the principle of total internal reflection. (B) The refractive index of the material of the core is less than that of the cladding. (C) an optical fibre can be used to act as an optical pipe. (D) there is no appreciable loss in the intensity of the light signal while propagating through an node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx optical fibre www.notesdrive.com Ans. (B) (ii) What is the condition for total internal reflection to occur? (A) angle of incidence must be equal to the critical angle. (B) angle of incidence must be less than the critical angle. (C) angle of incidence must be greater than the critical angle. (D) None of the above. Ans. (C) (iii) Which of the following is not an application of total internal reflection? (A) Mirage (B) Sparkling of diamond (C) Splitting of white light through a prism. (D) Totally reflecting prism. Ans. (C) (iv) Optical fibers are used extensively to transmit :- (A) Optical Signal (B) current (C) Sound waves (D) None of the above Ans. (A) E

44 Physics ALLENÒ CHAPTER-10 : WAVE OPTICS PREVIOUS YEARS QUESTION EXERCISE-I ONE MARK QUESTIONS : 1. Define the term ‘wavefront’ Sol. Locus of all the points vibrating in the same phase is called wave front. 2. Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double slit experiment. Sol. Two sources are perfectly coherent if their frequency is same and their phase difference is constant. EXERCISE-II www.notesdrive.com TWO MARK QUESTION : node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx 1. Draw the intensity pattern for single slit diffraction and double slit interference. Hence, state two differences between interference and diffraction patterns. Sol. I0 (Intensity) I0 (Intensity) –5l/2 –3l/2 –l/2 O l/2 3l/2 5l/2 I0/22 Path diff(Dx) I0/61 Interference –2l/a –l/a l/a 2l/a (1) Intensity remains constant most of the time (2) Fringes are of equal width. (sinq)® Diffraction (1) Intensity decreases rapidly (2) Fringes are of unequal width. EXERCISE-III THREE MARK QUESTIONS : 1. (a) If one of two identical slits producing interference in Young's experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern. (b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light. Sol. (a) Let the intensity through one of the slit I1 = I So intensity of light through the slit covered by glass I2 = 0.5 I ( )2 Imax = I1 + I2 ( )2 Þ Imax = I + 0.5I = 2.9I ( )Imin = I1 - I2 2 E

ALLENÒ CBSE 45 ( )2 Þ Imin = I - 0.5I = 0.0865 I So Imax = 2.9I ; 34 Imin 0.086I (b) The central fringe remains white. No clear fringe pattern is seen after a few (coloured) fringes on either side of the central fringe. 2. Define the term wave front. State Huygen’s principle. Consider a plane wave front incident on a thin convex lens. Draw a proper diagram to show how the incident wave front transveres through the lens and after refraction focusses on the focal point of the lens, giving the shape of the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx emergent wave front. www.notesdrive.com Sol. Wavefront : Locus of all the points vibrating in same phase is called wave front. Huygen’s Principle – It states that every point on a wavefront is a source of new disturbance which travels further in the form of secondary wavelets. These wavelets spread out in the forward direction at the same speed as the source wave. A new wavefront is a tangential surface to the secondary wavelets. emerging WF F incident Convex Lens WF 3. (i) In Young’s double slit experiment, two slits are 1 mm apart and the screen is placed 1 m (ii) away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. Sol. (i) What should be the width of each slit in order to obtain 10 maxima of the double slit (ii) pattern within the central maximum of the single slit pattern ? d = 1mm, D = 1m, l = 500 nm (given) b = lD = 500 ´10–9 ´1 m = 500 × 10–6 m = 500 mm d 10–3 In single slit diffraction, path diff. = a sinq @ aq = l Þ q = l/a Width of central maxima of single slit = 2l/a Width of 10 maxima = 10 × fringe spacing = 10 × l/d Width of central maximum of single slit = Width of 10 maxima of double slit \\ 10l = 2l Þ a = d = 0.2 mm d a 5 E

46 Physics ALLENÒ 4. Explain the following, giving reasons : (i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. (ii) When light travels from a rarer to a denser medium, the speed decreases, does this decrease in speed imply a reduction in the energy carried by the wave ? (iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light ? Sol. (i) Reflection & refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of emitted light by a charged oscillator equals to its freq. of oscillation. Thus, the frequency of scattered light equals the frequency of incident light. (ii) No, energy carried by a wave depends on the amplitude of the wave, not on the speed of wave. www.notesdrive.com (iii) For a given frequency, intensity of light in the photon picture is determined by the no. of node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx photons falling on unit area per unit time. 5. (i) Derive Snell’s law on the basis of Huygen’s Wave theory when light is travelling from a denser to a rarer medium. (Refer to Chapter-10, NCERT Q. No. 6) (ii) Draw the sketches to differentiate between plane wavefront and spherical wavefront. Sol. (ii) S Ray W1 Plane wavefront W2 W3 w1, w2, and w3 are various spherical wavefronts 6. Two harmonic waves of monochromatic light y1 = a coswt and y2 = a cos(wt + f) are superimposed on each other. Show that maximum intensity in interference pattern is four times the intensity due to each slit. Hence write the conditions for constructive and destructive interference in terms of the phase angle f. Sol. Given, y1 = a coswt, y2 = a cos(wt + f) (a) Resultant displacement is given as : y = y1 + y2 = a coswt + a cos(wt + f) = a coswt + a coswt cosf – a sinwt sinf = a coswt(1 + cos f) – a sinwt sinf Put R cosq = a (1 + cos f) .....(i) R sinq = a sinf ....(ii) By squaring & adding Eqs. (i) & (ii), we get R2 = a2(1+cos2f + 2cosf) + a2sin2f = 2a2(1 + cosf) = 4a2cos2f/2 ff \\ I = R2 = 4a2cos2 2 = 4I0cos2 2 E

ALLENÒ CBSE 47 (b) For constructive interference, ff cos 2 = ±1 or 2 = np or f = 2np For destructive interference, ff p cos 2 = 0 or 2 = (2n + 1) 2 or f = (2n + 1)p EXERCISE-IV FIVE MARK QUESTIONS : 1. (a) Define a wavefront. Using Huygen's principle, verify the laws of reflection at a plane surface. (Refer to Chapter-10, NCERT Q. No. 10) (b) In a single slit diffraction experiment, the width of the slit is made double the original node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx width. How does this affect the size and intensity of the central diffraction band ? Explain. www.notesdrive.com (c) 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 obstacle. Explain why ? Sol. (b) If the width of the slit is made double then the size of the central maxima reduces to half and intensity increases upto four times. (c) This is because of diffraction of light. [Alternatively : Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.] [Alternatively : There is a maxima, at the centre of the obstacle, in the diffraction pattern produced by it.] 2. (a) Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen. (Refer to NCERT Q. No. 8) (b) The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is l , l and l 6 4 3. Sol. (b) As we know, I = 4I0 cos2(f/2) f = Phase difference f = 2p×Dx (Dx = path diff.) l (i) From, f= 2p ×Dx = 2p × l l l 6 f = p/3 = 60o 3 So I = 4I0cos230o = 4I0× 4 (Q f = 60o) I = 3I0 E

48 Physics f= 2p×Dx = 2p × l = p = 90o ALLENÒ l l 4 2 (Q f = 90o) (ii) From, I = 4I0cos245o = 4I0 × 1 So, 2 I = 2I0 (iii) From, f = 2p × Dx = 2p × l = 2p = 120o So, l l 3 3 www.notesdrive.comI= 4I0cos260o = 4I0 × 1 (Q f = 120o) 4 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx I = I0 3. (i) In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position ‘x’ on the screen. (Refer to NCERT Q. No. 8) (ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features. Sol. (ii) Interference Diffraction 1. It is result of superposition between secondary wavelets coming from different parts of same source. 2. Secondary maxima I0 Central Second Second First maximum minimum First I0/61 minimum I0/22 –3—al –2—al – —la O —la 2—al 3—al Sinq 3. All bright fringes are of same These are of varying width and varying width and same intensity. intensity. 4. A large number of interference Only a few diffraction bands are seen. fringes are observable. E