{"id":37850,"date":"2019-02-15T12:04:24","date_gmt":"2019-02-15T12:04:24","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=37850"},"modified":"2020-11-19T15:26:02","modified_gmt":"2020-11-19T09:56:02","slug":"plus-two-physics-notes-chapter-10","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/plus-two-physics-notes-chapter-10\/","title":{"rendered":"Plus Two Physics Notes Chapter 10 Wave Optic"},"content":{"rendered":"

Plus Two Physics Notes Chapter 10 Wave Optic is part of Plus Two Physics Notes<\/a>. Here we have given Plus Two Physics Notes Chapter 10 Wave Optic.<\/p>\n\n\n\n\n\n\n\n\n\n
Board<\/strong><\/td>\nSCERT, Kerala<\/td>\n<\/tr>\n
Text Book<\/strong><\/td>\nNCERT Based<\/td>\n<\/tr>\n
Class<\/strong><\/td>\nPlus Two<\/td>\n<\/tr>\n
Subject<\/strong><\/td>\nPhysics Notes<\/td>\n<\/tr>\n
Chapter<\/strong><\/td>\nChapter 10<\/td>\n<\/tr>\n
Chapter Name<\/strong><\/td>\nWave Optic<\/td>\n<\/tr>\n
Category<\/strong><\/td>\nPlus Two Kerala<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Kerala Plus Two Physics Notes Chapter 10 Wave Optic<\/h2>\n

Introduction<\/span>
\nIn 1678, the Dutch physicist Christian Huygens put forward the wave theory of light. We will discuss in this chapter.<\/p>\n

Wavefront:
\nThe wavefront is defined as the locus of all points which have the same phase of vibration. The rays of light are normal to the wavefront. Wavefront can be divided into 3.<\/p>\n

    \n
  1. Spherical wavefront<\/li>\n
  2. Cylindrical wavefront<\/li>\n
  3. Plane wavefront.<\/li>\n<\/ol>\n

    1. Spherical Wavefront:
    \n\"Plus
    \nThe wavefront originating from a point source is spherical wavefront.<\/p>\n

    2. Cylindrical Wavefront:
    \n\"Plus
    \nIf the source is linear, the wavefront is cylindrical.<\/p>\n

    3. Plane wavefront:
    \nIf the source is at infinity, we get plane wavefront.
    \n\"Plus<\/p>\n

    Huygen\u2019s Principle<\/span>
    \nAccording to Huygen\u2019s principle<\/p>\n

      \n
    1. Every point in a wavefront acts as a source of secondary wavelets.<\/li>\n
    2. The secondary wavelets travel with the same velocity as the original value.<\/li>\n
    3. The envelope of all these secondary wavelets gives a new wavefront.<\/li>\n<\/ol>\n

      Refraction And Reflection Of Plane Waves Using Huygens Principle<\/span>
      \n1. Refraction of a plane wave. (To prove Snell\u2019s law):
      \nAB is the incident wavefront and c1<\/sub> is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c2<\/sub> is the velocity of the wavefront in the second medium. AC is a plane separating the two media.
      \n\"Plus
      \nThe time taken for the ray to travel from P to R is
      \n\"Plus
      \nO is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.
      \n\"Plus
      \nwhere 1<\/sup>n2<\/sub> is the refractive index of the second medium w.r.t. the first. This is the law of refraction.<\/p>\n

      2. Reflection of plane wave by a plane surface:
      \n\"Plus
      \nAB is the incident wavefront and CD is the reflected wavefront, \u2018i\u2019 is the angle of incidence and \u2018r\u2019 is the angle of reflection. Let c1<\/sub> be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
      \nThe time taken for the ray to travel from P to Q is
      \n\"Plus
      \nO is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero. AO
      \n\u2234 \\(\\frac{A O}{C_{1}}\\)(sin i – sin r) = 0
      \nsin i – sin r= 0
      \nsin i = sin r
      \ni = r
      \nThis is the law of reflection.
      \nBehavior of wave frond as they undergo refraction or reflection.<\/p>\n

      a. Wave frond through the prism:
      \n\"Plus
      \nConsider a plane wave passing through a thin prism. The speed of light waves is less in glass. Hence the lower portion of the incoming wave frond will get delayed. So outgoing wavefrond will be tilted as shown in the figure.<\/p>\n

      b. Wave frond through a thin convex lens:
      \n\"Plus<\/p>\n

      Consider a plane wave passing through a thin convex lens. The central part of the incident plane wave travels through the thickest portion of lens.<\/p>\n

      Hence central part get delayed. As a result the emerging wavefrond has a depression at the centre. Therefore the wave front becomes spherical and converges to a point F.<\/p>\n

      c. Plane wave incident on a concave mirror:
      \n\"Plus
      \nA plane wave is incident on a concave mirror and on reflection we have spherical wave converging to the focul point F.<\/p>\n

      3. The Doppler Effect:
      \nThere is an apparent change in the frequency of light when the source or observer moves with respect to one another. This phenomenon is known as Doppler effect in light.<\/p>\n

      When the source moves away from the observer the wavelength as measured by the source will be larger. The increase in wavelength due to Doppler effect is called as red shift.<\/p>\n

      When waves are received from a source moving towards the observer, there is an apparent decrease in wavelength, this is referred to as blue shift.<\/p>\n

      Mathematical expression for Doppler shift:
      \nThe Doppler shift can be expressed as
      \n\"Plus
      \nVradial<\/sub> is the component of source velocity along the line joining the observer to the source.<\/p>\n

      Coherent And Incoherent Addition Of Waves<\/span>
      \nSuperposition principle:
      \nAccording to the superposition principle, the resultant displacement produced by a number of waves at a particular point in the medium is the vector sum of the displacements produced by each of the waves.<\/p>\n

      Coherent sources:
      \nTwo sources are said to be coherent, if the phase difference between the displacements produced by each of the waves does not change with time.<\/p>\n

      Incoherent sources:
      \nTwo sources are said to be coherent, if the phase difference between the displacements produced by each of the waves changes with time.<\/p>\n

      Constructive interference:
      \nConsider two light waves meet together at a point. If we get maximum displacement at the point of meeting, we call it as constructive interference.<\/p>\n

      Destructive interference:
      \nConsider two lightwaves meet together at a point. If we get minimum displacement at the point of meeting, we call it as destructive interference.<\/p>\n

      Mathematical condition for Constructive interference and Destructive interference:
      \n\"Plus
      \nConsider two sources S1<\/sub> and S2<\/sub>. Let P be point in the region of s1<\/sub> and s2<\/sub>. The displacement produced by the source s1<\/sub> at P.
      \ny1<\/sub> = a cos \u03c9t
      \nSimilarly, the displacement produced by the source s2<\/sub> at P
      \ny2<\/sub> = a cos (\u03c9t + \u03a6)
      \nWhere \u03a6 is the phase difference between the displacements produced by s1<\/sub> and s2<\/sub>
      \nThe resultant displacement at P,
      \nY = y1<\/sub> + y2<\/sub>
      \n= a cos \u03c9t + a cos (\u03c9t + \u03a6)
      \n= a (cos \u03c9t + cos (\u03c9t + \u03a6))
      \n\"Plus<\/p>\n

      Therefore total intensity at P,
      \n\"Plus<\/p>\n

      Constructive interference:
      \nIf we take phase difference \u03a6 = 0, \u00b12\u03c0, \u00b14\u03c0……., we get maximum intensity (4I0<\/sub>) at P. This is the mathematical condition for constructive interference. The condition for constructive interference can be written in the form of path difference between two waves.
      \n\"Plus
      \nWhere n = 0, 1, 2, 3……..<\/p>\n

      Destructive interference:
      \nIf we take phase difference \u03a6 = \u00b1\u03c0, \u00b13\u03c0, \u00b15\u03c0………., we get zero intensity at P. This is the mathematical condition for destructive interference. The condition for destructive interference can be written in the form of path difference between two waves.
      \n\"Plus
      \nWhere n = 0, 1, 2, 3……..<\/p>\n

      Interference Of Light Waves And Youngs Double Slit Experiment<\/span>
      \nYoung\u2019s double-slit experiment:
      \n\"Plus
      \nThe experiment consists of a slit \u2018S\u2019. A monochromatic light illuminates this slit. S1<\/sub> and S2<\/sub> are two slits in front of the slit \u2018S\u2019. A screen is placed at a suitable distance from S1<\/sub> and S2<\/sub>. Light from S1<\/sub> and S2<\/sub> falls on the screen. On the screen interference bands can be seen.<\/p>\n

      Explanation:
      \nIf crests (ortroughs) from S1<\/sub> and S2<\/sub> meet at certain points on the screen, the interference of these points will be constructive and we get bright bands on the screen.<\/p>\n

      At certain points on the screen, crest and trough meet together. Destructive interference takes place at those points. So we get dark bands.<\/p>\n

      Expression for band width:
      \n\"Plus
      \nS1<\/sub> and S2<\/sub> are two coherent sources having wave length \u03bb. Let \u2018d\u2019 be the distance between two coherent sources. A screen is placed at a distance D from sources. \u2018O\u2019 is a point on the screen equidistant from S1<\/sub> and S2<\/sub>.
      \nHence the path difference, S1<\/sub>O – S2<\/sub>O = 0
      \nSo at \u2018O\u2019 maximum brightness is obtained.
      \nLet \u2018P\u2019 be the position of nth<\/sup> bright band at a distance xn<\/sub> from O. Draw S1<\/sub>A and S2<\/sub>B as shown in figure. From the right angle \u2206S1<\/sub>AP
      \nwe get, S1<\/sub>P2<\/sup> = S1<\/sub>A2<\/sup> + AP2<\/sup>
      \nS1<\/sub>P2<\/sup> = D2<\/sup> + (Xn<\/sub> – d\/2)2<\/sup> = D2<\/sup> + Xn<\/sub>2<\/sup> – Xn<\/sub>d + \\(\\frac{d^{4}}{4}\\)
      \nSimilarly from \u2206S2<\/sub>BP we get,
      \nS2<\/sub>P2<\/sup> = S2<\/sub>B2<\/sup> + BP2<\/sup>
      \nS2<\/sub>P2<\/sup> = D2<\/sup> + (Xn<\/sub> + d\/2)2<\/sup>
      \n\"Plus
      \nS2<\/sub>P2<\/sup> – S1<\/sub>P2<\/sup> = 2xn<\/sub>d
      \n(S2<\/sub>P + S1<\/sub>P)(S2<\/sub>P – S1<\/sub>P) = 2xn<\/sub>d
      \nBut S1<\/sub>P \u2248 S2<\/sub>P \u2248 D
      \n\u2234 2D(S2<\/sub>P – S1<\/sub>P) = 2xn<\/sub>d
      \ni.e., path difference S2<\/sub>P – S1<\/sub>P = \\(\\frac{x_{n} d}{D}\\) ____(1)
      \nBut we know constructive interference takes place at P, So we can take
      \n(S2<\/sub>P – S1<\/sub>P) = n\u03bb
      \nHence eq(1) can be written as
      \n\"Plus
      \nLet xn+1<\/sub> be the distance of (n+1)th<\/sup> bright band from centre o, then we can write
      \n\"Plus
      \nThis is the width of the bright band. It is the same for the dark band also.<\/p>\n

      Diffraction<\/span>
      \nThe bending of light round the comers of the obstacles is called diffraction of light.<\/p>\n

      1. The single slit diffraction:
      \n\"Plus
      \nConsider a single slit AC having length \u2018a\u2019. A screen is placed at suitable distance from slit. B is midpoint of slit, A straight line through B (perpendicular to the plane of slit), meets the screen at O. AD is perpendicular CP.<\/p>\n

      Calculation of path difference:
      \nConsider a point P on the screen having a angle \u03b8 with normal AE. The path difference between the rays (coming from the bottom and top of the slit) reaching at P,
      \nCP – AP = CD
      \n(CP – AP) = a sin \u03b8
      \npath difference, (CP – AP) = a \u03b8______(1)
      \n[for small \u03b8. sin \u03b8 \u2248 \u03b8]<\/p>\n

      (I) Position of maximum intensity:
      \nConsider the point \u2018O\u2019, the path difference between the rays (coming from AB and BC) reaching at O is zero. Hence constructive interference takes place at ‘O’. Thus maximum intensity is obtained. This point is called central maximum or the principal maximum.<\/p>\n

      (II) Position of secondary minima:
      \nLet P be a point on the screen such that the path difference between the rays AP and CP be \u03bb.
      \nie, CP – AP = \u03bb______(2)
      \nSubstituting eq (1) in eq (2) we get
      \n\u03b8 = \u03bb
      \n(or) \u03b8 = \\(\\frac{\\lambda}{a}\\)______(3)
      \nLet the slit AC be imagined to be split into two equal halves AB and BC. For every point in AB, there is a corresponding point in BC such hat the distance between the points are equal to a\/2 Consider two points K and L such that, KL = a\/2. There fore, the path difference between the rays (coming form K and L) at P is,.
      \nLP – KP = \\(\\frac{a}{2}\\)\u03b8_______(4)
      \nSubstituting (3) in (4) we get
      \n\"Plus<\/p>\n

      This means that the rays (coming from K and L) reaching at P are out of phase and cancel each other. Hence the intensity at P becomes zero.
      \nIn otherwards, at angle \u03b8 = \\(\\frac{\\lambda}{\\mathrm{a}}\\)
      \nThe intensity becomes zero.
      \nSimilarly on the lower half of the screen, the intensity is zero for which \u03b8 = – \\(\\frac{\\lambda}{\\mathrm{a}}\\)
      \nThe general equation for zero intensity can be written as
      \n\u03b8 = \\(\\pm \\frac{n \\lambda}{a}\\)
      \nWhere n = 1, 2, 3,…
      \nFor first minima n = 1, and second minima n = 2.<\/p>\n

      (III) Position of Secondary maxima:
      \nLet P be a point on the screen, such that
      \nCP – AP = \\(\\frac{3}{2}\\)\u03bb
      \nFrom eq (1),we know (CP – AP) = a\u03b8
      \nTherefore a\u03b8 = \\(\\frac{3}{2}\\)\u03bb
      \nThe wave front AC can be divided into three equal parts.<\/p>\n

      The rays from first and second parts will cancel each other and the rays from third part will reach at P. Hence the point P becomes bright.<\/p>\n

      Similarly the next maximum occurs at \u03b8 = \\(\\frac{5}{2}\\)\\(\\frac{\u03bb}{a}\\)
      \nThe general equation for maximum can be written
      \n\\(\\theta=\\pm \\frac{(2 n+1) \\lambda}{2 a}\\)<\/p>\n

      1. (a) Intensity Distribution on the screen of diffraction pattern:
      \n\"Plus<\/p>\n

      (b) Comparison between interference and diffraction bands:
      \nInterference:<\/p>\n