OPTICS-02- Theory (26-38)

Illustration : An object is placed at a distance of 45 cm from a converging lens of focal length 30 cm. A mirror of radius of curvature 40 cm is to be placed on the other side of lens so that the object coincides with its image. Find the position of the mirror if it is (a) convex ? (b) concave ? Solution: (a) The object and image will coincide only if the light ray retraces its path and it will occur only when the ray normally strike at the mirror. In other words, the centre of curvature of the mirror and the rays incident on the mirror are collinear. The rays after refraction from lens must be directed towards the centre of curvature of mirror at C. If x is the separation, then for the lens u  45cm, v  x  40, f  30 cm Using lens formula 1  1  1 v u f or 1  1  1 x  40 45 30 or x  45(30)  40  50 cm 45  30 (b) 45cm x In case of concave mirror, the refracted rays through the lens meets at C, the centre of curvature (C) of the mirror. Using lens formula 1  1  1 v u f u = - 45 cm, v = x – 40, f = 30 cm 1  1  1 x 10 45 30 or x 10  (45)  30 45  30 or x  90  40  130 cm . Power of Lens Power of a lens is defined as the ratio of the refractive index of the surrounding medium and the focal length of the lens in that medium f. The unit of power is dioptre. Mathematically it can be written as P   . f Displacement Method to Determine the Focal Length of a Convex Lens: In this method, an object and a screen are placed at a distance a apart. A convex lens of focal length f (f  a ) can be placed in two position between the object and the screen such that a real image of the same 4 object is formed on the screen in both the positions. B Screen A' A B" B' u v v u d a Let a be the distance between the object and the screen and let d be the distance between the two position of the lens. From the figure : u + v = a v – u = d Solving, we get v  a  d , u  a  d 2 2 Using the lens formula and simplifying, we get a 2  d2 f  ...(i) 4a Let I1  A ' B'; I2  A ' B; O  AB m  I1  a  d 1 O a  d m  I2  a  d 2 O a  d i.e. the one image is larger than the object i.e. the other image is shorter than the object  O  ...(ii) i.e. the size of the object is the geometric mean of the sizes of the two images. Illustration : For two positions of a converging lens between an object and a screen which are 96 cm apart, two real images are formed. The ratio of the lengths of the two images is 4.84. Calculate the focal length of the lens. Solution: Here m1  4.84 m2  a  d 2 a  d 2.2   a  d   4.84  a  d 1 Putting a = 96 cm gives d = 36 cm a2  d2  f  4a  20.625 cm . Equivalent Focal Length of Combination of Lenses: If n number of lenses of focal lengths f1, f2 f3 fn are placed coaxially in contact then the equivalent focal length of the combination is given by 1  1  1  .....  1 ...(iii) F f1 f2 fn In terms of power P  p1  p2   pn Sign Convention Focal length of converging lens is taken as positive and that of the diverging lens is taken as negative. Illustration : A convergent lens of 6 diopters is combined with a diverging lens of –2 diopter. Find the power and focal length of the combination. Solution: Here P1  6 diopter, P2  2 diopter Power of the combination is given by Using the formula P  P1  P2  6  2  4 diopters f  1  1  0.25 m  25cm P 4 Equivalent Focal Length of Combination of Lens & Mirror: When one face of a lens is silvered as shown in figure it acts like a lens-mirror combination. Lenses with one face silvered act like lens-mirror combination It is obvious from the ray diagram as shown in figure that the incident ray of light is refracted through the lens twice (i.e., once when light is incident on the lens and second time when reflected by the mirror) and reflected from the mirror once. The combination acts like a mirror whose equivalent focal length is given by 1  1  1  1  2  1 ...(iv) F f1 fm f1 f1 fm where f1 = focal length of lens and f2 = focal length of mirror . Sign Convention : Focal length of converging lens or mirror is positive; and that of diverging lens or mirror is negative. Illustration : One face of an equiconvex lens of focal length 60 cm made of glass (  1.5) is silvered. Does it behave like a concave mirror or convex mirror? Solution: 1  2  1 F fl fm here fl  60 cm (converging lens) f  R  30 (converging mirror) m 2 1  2  1  2 F 60 30 30 The positive sign indicates that the resulting mirror is converging or concave. Illustration : The plane surface of a plano-convex lens of focal length 60 cm is silvered. A point object is placed at a distance of 20cm from the convex face of lens. Find the position and nature of the final image formed. Solution: Using equation 1  2  1 F fl fm here, fl  60cm; fm    1  2  1  1 F 60  30 or F  30cm The problem is reduced to a simple case where a point object is placed in front of a concave (converging) mirror. Using mirror formula 20 cm 1  1  1 v u f here u = - 20 cm; f = - 30 cm 1  1  v 20 1 30 Thus v = 60 cm., The image is erect and virtual. (a) For Convergent or Convex Lens (b) For Divergent or Concave Lens S. No. Position of Object Ray-Diagram Details of Image Position and nature of image formed for a given position of object. WAVE OPTICS: Wavefronts and Rays: A wavefront is defined as a surface joining the points of same phase. The speed with which the wavefront moves onwards from the source is called the phase velocity or wave velocity. The energy of the wave moves in a direction perpendicular to the wavefront. Figure shows light waves emitting out from a point source forming a spherical wavefront in three dimensional space. The energy travels outwards along straight lines emerging from the source, that is, radii of the spheri- cal wavefront. These lines are called the rays. Important: Huygen’s Principle: 1. Every point on a wavefront vibrates in same phase and with same frequency 2. every point on a wavefront acts like a secondary source and sends out a spherical wave, called a secondary wave. 3. Wavefronts move in space with the velocity of wave in that medium. Coherent Sources: Two source are said to be coherent if they have the same frequency and the phase realationship remains independent of time. In this case, the total intensity I is not just the sum of individual intensities I1 and I2 due to two sources but includes an interference term whose magnitude depends on the phase difference at a given point. I  I1  I2  2 I1I2 cos  –– ...(v) Interference Term where  is the phase difference. The interference term averaged over cycles is zero if (i) the sources have different frequencies (ii) the sourced have the same frequency but no constant phase difference For such incoherent sources : I  I1  I2 Unlike sources of sound waves, two independent sources of light can not be coherent. Because sound is a bulk property of matter while light is a characteristic of individual atoms which emit light randomly and independent of each other. In practice coherent sources are produced either by dividing the wavefront (as in the case of Young’s Double Slit Experiment, Fresnel’s biprism, Lloyd mirror etc.) or by dividing the amplitude (as in the case of thin films, Newton Rings, etc.) of the incoming waves. INTERFERENCE : YOUNG’S DOUBLE SLIT EXPERIMENT : It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light. Two slits S1 and S2 are made in an opaque screen, parallel and very close to each other. These two are illuminated by another narrow slits S and fall on both S1 and S2 which behave like coherent sources. Note that the coherent sources are derived from the same source. In this way, any phase change which occurs in S will occur in both S1 and S2 .The phase difference (1  2 ) between S1 and S2 is unaffected and remains constant. Light now spreads out from both S1 and S2 and falls on a screen. It is essential that the waves from the two sources overlap on the same part of the screen. If one slit is covered up, the other produces a wide smoothly illuminated patch on the screen. But when both slits are open, the patch is seen to be crossed by dark and bright bands called interference fringes. This redistribution of intensity, pattern is called interference pattern. Condition for bright fringes or maxima   2n or path difference, p  n where n = 0, 1, 2, ……. Imax  (  ) ...(vi) Condition for dark fringes or minima   (2n 1) or path difference, p   n  1   2  Imin  ( where n = 1, 2, 3, …….  I )2   ...(vii) The relation between phase difference () and path difference (p) is given by Let us see   2 p  ...(viii) how to find the position of the nth maxima or minima on the screen ? P yn O Schematic arrangement of YDSE Let P be the position of the nth maxima on the screen. The two waves arriving at P follow the path S1P and S2P , thus the path difference between the two waves is p  S1P  S2P  dsin  From experimental conditions, we know that D >> d, therefore, the angle  is small, Thus sin   tan   yn D p  dsin   d tan   d  yn     yn  pD / d For nth maxima p  n  yn  n D d where n = 0, 1, 2, (ix) For nth minima p   n  1   2     y   n  1  D where n = 1, 2, 3, (x) n  2  d   Note that the nth minima comes before the nth maxima. Fringe Width: It is defined as the distance between two successive maxima or minima.    y n1  yn  (n 1) D  nD d d or   D d ...(xi) Illustration : In a YDSE, if D = 2 m; d = 6 mm and  = 6000 A, then (a) find the fringe width (b) find the position of the 3rd maxima (c) find the position of the 2nd minima Solution: (a) Frindge width,   D  (60001010 )(2)  d 6103 0.2 mm (b) Position of 3rd maxima ...(xii), y  3D  3  3(0.2)  0.6 mm 3 d (c) Position of 2nd minima ...(xiii), y   2  1  D  3   3 (0.2)  0.3 mm 2  2  d 2 2   Illustration : White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is b and the screen is at a distance D >> b from the slits. At a point on the screen directly in front of one of the slits find the missing wave- lengths. Solution: According to theory of interference, position y of a point on the screen is given by y  D (x) d and as for missing wavelengths intensity will be min. (=0) if x  (2n 1) / 2 (2n 1) So, y  D 2d However, here d  b and y  (b / 2) . So, 2   (2n 1)D with n  1, 2, 3,.... i.e., wavelengths (b2 / D).(b2 / 3D), (b2 / 5D) , etc., will be absent (or missing) at point P. Optical Path It is defined as distance travelled by light in vacuum taking the same time in which it travels a given path length in a medium. If light travels a path length d in a medium at speed c, the tame taken by it will be (d/c). So optical path length L  c  d   d (because   c0 ) ...(xiv) 0  c  c Since for all media   1, optical path length is always greater then geometrical path length. When two light waves arrive at a point by travelling different distances in different media, the phase difference between the two is related by their optical path difference instead of simply path difference. Phase Difference  2  (optical path difference)  2 L .  Fringe Shift When a transparent film of thickness t and refractive index  is introduced in one of the slits, the fringe pattern shifts in the direction where the film is placed. P How much is the fringe shift ? Consider the YDSE arrangement shown in the figure. yn A film of thickness t and refractive index  is placed in front of the lower slit. O The optical path difference is given by p  [(S2P  t)  t]  S1P or Since p  (S2P S1P)  t( 1) S2P S1P  dsin   p  dsin   t( 1) As sin   tan   y 'n D p  d.y 'n  t( 1) D or y 'n  nD  ( 1) tD d d In the absence of film the position of the nth maxima is given by equation (xv) y  nD n d Therefore, the fringe shift is given by FS  y n  y 'n  ( 1) tD d ...(xvi) Note that the shift is in the direction where the film is introduced. Illustration : In a YDSE  =6000A D = 2 m, d = 6 mm. When a film of refractive index 1.5 is introduced in front of the lower slit, the third maxima shifts to the origin. (a) find the thickness of the film (b) find the positions of the fourth maxima Solution: (a) Since 3rd minima shifts to the origin, therefore, the fringe shift is given by FS  y3  3D d From equation (xxxvii), we know FS  ( 1) tD d ( 1) tD  3 D d d t  2  1 Here   0.6 106 m; 1.5 (3)(0.6 106 )  t  1.5 1  3.6m (b) There are two positions of 4th maxima; one above and the other below the origin. y  1  D  0.2 mm 4 d D y4 n=4 n=3 n=2 y '4  7  d  1.4 mm y'4 D n=0 n=4 Intensity Distribution : When two coherent light waves of intensity I1 pose, then the resultant intensity is given by and I2 Bright Fringe with a constant phase difference  superim- I  I1  I2  2 cos  In YDSE, usually the intensities I1 and I2 are equal i.e., Then or I1  I2  I0 I  2I0 (1 cos ) I  4I cos2 ( / 2) ...(xvii) For maxima :   2n  Imax  4I0 For minima :   (2n 1)  Imin  0 B D B D B D B D B D B y Intensity distribution on the screen as a function of y in YDSE. Imax  4I0 for bright fringe and Imin  0 for dark fringe. Illustration : The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. I  I   Solution: max   1  Imin  I1   Since I1  2I0 and I2  I0 , therefore I  2  1 2 max     34 . Imin  2 1  Illustration : In figure(a) shown, S is a monochromatic point source emitting light of wavelength   500 nm. A thin lens of circular aperature and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 m. the screen at O is normal to SO. (a) If the third intensity maximum occurs at the point A on the screen, find distance OP. (b) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OP increase, decrease or remain the same? Solution: As shown in figure (b) each part of the lens will from image of S which will act as coherent sources. So from lens theory if v is the distance of image from L1 and L2 , S O (a) 0.15m (b) 1  1  1 i.e., v  30cm v 15 10 So in sSL L and SS S , 1 2 d  u  v  1  v  1  30  3 L u u 15 i.e., d  3L And from figure (b) D = 1.30 – 0.30 = 1m. Now from the theory of interference the position y of a point P on the screen is given by y  D (x) d and as here point P is third maximum, i.e., x  3 , So, y  D (3)  D    …(i) 3L 5 107  L  3 i.e., y  0.5 103  10 m = 1 mm. (b) As from equation (i) it is clear that y  (1/ L) , with decrease in gap width L the dis- tance y, i.e., OP will increase.