OPTICS Interference of Light: Many theories were put forward to explain the nature of light: Newton s corpuscular theory Huygens wave theory

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1 OPTICS Interference of Light: Many theories were put forward to explain the nature of light: Newton s corpuscular theory Huygens wave theory Electromagnetic theory Quantum theory The above-mentioned theories explain that some of the properties like refraction, reflection, interference etc. can be explained by considering the light as a wave while the phenomenon like photoelectric effect, Compton Effect etc. may be described by considering the light as particle. Interference: It is the phenomenon of superposition of two coherent* waves in the region of superposition. At some points in the medium, the intensity of light is maximum while at some other points the intensity is minimum. The positions of maximum intensity are called maxima, while those of minimum intensity are called minima. Principle of Superposition: In a medium, the resultant displacement of a particle acted upon by two or more waves simultaneously is equal to the algebraic sum of the displacements of the particle due to individual waves. Thus if the displacement due to a single wave train at a point be y 1 and y be the displacement due to the another wave train in the same direction, then the resultant displacement y may be written as, 1

2 y y 1 y 1 a1 sin t and y a sin t If, y Where δ is the phase difference between two waves. y a1 sint a sint a1 a cos sint a sin cos t Let a a cos cos and a sin sin ; we have, 1 A y Acos sin t Asin t Where, A a a a a 1/ 1 1 cos a Asin cos t a sin and tan a1 a cos The intensity at any point is proportional to the square of the amplitude, i.e. I A, then I A a1 c a1a cos (1) Since a1 and a are the intensities of the two interfering waves, the resultant intensity at any point is not just the sum of the intensities due to the separate waves a. 1 a Condition for maximum and minimum intensities: From (1) it is clear that I will be maximum at points where cos 1or n ; n 0,1,... Therefore, the intensity will be maximum when phase difference is even multiple of, but we know that, Path difference phase difference

3 n or Therefore, I max a1 a n Similarly I will be minimum at points where cos δ=-1and therefore, n 1 n 1 And, I min a1 a n=0,1,. *Coherent Sources: The light waves of practically the same intensity, of exactly the same wavelength and of exactly a constant initial phase difference are known as Coherent Sources. For Interference of light to take place the two light waves must have the same plane of polarization. Two light waves from independent sources can never be coherent. Thus a pair of coherent waves may be obtained from a single source of light. Young s Double Slit experiment: In 1801, Young first demonstrated experimentally the phenomenon of interference. 3

4 X S 1 S I S In his experiment light from a monochromatic source is allowed to fall on a slit S and then through a double slit S 1 and S. The interference pattern was observed on a screen XY where few dark and bright bands or fringes are observed. These fringes are of equal distances. The bold lines show the amplitude in positive direction (crest) where the dotted lines show the amplitude in the negative direction (troughs). At points where crests fall on crests or troughs fall on troughs the amplitudes add up and the resultant intensity increases (I α A ) and thus we obtain constructive interference. At points where crests fall on the troughs or the troughs fall on the crests, the amplitudes are reduced (the resultant amplitude will be zero if the two lights from S 1 and S have equal amplitudes and so the intensity is also zero). This is known as destructive interference. Y 4

5 Thus on the screen alternate bright and dark regions of equal width, called interference fringes are observed. Analytical treatment for Young s double Experiment: P d S 1 C S Q y d/ O d/ R D Let S 1 and S be two coherent sources separated by a distance d and made from a monochromatic source S. Let a screen be placed at a distance D from the coherent sources. O is a point on the screen which is equidistant from S 1 and S. Therefore, path difference at O will be zero and the intensity at O will be maximum. Let P be a point on the screen such that, OP= y Therefore, PQ d y, PR y d 5

6 6 D d y D d y D P S, 1/ 1 and D d y D P S The path difference of two light rays emerging from S 1 and S and reaching to a point P on the screen is, D yd P S P S 1 Therefore the phase difference, path difference= D yd Position of bright fringes: As mentioned earlier that for bright fringes path difference is even multiple of λ/, i.e. n n D yd Thus the distance of nth bright fringe from point O is given by: d D n y n. Position of dark fringes: For dark fringe or the minimum intensity at P, the path difference must be an odd multiple of half wavelength, i.e. 1 n D yd

7 Thus the distance of the nth dark fringe from point O is given by, n 1 D y n d. Fringe Width: The distance between any two consecutive dark fringes or bright fringes is same and known as fringe width. It is given by: yn1 y n D d Conclusions: 1. There is a bright fringe at the centre of the screen.. Alternatively dark and bright fringes on both the sides of the sides of the central maximum occur. 3. Fringe widths of dark and bright fringes are same. 4. Fringe width depends upon the wavelength of light, separation of sources and source to screen separation. Condition for Sustained Interference: To obtain a well defined and observable interference the condition for interference of light are given as under: The beams from two sources must propagate along the same line; otherwise the vibrations cannot be along a common line. The two sources must be very narrow because a broad source is equivalent to a no. of sources and so positions of darkness and brightness cannot be well defined. The two sources must have equal intensities. The two sources should have equal amplitudes and frequencies. In case of unequal frequencies the phase 7

8 difference between the interfering waves will vary continuously and as a result, intensity at any point will not be constant but vary with time. If the amplitudes are different then a clear cut distinction between maximum (a 1 + a ) and minimum (a 1 - a ) is not possible. The common original source must be monochromatic, i.e., emitting light of single wavelength. The light from the two sources must have either zero phases or a constant difference of phase. This condition is very well observed in the case of coherent sources. The separation between source and screen should be large to get wide fringes. If the interference pattern has to be obtained by polarised light then the polarised waves must be in the same state of polarization. Two Classes of Interference The interference is divided into two classes on the basis of the way of obtaining the two coherent sources: 1. Division of wavefront: S 1 S S In this case the wavefront originating from one source is divided into two parts. This division can be made with the help of any optical system like Fresnel s biprism, Fresnel s 8

9 mirror etc. The two wavefronts after traveling certain distances are brought together to give interference pattern.. Division of amplitude: S Reflected System Transmitted system In this process the amplitude is divided into two or more parts either by partial reflection or refraction and which after traveling different paths produce interference at a point. Newton s rings, Michelson interferometer are the examples of this class. Fresnel s Biprism: However Young demonstrated the phenomenon of interference, but it was doubted that the fringes are not due to interference of light waves but due to some modification of light at the slits. These doubts are removed by the Fresnel s biprism experiment. By using biprism, Fresnel obtained two coherent sources from a single source by refraction. Construction: It consists of two acute angle prisms with their bases in contact forming a single obtuse angle prism ABC. In actual practice the prism is grounded from a single optically 9

10 true glass plate. The obtuse angle of the prism is nearly 179 0, so that each of acute angles is nearly 30. Working: A narrow vertical slit S is illuminated by a source of monochromatic light. The prism is placed with its refracting edge parallel to the line joining the sources. The light from the slit S is allowed to fall symmetrically on the biprism ABC with its refracting edge vertical (i.e., parallel to the line source S). When the light from S falls on the upper half of the prism, after refraction it appears to come from virtual source S 1. Similarly after refraction from the lower half of the prism it appears to be coming from virtual source S. The distance between the biprism and the source S is so adjusted that the sources S 1 and S are very close to each other. The interference fringes are observed in the region EF of the screen. G S A 1 E Screen O d S B O S a C D b F H 10

11 The analytical treatment is same as for Young s Double Slit experiment. If d be the distance between the sources S 1 and S and D between the source and the screen, then the fringe width is given by, d D D And thus d a d Moreover the distance of n th bright fringe from O. ( y n ) Where bright d nd d, Similarly a( 1) b ( y Applications: a) Determination of wavelength of light: d D ) n dark n 1 D d b) Determination of thickness of a thin sheet of transparent material: a) Biprism can be used to determine the thickness of a given thin sheet of transparent material e.g., glass, mica.. 11

12 A t G P x d C B D b) Suppose A and B are two virtual coherent sources. c) The point C is equidistant from A and B. d) When a transparent plate G of thickness t and refractive index µ is introduced in the path of one of the beams, the fringe which was originally at C shifts to P. e) The time taken by the wave from B to P in air is the same as the time taken by the wave from A to P partly through air and partly through the plate. f) Suppose c 0 is the velocity of light in air and c its velocity in medium then, BP AP t t c0 c0 c c0 c BP AP t t 0 c, But c BP AP t t 1t If P is the point originally occupied by the n th fringe, then the path difference BP AP n t n 1..(i) 1

13 Also the distance x through which the fringe shifted nd d D Where d, the fringe width. Also, Or, nd x d xd n D xd 1 t D g) Therefore, knowing x, the distance through which the central fringe is shifted, D, d and µ, the thickness of the transparent plate can be calculated. h) If a monochromatic source is used, the fringes will be similar and it is difficult to locate the position where the central fringe shifts after the introduction of the transparent plate. Hence white light is used. The fringes will be coloured but the central fringe is white. Interference in Thin films: It has been observed that interference in the case of thin films takes place due to (1) reflected light and () transmitted light. Newton was able to show the interference rings when a convex lens was placed on a plane glass-plate. Interference Due To Reflected Light(Thin films): Consider a transparent film of thickness t and refractive index µ. 13

14 A ray SA incident on the upper surface of the film is partly reflected along AT and partly refracted along AB. At B part of it is reflected along BC and finally emerges out along CQ. The difference in path between the two rays AT and CQ can be calculated. CN is normal to AT and AM is normal to BC. The angle of incidence is i and the angle of refraction is r. CB and AE are extended to meet at P. APC r The optical path difference x AB BC AN sin i AN Here, sin r CM AN.CM AB BC CM AB BC CM PC CM x. x..pm In the ΔAPM, cos r PM AP or PM AP. cos r AE EPcos r t cos r AE EP t x. PM t. cos r (1) This eq. (1), in the case of reflected light does not represent the correct path difference but only the apparent. It has been established on the basis of electromagnetic theory that, when 14

15 light is reflected from the surface of an optically denser medium (air-medium interface) a phase change equivalent to a path difference occurs. Therefore, the correct path difference in this case, t cos r x..() 1) If the path difference x = n where n = 0,1,,3.etc., constructive interference takes place and the film appears bright. t cos r n or, t cos r n 1..(3) ) If the path difference is x n 1 where n = 0,1,, 3.etc., destructive interference takes place and the film appears dark. t cos r n 1 or, t cos r n 1.(4) Here n is an integer only, therefore (n+1) can also be taken as n. t cos r n.(5) Where n = 0,1,, 3 etc. t S T i Q N AIR i A C r M µ E r B P AIR F 15

16 Interference Due To Transmitted light (Thin Film): S t i A r B P r r r i C r i N M D R AIR μ AIR Q Let there be a thin transparent film of thickness r and refractive index μ. A ray SA after refraction goes along AB. At B it is partly reflected along BC and partly refracted along BR. The ray BC after reflection at C, finally emerges along DQ. Here at B and C reflection takes place at the rarer medium (medium-air interface). Therefore, no phase change occurs. BM is normal to CD and DN is normal to BR. The optical path difference between DQ and BR is given by x BC CD also sin i sin r BN MD BN or BN. MD from fig. BPC r and CP = BC = CD 16

17 BC CD PD x PD MD In the But, PD MD PM PM BPM, cos r or PM BP. cos r BP BP t PM t cos r x. PM t cos r 1) For bright fringes, the pat difference x n t cos r n where n = 0, 1,, 3...etc. ) For dark fringes, the path difference t cos r n 1 17 x n 1 where n = 0, 1,, 3...etc. In the case of transmitted light, the interference fringes obtained are less distinct because the difference in amplitude between BR and DQ is very large. Important points: When the film is exposed to white light then only those wavelengths will be present for which the condition of maxima is satisfied or which have the path difference satisfying the condition for bright fringe. Newton s Rings: When a Plano convex lens of long focal length is placed on a plane glass plate, a thin film of air is enclosed between the lower surface of the lens and the upper surface of the plate. The thickness of the air film is very small at the point of contact and gradually increases from the centre outwards. The fringes produced with the monochromatic light are circular. The fringes are concentric circles, uniform in

18 thickness and with the point of contact as the centre. These fringes are known as Newton s Rings. M 45 0 B L 1 S L G AIR FILM Construction: S is a source of monochromatic light at the focus of the lens L 1. A horizontal beam of light falls on the glass plate B at The glass plate B reflects a part of the incident light towards the air film enclosed by the lens L and the plane glass plate G. The reflected beam from the air film is viewed with a microscope. Working: Interference takes place and dark and bright circular fringes are produced. 18

19 This is due to the interference between the light reflected from the lower surface of the lens and the upper surface of the glass plate G. Theory: 1. Newton s ring by reflected light: Let the radius of curvature of the lens is R and the air film is of thickness t at a distance OQ= r, from the point of contact O. REFLECTED LIGHT R L G Here interference is due to reflected light. Therefore for the bright rings t cos r n 1.(i) where n = 1,,3.etc. For normal incidence r = 0 0 therefore, cos r = 1 For air, μ = 1 Hence t n 1..(ii) For the dark rings, t cos r n Or, t n (iii) where n = 0,1,,3.etc. From fig. in ΔCEP CP CE EP CP CO OE Or EP AIR FILM C H E P t O Q r n 19

20 R R t r n or or r n Rt (ignoring t ) rn hence t R substituting the value of t in (ii) and (iii) we get, n 1 R for bright rings r n..(iv) n 1 R or r n..(v) for dark rings r n nr..(vi) or r n nr..(vii) Result: The radius of nth dark ring is proportional to i. n ii. iii. R Similarly the radius of nth bright ring is proportional to ( n 1) i. ii. iii. R If D n is the diameter of the nth dark ring, Dn rn 4nR..(viii) If D n is the diameter of the nth bright ring, 0

21 n 1 R Dn rn..(ix). Newton s ring by transmitted light: L G AIR FILM TRANSMITTED LIGHT In case of transmitted light the interference fringes are produced such that for bright rings, t cos r n (x) For air, μ = 1and cos r = 1 then t n.(xi) For dark rings, t n 1 (xii) Hence in this case the radius of nth bright is r n nr.(xiii) And the radius of the nth dark ring is n 1 R r n.(xiv) Applications of Newton s Ring: 1

22 Determination of wavelength of light: For nth dark ring we have D n 4nR..(i) D n or n 4R.(ii) again for (n+p)th dark ring, D n p n p 4R (iii) D n p D 4 pr n (iv) Determination of Refractive Index of a liquid: The refractive index of a liquid, forming film between lens and glass plate, can be obtained from the following eq. D n p D n 4 pr liquid..(i) D n p D 4 pr Since for air μ= 1 n air Hence Dn p Dn 4 pr air Dividing (ii) by (i) we get D n p Dn air D D n p n liquid..(ii)

23 Newton s Rings with white light: In case of white light the diameter of the rings of different colours will be different and there will be an overlapping of the rings of different colours over each other. The only first few rings will be clear while other rings cannot be viewed. Michelson s Interferometer: Principle: The amplitude of light beam from a source is divided into two parts of equal intensities by partial reflection and transmission. These beams are sent in two directions at right angles and are brought after they suffer reflection from plane mirrors to produce interference fringes. Construction: M 1 C Reflected G 1 G S A Transmitted B M T 3

24 It consists of two highly polished plane mirrors M 1 and M two plane parallel glass plates G 1 and G of exactly the same thickness. The plate G 1 is semi silvered at the back so that the incident beam is divided into reflected and transmitted beams of equal intensities. M 1 and M are mutually at right angles to each other. The plates G 1 and G are parallel to each other and are at an angle of 45 0 to M 1 M. The mirrors M 1 and M can be adjusted exactly perpendicular to each other with the help of the screws on their backs. The mirror M 1 can be moved exactly parallel to itself with a carriage on which it is moved. Working: The light falling on the glass plate G 1 is partly reflected and partly transmitted. The reflected ray AC travels normally towards plane mirror M 1 reflected back to the same path and comes out along AT. The transmitted ray after reflection from mirror M follows the same path and then moves along AT after reflection from the back surface of G 1. So the rays received at T are produced from a single source by the division of amplitude and we get an interference pattern. The reflected ray AC travels through glass plate G 1 twice and in the absence of plate G the transmitted ray does not cross plate G 1. To compensate it a second plate G is introduced in the path of transmitted ray. 4

25 Adjustment: Distance of the mirrors M 1 and M are adjusted to be nearly equal from glass plate G 1. In order to obtain a parallel ray of light a lens is adjusted between the source S and plate G 1. Formation of fringes: 1. Circular fringes: If eye is placed to see mirror M 1 directly then in addition it will see virtual image of M formed by reflection in glass plate G 1. Thus one of the interfering beams cones from M 1 and other appears to come from the virtual image of M i.e. M. If M is exactly at right angles to M 1 and the plate G 1 is at 45 0 to each of then, the air film formed between M 1 and M will have uniform thickness. Let the two interfering beams appear to come to the eye from two virtual images S 1 and S of the original source and let d be the distance between them. Now since reflected ray from M suffers reflection at the silvered surface of plate G 1, an additional path difference λ/ is introduced. Therefore, the total path difference d cos /. Therefore, for brightness, d cos / n, n 0,1,... etc. If M coinsides with M 1 then total path difference will be λ/ which is the condition for destructive interference. 5

26 If the light is obtained from an extended monochromatic light source then for a given value of n, the value of α is constant and so the locus of the fringes is circle. d cos α S α d S 1 M 1 M. Localized fringes: If M 1 and M are not exactly parallel the path difference is very small; we observe fringes as in the case of a wedge shaped film. In this case the locus of points of equal thickness is a straight line parallel to the edge of the wedge. S 6

27 For small path difference, the fringes are nearly straight but for larger path difference the fringes are generally curved with the convex surface of the film towards the edge of the wedge. M 1 M M M 1 M M 1 Important points: 1. As the circular fringes are formed due to two parallel interfering waves, hence they are formed at infinity.. Here the circular fringes are formed at same inclination hence they are called equal inclination fringes or Haidinger fringes. Applications: 1. Determination of wavelength of Monochromatic light: Mirror M 1 and M are adjusted so that circular fringes are visible in the field of view. If M 1 and M are equidistant from the glass plate G 1, the field of view will be perfectly dark. The mirror M is kept fixed and the mirror M 1 is moved with the help of the handle of the micrometer screw and the number of fringes that cross the field of view is counted. 7

28 If mirror is moved through a distance of d for a monochromatic light of wavelength λ and no. of fringes n that cross the centre of the field of view is n then d, because for one fringe shift the mirror moves through a distance equal the half the wavelength. Hence λ can be determined.. Determination of the difference in wavelength between two neighbouring spectral lines: Set the apparatus for circular fringes. Let the source has two wavelengths λ 1 and λ (λ 1 > λ ) which are very close to each other. The two wavelengths form their separate fringe patterns, but because of very small change in wavelength, the two patterns overlap. As the mirror M 1 is moved slowly the two patterns separate out slowly and when the path difference is such that the dark fringe of λ 1 falls on the bright fringe of λ maximum indistinctness occurs. Again when the path difference is such that the bright fringe of λ 1 falls on bright fringe of λ maximum two successive positions of maximum distinctness occurs. Let the mirror M 1 is moved through a distance of d between distinctness. This will happen only when n th fringe of λ 1 coincides with n+1 th fringe of λ then d n1 ( n 1) d d n 1 n and 1 8

29 Subtracting we get, 1 d d d = Or 1 d 1 hence 1 hence 1 d Thus by measuring the distance d moved by M 1 the difference between two wavelengths can be determined. 3. Determination of thickness of a thin plate: The apparatus is set for parallel fringes. White light source is used instead of monochromatic source. The cross wire is set on the central fringe. Now the thin plate whose thickness is to be determined is introduced in the path of one of the interfering beams. Due to the insertion of the plate path difference is changed to (μ-1)t. Thus a shift in the fringe system occurs. Now the mirror M 1 is moved till the central fringe coincides with the cross wire. The distance x moved by mirror M 1 is measured using micrometer screw. Hence we have, x ( 1)t 9

30 t x 1 Using above equation t can be determined. Diffraction: The bending of light beam round the corners or edge and spreading of light into the geometrical shadow of an opaque obstacle placed in its path is known as diffraction. In other words the diffraction may be defined as the encroachment of light into the geometrical shadow region of small opaque obstacle or aperture placed in the path of light. A A S B B Types of diffraction: There are two classes of diffraction: 1. Fresnel Diffraction. Fraunhofer Diffraction 1. Fresnel Diffraction: In this case diffraction is considered to take place from different parts of the aperture. Here the source or screen or both are at finite distance from obstacle. In this class the incident wave front is divergent, either spherical or cylindrical. 30

31 Here the observed pattern is a projection of diffracting element modified by diffracting effects and the geometry of the source. Here the diffraction centre of the pattern may be either bright or dark.. Fraunhofer Diffraction: Here the source of light and screen are at infinite distance from the obstacle. Here the inclination is important. Here the incident wave front is generally plane and the centre of diffraction pattern is always bright. Fraunhofer Diffraction At A Single Slit: W L 1 W 1 W e O θ K B θ W B L 1 W 1 W L Y The light from a monochromatic point source S is converted into parallel beams of light by a convex lens L 1 L 1. Now this beam is incident normally on a slit AB of width e. 31 A θ L Y A P

32 The incident light wave fronts are shown by W 1 W 1 and W W. Every point on the wave front incident on the slit and lying within the width of the slit emits secondary waves which superimpose to give diffraction pattern on the screen YY. In this diffraction pattern a central bright band is obtained. On either side of this band alternate dark and bright bands of decreasing intensity are observed. Analysis: AK is perpendicular from A on BB. The path difference of the rays meeting at P is clearly BK. Here, BK AB sin esin Therefore, the corresponding phase difference pathdiffer ence esin If the slit is assumed to be made up of n equal parts, the amplitude of the secondary waves originating from these parts will be equal. Let the amplitude of each secondary wave be a and then the phase difference between any two successive waves will be 1 esin d,( say) n Thus the resultant in a direction θ, due to the superposition of these n secondary waves can be calculated in the following way: Let there be n-harmonic waves having equal amplitudes a and common phase difference δ between successive vibrations. To find the resultant we construct a polygon, the closing side of which gives the resultant amplitude R and a resultant phase Ф. 3

33 R a 3δ a δ O Ф a a 1δ Now resolving the amplitudes parallel and perpendicular, we get, Rcos a acos acos... acos( n 1)...(1) Rsin asin asin... asin( n 1)...(),we R cos sin a sin sin cos sin cos... sin cos( n 1) Multiplying eq. (1) by sin From the trigonometrical relation cos Asin B sin( A B) sin( A B) we have, Rcos sin 3 a[sin sin sin sin n sin n ] n 1 n a sin sin sin cos a n sin sin 33

34 or R cos n asin cos sin n 1 (3) similarly, multiplying () by sin and applying sin Asin B cos( A B) cos( A B) and we get, C D D C cos C cos D sin sin n n 1 asin sin Rsin sin from (3) and (4) we get n asin R sin and 1 n Hence for the present case R Let nd asin d sin 1 n esin n asin 1 esin n sin esin then,..(4) esin asin esin sin n 34

35 R As asin sin asin / n / n n n / n is small, sin Again since n is infinitely large and a is negligibly small, Let na =A, a finite no. Hence, Asin R...(5) The secondary waves diffracted by the slit in a direction θ are focused by the lens L L at a point P on the screen. Therefore, the resultant intensity at P, I R. i.e. A sin I...(6) Conditions for maxima and minima: di 0 d d A sin or 0 d sin cos sin or A 0...(7) sin Case I: When 0 or sin α =0 or n ( o = 0 then (sin α)/α = indeterminate) Therefore, n 0, n 1,,3... esin or n or esin n...(8), if α 35

36 y=tanα y=tanα y=tanα which gives the condition for nth minima. Case II: The second bracketed term in eq.(7) will give the condition for maxima. cos sin Thus 0 or tan...(9) y tan and y The intersection of the curves will give the solution of eq.(9). The points of intersection will give the values of α as ,,,,... Here α = 0 corresponds to central maximum. Hence for central maximum the intensity is given by, A sin I A I 0 y = α y O 45 0 π/ 3π/ 5π/ α 36

37 e e e e e e 37

38 Important Results: 1. The width of central maximum is directly proportional to the wavelength of light used and as the wavelength of red light is more than the wavelength of violet light hence the width of the central is more for red light than for violet light.. The width of the central maximum is inversely proportional to the width of slit e hence width of central maximum is greater for narrow slits. Diffraction At Double Slit: L A S 1 θ L Y P S e+d θ B C M O D S θ P Let there be two slits AB and CD of equal width e, separated by a distance d. The incident light gets diffracted from these two slits and focused on the screen XY. The diffraction by two parallel slits is a case of diffraction as well as interference. 38 Y

39 Thus the pattern obtained on the screen consists of equally spaced interference fringes due to both the slits and their intensity being modulated by the diffraction phenomenon occurring due to individual slits. The two slits AB and CD may be considered equivalent to two coherent sources S 1 and S. Now due to individual slit, the resultant amplitude due to all secondary waves diffracted in the direction θ is given by: Where, R Asin esin...(1) For simplicity, the resultant of all the secondary waves may be taken as a single wave of amplitude R. The resultant at point P on the screen will be the result of interference between these two waves, of same Asin amplitudes, starting from S 1 and S. The path difference between the wavelets from S 1 and S in a direction θ is, SM e d)sin and therefore the phase. difference SM e d sin The resultant amplitude at P may be given as, R r R 1 R Asin R1R Asin cos Asin cos 39

40 4A Hence resultant intensity, or I r sin cos where e d sin 40 R sin cos r 4A 4 sin cos R r A...() From the above expression, it is clear that the resultant intensity depends on the two factors. (a). Diffraction term sin A :This factor is same as in the case of single slit. Thus this term corresponds to diffraction pattern due to secondary waves from the two slits individually. This term gives a central maximum having alternately minima and subsidiary maxima of decreasing intensity on either side. The direction of minima are given by, esin sin 0 or m or m esin m, m 1,, (3) and the position of secondary maxima approach to 3 /, 5 /, 7 /... cos (b). Interference term :This term corresponds to the two diffraction patterns coming out from the two slits. This term gives a set of equidistant dark and bright fringes. The directions of the maxima are given by

41 or n or e d sin n cos 1 ore dsin n where n=0,1,,3,...(4) Thus in a direction θ = 0, the central maxima due to interference and diffraction coincide. The minima due to interference term is given by, cos 0 e d sin or n 1 / or n 1 / Missing Orders: In the double slit arrangement, we find that some of the interference maxima are missing. Since for the same value of θ, the following two relations hold true, e dsin n, Interference Maxima...(5) esin m, Diffraction Minima...(6) When both the above conditions are satisfied simultaneously, then the interference maxima will be absent in the direction for which θ is common. From eq. (5) and (6) we have, e d e n m Now the following conditions may be considered: 1. If d = e then n = m =,4,6,... Hence the nd,4 th,6 th etc. order interference maxima will be absent.. If d = e, then n = 3m = 3,6,9... Thus 3 rd, 6 th, 9 th order of the interference maxima will be absent. 3. If e +d =e, i.e. d = 0, then the diffraction pattern will similar to as observed to a single slit of width equal to e. 41

42 Cos β -π -π 0 β π π 4A sin αcos β/α θ Effect of Increasing the slit width e: On increasing the slit width e, the central peak will become sharper, but the fringe spacing remains unchanged. Hence less interference maxima fall within the central diffraction maximum. Effect of increasing the distance between slits d: On increasing the separation between the slit d keeping the slit width constant, the fringes become closer together but the 4

43 envelope of the pattern remains unchanged. Hence more interference maxima fall within the central envelope. Effect of increasing the wavelength : When the wavelength of monochromatic light falls on the slit increases, the envelope becomes broader. As a result, the fringes move farther apart. Diffraction Due to N-Slits (Plane Transmission Grating): A diffraction grating consists of a large no. of parallel slits of equal width and equal separation. It generally consists of to lines per inch. Theory: The diffraction grating is equivalent to N-slits arrangements and the diffraction pattern we obtain will be the combined effect of all such slits. Let e be the width of each slit and d that of opaque part between them, then (e + d) is known as grating element. X L (e+d) S 1 S θ M 1 P M S N 43 M N - 1 Y

44 The diffracted rays from each of the slits are allowed to fall on a convex lens which focuses all of them at a point P on the screen. As in the single slit, the waves diffracted from each slit are equivalent to a single wave of amplitude, where R Asin esin.(1) () The path difference between any two consecutive waves from two slits ( e d)sin. Therefore, the corresponding phase difference will be ( e d)sin. Since the phase difference is constant between any two consecutive waves it can be taken as ( e d)sin (3) Thus we have to find out the resultant of N vibrations in a direction θ and each vibration is of amplitude Asin Now similar to the resultant of n-harmonic waves, the resultant of N-slits may be given as, N Rsin Rsin N Asin sin N R'. sin sin sin Thus the resultant intensity at P may be given as,. 44

45 I R' A sin sin. sin N..(4) A sin Here the factor is the intensity factor due to a single sin N slit while sin is due to the interference from all the N-slits. Principle Maxima: for maximum intensity, we have, sin 0or n, n 0,1,... sin N 0 But in this condition sin 0 is in indeterminate form. Thus to find its value we adopt the differential calculus method and then we get, lim n I sin N sin A sin N 45 N. Thus, (5) The intensity at these maxima is maximum and that is why it is principal maxima. Therefore, the condition for principal maxima is sin 0 or n or ( e d) sin n or e d n sin.(6) Now for n = 0, we get θ = 0, which gives the direction of the central order maxima. The values n = 1,,3.. correspond to the first, second, third.order maxima.

46 Minima: For sin N β =0. But sin 0. We get the minimum intensity. A series of minima, thus, obtained for m N m or N or N ( e d) sin m N( e d) sin m.(7) Thus for all integral values of m except 0,N,N,3N.we get a minima, because for these values sin β = 0 and this will give the position of principal maxima. Between m = 0 and N, i.e. two maxima the no. of minima exist for m= 1,,3,,(N-1). Since maxima and minima are obtained alternately there will be (N-) other maxima. These (N-),maxima are known as secondary maxima and the position of these maxima can be obtained by differentiating eq. (4) w.r.t. β and equating it with zero. Thus, di d Asin sin N N cos N sin sin N cos. 0 sin sin or N cos N sin sin N cos 0 N tan tan N..(8) or Therefore the roots of this equation except for which n (central maxima) correspond to the positions of secondary maxima. 46

47 If we construct a right angle triangle with its sides as 1, N tan β and sin N 1 N 1 tan then we have N tan N tan.(9) 1 N tan Nβ N tan β Hence from eq. (9) the intensity of secondary maxima may be given as A sin N tan I'. 1 N tan sin A I' A I' sin sin N cos N sin sin N N 1sin (10) Hence from (5) and (10) we obtain I' N Intensityof secondary max ima I 1 N 1 sin Intensityofprincipal max ima 47

48 Intensity Central Maxima Intensity Distribution Curve Condition For Absent Spectra With a diffraction grating: For a grating the direction of n th order principal maxima is given by ( e d) sin n...(1) And the direction of minima in the diffraction pattern due to a single slit is given by the eq. esin m...() The resultant intensity in the direction θ is given by A sin sin N I. sin...(3) 48

49 esin ( e d)sin Where, and A sin In eq. (3), is the diffraction term and represents the sin N intensity due to diffraction at a single slit while sin is interference term representing the intensity of waves interfering from all the N slits. Therefore, if in any direction, diffraction term is zero and interference term has maximum value, the principal maxima will not be present in that direction. Form (1) and (3) we get, e d n e d e m or n m e...(4) Eq (4) represents the condition for the n th order to be absent from the grating spectra. Thus, 1. If d = e then n = m =,4,6,... Hence the nd,4 th,6 th etc. order interference maxima will be absent.. If d = e, then n = 3m = 3,6,9... Thus 3 rd, 6 th, 9 th order of the interference maxima will be absent. Maximum No. Of Orders Of The Spectra With A Grating: In a diffraction grating the direction of the nth order principal maxima for wavelength λ is given by or ( e d) sin n ( e d)sin n Thus the maximum no. of possible orders n max. is given by ( e d) nmax (sin ) max 1 49

50 Dispersive Power Of A Diffraction Grating: The dispersive power of diffraction grating is defined as the rate of change of the angle of diffraction with the wavelength of light. It is expressed as d d. ( e d) sin For a grating we have Differentiating it w.r.t. λ we get, d ( e d)cos n d d n d ( e d)cos or n Linear Dispersive Power: Linear dispersive power is defined dx as, where dx is the linear separation between two d wavelengths differing each other by dλ. If the focal length of the lens used is f then. Linear dispersive power dx d f Angular dispersive power n f ( e d)cos Characteristics Of Grating Spectra: 1. Spectra of different orders are situated symmetrically on both sides of zero order image.. Spectral lines are almost straight and quite sharp. 3. Spectral colours are in order from violet to red. f d d 50

51 4. Most of the intensity goes to zero order and rest is distributed among the other orders. 5. The lines are more and more dispersed as we go to higher orders. Resolving Power Of An Optical Instrument: The ability of an optical instrument to form separate images of two objects, placed very close to each other, is known as its resolving power. The minimum separation between two objects up to which they can be seen as distinct objects by an optical instrument is called the limit of resolution of that instrument. Rayleigh s Criterion Of Resolving: According to Rayleigh Two nearby point objects are just resolved by an optical instrument when the principal maxima in the diffraction pattern of one falls over the first minimum in the diffraction pattern of the other and vice versa. The same thing happens in case of spectral lines of two different wavelengths, the lines are resolvable when the principal maximum due to one wavelength falls over the first minimum due to the other or vice versa. 51

52 Principle Maxima Principle Maxima λ 1 λ Easily Resolvable DIP Resultant Intensity curve λ 1 λ Just Resolvable λ 1 λ Not Resolvable 5

53 Resolving Power Of A Grating: The resolving power of a grating is defined as its ability to form separate diffraction maxima of two wavelengths which are close to each other. If d λ is the smallest difference in two wavelengths, which can be just resolved by a grating and λ is the wavelength of either of them or mean wavelength, then λ /d λ is known as the resolving power of the grating. Expression for resolving power: A X P dθ n P 1 θ n Central Image B Y Let (e + d) be the grating element and N the total number of slits. Let P 1 be position of the nth principal maximum of spectral line of wavelength λ while nth principal maximum due to wavelength (λ + d λ) be at P. 53

54 Now according to Rayleigh s criterion of resolution the two wavelengths can be resolved if the position of P coincides with the first minimum due to wavelength λ. In fig. the dotted line represents the diffraction pattern due to wavelength (λ + d λ) and the solid line represents the diffraction pattern due to wavelength λ. Now the principal maximum of wavelength λ in a direction θ n is given by ( e d ) sin n n...(1) And the equation of minimum is given by N ( e d ) sin n m...() Where m has all integral values except 0, N, N...nN, because for these values of m the condition for maxima is satisfied. Therefore, the first minimum adjacent to nth principal maximum in the direction ( n d n ) may be obtained by putting m=nn +1. Thus the first minimum in the direction ( n d n ) is given by N( e d)sin( n dn ) ( nn 1)...(3) And the principal maximum of (λ + d λ) in the direction ( ) n d n may be given as ( e d)sin( n dn) n( d)...(4) or N( e d)sin( n dn ) nn( d)...(5) comparing (3) and (5), we get ( nn 1) nn( d) or also from eq. (1) we have / d nn...(6) 54

55 n e Hence (6) can be modified as d sin n N e d sinn / d...(7) Here N(e + d) is the total width of the grating. Resolving Power Of A Telescope: A telescope is used to see distant objects. The ability of telescope to form separate images of two distant point objects situated close to each other is known as its resolving power. The resolving power of a telescope is measured by means of the angle subtended by two nearby point objects at its objective; it does not depend on the linear separation between the objects. When the images of the nearby distant objects are just resolved by the telescope then the angle subtended by these two objects at the telescope objective is called the limit of resolution of the telescope. Thus the resolving power of telescope is defined as the reciprocal of the smallest angle subtended at the objective by the two distant objects, the images of which are just seen as separate ones through a telescope. Expression for resolving power: dθ A dθ O 55 C P P 1

56 Let a be the diameter of the objective of the telescope. Consider the incident ray of light from two neighbouring points of a distant object. The image of each point is a Fraunhofer diffraction pattern. According to Rayleigh, these two images are said to be resolved if the position of the central maximum of the second image coincides with the first minimum of the first image and vice versa. The path difference between the secondary waves traveling in the directions AP 1 and BP 1 is zero and hence they reinforce with one another at P 1. Similarly, all the secondary waves from the corresponding points between A and B will have zero path difference. Thus, P 1 corresponds to the position of the central maxima of the first image. The secondary waves traveling in the directions AP 1 and BP will meet at P on the screen. Let the angle P AP 1 be dθ. The path difference between the secondary waves travelling in the directions BP and AP 1 is BC. From fig. ΔABC, BC AB sin d ABd a. d 56

57 a. d, the position of P If this path difference corresponds to the first minimum of the first image. But P is also the position of the central maximum of the second image. Thus, Rayleigh s condition of resolution is satisfied if a. d or d a...(1) The whole aperture AB can be considered to be made up of two halves AO and OB. The path difference between the secondary waves from the corresponding points in the halves will be. All the secondary waves destructively interfere with one another and hence P will be the first minimum of the first image. The eq. d a holds good for rectangular apertures. For circular aperture, this eq., according to Airy, can be written as 1. d a...() where λ is the wavelength of light and a is the aperture of the telescope objective, which is equal to the diameter of the metal ring in which the objective lens is mounted. Here dθ refers to the limit of resolution of the telescope. The reciprocal of dθ measures the resolving power of the telescope. 57

58 1 a d 1....(3) Resolving Power Of Microscope: The ability of a microscope to form separate distinct images of two nearby small objects, which cannot be seen by necked eye, is known as its resolving power. The limit of resolution of a microscope is the smallest separation between the two objects when their images are just resolved. B M A α O A N B Expression for resolving power: In fig. MN is the aperture of the objective of a microscope and A and B are two object points at a distance d apart. A and B are the corresponding Fraunhofer diffraction patterns of the two images. A is the position of the central maximum of A and B is the position of the central maximum of B. A and B are surrounded by alternate dark and bright diffraction rings. The two images are said to be just resolved if the position of the central maximum B also corresponds to the first minimum of the image of A. The path difference between the extreme rays from the point B and reaching A is given by 58

59 But NA = MA path ( BN NA') ( BM MA') difference BN MN D α B A α C α α M O In fig. AD is perpendicular to DM and AC is perpendicular to BN BN BM ( BC CN) ( DM DB) But CN AN AM DM path difference BC DB From Δs ACB and ADB and Path difference BC DB d sin AB sin d sin ABsin d sin If the path difference d sin Then limit of resolution d sin, for self- Thus resolving power luminous object. R. P d sin 1. ( N. A.) 1.

60 If a medium of refractive index μ is introduced between the 1 sin ( N. A.) object and objective then, R. P. d Resolving Power Of A Prism: Resolving power of prism is defined as its ability to form separate images of two neighbouring wavelengths. L 1 α α A α α a δ dδ L α S α t I I 1 α α Expression for resolving power: Let S is a source of light; L 1 is a collimating lens and L in the telescope objective. As the two wavelengths, λ and λ + d λ are very close, if the prism is set in the minimum deviation position it would hold good for both the wavelengths. 60

61 The final image I 1 corresponds to the principal maximum for wavelength λ and image I corresponds to the principal maximum for wavelength λ +d λ. I 1 and I are formed at the focal plane of the telescope objective L. The face of the prism limits the incident beam to a rectangular section of width a. Hence, the Rayleigh criterion can be applied in the case of a rectangular aperture. In the case of diffraction at a rectangular aperture, the position of I will corresponds to the first minimum of the image I 1 for wavelength λ provided ad or d...(1) Here δ is the angle of minimum deviation for wavelength λ. From fig. A A or But a sin sin A sin cos a sin l A a cos l A...() 61

62 Also sin A t l...(3) A sin In the case of a prism A sin A A sin sin...(4) Here μ and δ are dependent on wavelength of light λ. Differentiating eq. (4) w.r.t. λ 1 A d d cos sin A d d Substituting the values from eq. () and (4) or 1 a d d t l d d l d d a t. d d....(6) substituting the values of dδ from eq. (1) The expression d d t. d d measures the resolving power of the prism. Polarisation Of Light: In transverse waves there can be a no. of directions perpendicular to the direction of propagation in which the particles of the medium can execute periodic vibrations. Transverse waves in which the vibrations are restricted to one particular direction are known as polarised waves and the phenomenon is as polarisation. 6

63 Unpolarised Light: The light having vibrations along all possible straight lines perpendicular to the direction of propagation of light is known as Unpolarised Light. Polarised Light: The light, which has acquired the property of one-sidedness, is called polarised light. Therefore, the polarised light is not symmetrical about the direction of propagation but its vibrations are confined only to a single direction perpendicular to the direction of propagation. The crystal which makes the light polarised is known as polariser and the crystal, which analyses the incoming polarised light, is called analyser. 1. Plane Polarised Light: Light is said to be plane polarised when vibrations take place only in one direction parallel to the plane through the axis of a beam.. Circularly Polarised Light: Light is said to be circularly polarised when the vibrations in transverse plane are circular having constant period. Here the amplitude of vibrations remains constant but the direction changes only. 63

64 3. Elliptically Polarised Light: Light is said to be elliptically polarised when the vibrations are elliptical and having a constant period and takes place in a plane perpendicular to the direction of propagation. Here the amplitude of vibrations changes in magnitude as well as in direction. Plane of Polarisation and Plane of Vibration: The plane in which the vibrations occurs known as plane of vibration and a plane perpendicular to plane of vibration in which no vibration occurs is known as plane of polarisation. A E D F G G B B C C H H 64

65 ABCD EFGH Plane of Vibration Plane of Polarisation Brewster s law: In 1811, Brewster discovered a simple relation between angle of polarisation and refractive index μ of the medium. He found that the refractive index of the material medium is equal to tangent of the angle of polarisation. i.e. tani p...(1) A B i p i p X O r θ Y Thus when angle of incidence becomes equal to the angle of polarisation, then from Snell s law, sin i p sin r...() Thus from (1) and (), cos i p sin r or sin( 90 i p ) sin r 65 C

66 0 or r i p 90 Also from fig. if θ be the angle between reflected and refracted rays then, ( r i ) p Therefore, the reflected and refracted rays are at right angle to each to other. Double Refraction: When an ordinary light beam after passing through a crystal splits up into ordinary and extraordinary light the crystal is known as double refracting crystal and the phenomenon is known as Double Refraction. Incident Ray r e r o E-Ray O-Ray Ordinary and Extraordinary Rays: When ordinary light beam when passed through a uniaxial crystal splits up into two refracted rays. One of them obeys laws of refraction known as (o-ray) ordinary ray while the other does not obey laws of refraction known as (E-ray) extraordinary ray. These two rays have the following characteristics: 1. The plane of polarisation for O-ray is same as the principle plane. Therefore, it has vibrations perpendicular to the principal plane. For E-ray the case is opposite.. The two rays emerge from the crystal along the parallel directions. 66

67 3. The refractive index of the material for the two rays is given as: sin i sin i o sin r and e o sin re The value of μ o is constant while μ e varies with the angle of incidence. 4. In a crystal if μ o >μ e the crystal is known as negative crystal in which the velocity of O-ray is less than E- ray. 5. In a crystal if μ o <μ e the crystal is known as positive crystal in which the velocity of E- ray is less than O-ray. 6. Since the refractive index of O- ray is constant, hence it travels with same speed in all directions. 7. Unpolarised light incident along optic axis does not split into O- ray and E- ray. 8. The difference between the refractive indices of O- ray and E- ray is known as birefringence i.e. birefringence = o e. NICOL PRISM: It was invented by William Nicol in 188. It is an excellent optical device used for producing and analyzing the plane polarised light. Principle: 1. It works on the phenomenon of double refraction.. When Unpolarised light is passed through uniaxial crystal, it splits up into ordinary and extraordinary light. 3. If by some optical method one of the two rays is eliminated the ray, emerging through the crystal will be plane polarised. 4. In Nicol prism O- ray is eliminated by total internal reflection and E- ray is transmitted through the crystal. 67