Polarisation PDF - B.Sc IV Sem Physics

B.Sc IV Sem POLARISATION A D B F G C, H HALLI Subject: PHYSICS B.Sc. IV Semester Paper 4: Physical Optics, F...

B.Sc IV Sem POLARISATION A D B F G C, H HALLI Subject: PHYSICS B.Sc. IV Semester Paper 4: Physical Optics, Fibre Optics and Computational Physics Polarization of light (10 hrs): Polarization, methods for obtaining polarized light. Double refraction in uniaxial crystals, Huygens’s theory, positive and negative crystals and principal refractive indices. Huygens’ construction of O and E rays in uniaxial crystals for plane wave front. Quarter and half wave plates. Production and detection of plane, circularly and elliptically polarized light. Babinet compensator (qualitative). Optical activity; specific rotation, Fresnel’s theory and Laurent’s half shade polarimeter. Polarization Fundamentals of polarization: The experiments on interference and diffraction have shown that light is form of wave motion. These effects do not tell us about the type of wave motion that is whether the light waves are longitudinal or transverse or whether the vibrations are linear or circular. The phenomenon of polarization has helped to establish beyond doubt that light waves are transverse. Transverse nature of light waves: The transverse nature of light can be easily understood by the following experiment. Let A and B are two tourmaline crystals cut parallel to their crystallographic axis. When light from a source falls on a crystal A which passes through the crystal. On rotating the crystal A, no change in the intensity of light is observed that is the transmitted light remains same. If the emergent beam of light is further passed through a similar crystal B with its axis parallel to the first, light is almost transmitted through the second crystal also. Suppose both the crystals are rotated simultaneously so that their axis is always parallel to each other. No change is observed in the light coming out of B (fig b). Keeping the crystal A fixed and rotate the crystal B about the beam as axis. It will be seen that the intensity of the emergent beam decreases (fig c), when the axes of both the crystals are at right angles to each other no light comes out of the crystal B (fig d). If the crystal B is further rotated the intensity gradually increases and becomes maximum, when A and B are parallel again. B.Sc IV Sem POLARISATION A D B F G C, H HALLI The variation in intensity of the emergent light shows that light waves are transverse in nature. If the wave is longitudinal no change in the intensity of the emergent light would be observed when the crystal B is rotated. Light from the source has vibrations in all directions it is called “unpolarised light”. When this unpolarised light is incident on the crystal A which absorbs all the vibrations except those vibrations parallel to its axis are transmitted. Thus the light after emerging from A contains the vibrations only in one direction. “Therefore the light which has acquired the property of one sidedness is called Polarized light. The phenomenon is known as Polarization”. The plane in which vibrations occurs is called plane of vibration, the plane in which no vibration occurs is called plane of polarization. Methods for obtaining polarized light: The following are the methods of producing plane polarised or linearly polarised light (1) Reflection (2) Refraction (3) Double Refraction and (4) Selective absorption Polarisation by reflection: This method of producing polarised light was discovered by Malus in 1808. He found that light reflected from the surface of glass was either partially polarised or completely polarised, the degree of polarisation depends upon the angle of incidence. Polarisation by refraction: When unpolarised light is incident on the first glass of pile of plates consisting of about 20 thin glass plates placed parallel to one another, 15% of the incident vibrations reflects, and transmits 85% of them. This process is repeated at each glass plate and finally the beam emerging out of the last plate will be plane polarised. Polarisation by double refraction: When a beam of unpolarised light is made to pass through certain crystals like calcite or quartz, on passing through the crystal the beam splits up into two refracted rays one is called ordinary ray and the other is extraordinary ray. Both the rays are plane polarised with their plane vibrations are mutually perpendicular to each other. B.Sc IV Sem POLARISATION A D B F G C, H HALLI Polarisation by selective absorption: There are certain crystals like tourmaline which are doubly refracting having a special property of absorbing the ordinary ray and the extra ordinary ray to different extents. This property is called selective absorption or dichroism and the crystal exhibiting this property are called dichroic crystals. This property can be used to produce plane polarised light. Double refraction in uniaxial crystals: When a beam of ordinary unpolarised light is allowed to pass through a calcite or quartz crystal there are two refracted beams. The crystal having this property are said to be doubly refracting and the phenomenon is called double refraction. One of the ray is called ordinary ray and the other ray is called extra-ordinary ray. “The ray which obeys ordinary laws of refraction is known as ordinary ray”. “The ray which does not obey the ordinary laws of refraction is called extraordinary ray”. Optic axis: In calcite crystal at the two diametrically opposite corners A and B the angles of the three faces meeting there are all obtuse (1020) while at its remaining six corners one angle is obtuse and two are acute (780). The corners A and B are known as blunt corners if a line is drawn through one of the corners (A or B) of the crystal so that it makes equal angles with all faces. This line gives the direction of the optic axis. If light incidents along this direction both the refracted rays travels with same velocity and therefore double refraction does not takes place. “Therefore the direction along which no double refraction take place is called optic axis”. Uniaxial and Biaxial crystals: A class of crystal in which only one optic axis along which no double refraction take place is called uniaxial crystal. Examples: Calcite, Quartz, Ice, Nitrate of Soda etc. A class of crystal in which there are two optic axis directions along which no double refraction take place is called biaxial crystal. Example: Mica, Borax, Topaz, Aragonite, Selenite etc. Huygen’s wave theory of double refraction: Huygens assumed that a wave which is propagated in a uniaxial crystal has two wave surfaces one within the other. He found that the wave front of the ordinary ray which obeys ordinary laws of refraction and travels with the same velocity in all directions in a crystal is a sphere and that of the extra-ordinary ray which does not obey the laws of refraction and travels with different velocities in different directions is an ellipsoid. Since no double refraction occurs along the optic axis the surfaces of the sphere and ellipsoid touch in two points the line joining which gives the direction of the optic axis of the crystal. B.Sc IV Sem POLARISATION A D B F G C, H HALLI Positive and Negative Crystals: In some crystals like quartz in which the E-ray travels slower than O- ray the sphere is outside the ellipsoid (fig A). Such crystals are known as positive crystals in which 𝜇𝑒 > 𝜇𝑜. There some crystals like calcite, in which the E-ray travels faster than O-ray, the sphere is inside the ellipsoid (fig B). Such crystals are known as negative crystals in which 𝜇𝑒 < 𝜇𝑜. Principle refractive indices: From the preceding sections we find that for any crystal the velocity of the ordinary ray is the same in all directions while that of the extra-ordinary ray is different in different directions but along the optic axis both have the same velocity. From this we conclude that in uniaxial crystals, there are two principal refractive indices, one corresponding to the velocity of the ordinary wave and the other to the velocity of the extra-ordinary wave, when travelling normal to the optic axis. In all types of uniaxial crystals, the refractive index for the ordinary ray is defined as 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑎𝑖𝑟 𝜇0 = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑜𝑟𝑑𝑖𝑛𝑎𝑟𝑦 𝑤𝑎𝑣𝑒 The principal refractive index of extra-ordinary wave in negative uniaxial crystals is expressed as 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑎𝑖𝑟 𝜇𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐸 𝑤𝑎𝑣𝑒 The principal refractive index of extra-ordinary wave in positive uniaxial crystals is expressed as 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑎𝑖𝑟 𝜇𝑒 = 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝐸 𝑤𝑎𝑣𝑒 Huygens’s construction of O and E wave front: In all the following cases AB represents the trace of an incident plane wave front perpendicular to the plane of the paper. It is incident on the refracting face XXl of a doubly refracting negative crystal (calcite). Case 1: Optic axis lying in the plane of incidence and inclined to the refracting surface. Oblique incidence:- AB is the incident wave front which meet obliquely at the surface XXl of the crystal (Fig 1) first at the point 𝐵𝐶 A and makes it the centre of ordinary and extra-ordinary wavelets. During the time 𝑡 = in which the 𝑉𝑎 disturbance from B reaches, the ordinary spherical wavelet has travelled a distance equal to B.Sc IV Sem POLARISATION A D B F G C, H HALLI 𝐵𝐶 𝐵𝐶 𝐵𝐶 𝑡 × 𝑉0 =. 𝑉0 = = 𝑉𝑎 𝑉𝑎 /𝑉0 𝜇0 Where 𝑉𝑎 the velocity of light in air , 𝑉0 is the velocity of light of O-ray and 𝜇0 is the R.I of O-ray. To find the 𝐵𝐶 position of the O-wave front in the crystal draw a sphere with A as the centre and as the radius. From C 𝑉𝑎 draw a tangent plane CO touching the sphere at O. A plane passing through CO represents the position of the O-wave front. Similarly for extra-ordinary ray the distance travelled is 𝐵𝐶 𝐵𝐶 𝐵𝐶 𝑡 × 𝑉𝑒 =. 𝑉𝑒 = = 𝑉𝑎 𝑉𝑎 /𝑉𝑒 𝜇𝑒 Where 𝑉𝑎 the velocity of light in air, 𝑉𝑒 is the velocity of light of E-ray and 𝜇𝑒 is the R.I of E-ray. The tangent plane CE touching the ellipsoid at E represents the position of the E-wave front. It is clear that O-ray and E- ray travel along different directions with different velocities. Normal incidence:- As in the figure 2 draw tangent planes OOl and EEl respectively to the spherical and ellipsoidal surface originating from the centers A and B where the incident waves strike the crystal simultaneously. The planes though OOl and EEl represent the O and E wave fronts in the crystal. O and E rays travel in different directions with different velocities. Case 2: Optic axis lying in the plane of incidence and perpendicular to the refracting surface. Oblique incidence:- In this case the two wave fronts of ordinary and extra-ordinary rays touch each other at P (Fig 1) the line joining AP represents the direction of the optic axis. Proceeding in the same manner as in case 1 above, the plane passing through CO and CE respectively represent the wave front of the O and E waves. The O and E rays travel with different velocities in different directions. Normal incidence:- O and E waves coincide at all instant (Fig 2) and travel in the same direction with the same velocity. Hence there is no double refraction in this case. Case 3: Optic axis lying in the plane of incidence and parallel to the refracting face. B.Sc IV Sem POLARISATION A D B F G C, H HALLI Oblique incidence:- In this case the optic axis lies along the line XXl as in Fig. The two secondary surfaces, the sphere and the ellipsoid touch on this line (Fig 1). The position of the O and E wave fronts CO and CE are drawn on exactly the same consideration as in case 1. The O and E refracted wave fronts travel with different velocities along different directions. Normal incidence:- In this case OOl and EEl are the refracted wave fronts at the same instant of time (Fig 2), they are parallel to each other and travel in the same direction but with different velocities, thus producing path difference between the O and E waves on emergence, but there is no separation into two beams. This property is made use in the construction of quarter wave and half wave plates. Theory of plane, elliptically and circularly polarized light: Suppose the amplitude of the incident plane polarized light in the crystal is A and it makes an angle 𝜃 with the optic axis (Fig).Therefore the amplitude of the ordinary ray vibrating along PO is 𝐴 sin 𝜃and the amplitude of the extraordinary ray vibrating along PE is 𝐴 cos 𝜃. Since a phase difference 𝛿 is introduces between the two rays, after coming out of the crystal can be represented in terms of two simple harmonic motions, at right angles to each other and having a phase difference. For extraordinary ray, 𝑥 = 𝐴 cos 𝜃. sin(𝜔𝑡 + 𝛿) For ordinary ray, 𝑦 = 𝐴 sin 𝜃. sin 𝜔𝑡 Taking 𝐴 cos 𝜃 = 𝑎 𝑎𝑛𝑑 𝐴 sin 𝜃 = 𝑏, we have, 𝑥 = 𝑎 sin(𝜔𝑡 + 𝛿) … … …. (1) 𝑦 = b sin 𝜔𝑡 … … … (2) From equation (2), 𝑦 = sin 𝜔𝑡 𝑏 𝑦2 cos 𝜔𝑡 = √1 − 𝑏2 𝑥 From equation (1), 𝑎 = sin 𝜔𝑡. cos 𝛿 + cos 𝜔𝑡. sin 𝛿 Substituting the values of 𝑠𝑖𝑛𝜔𝑡 𝑎𝑛𝑑 𝑐𝑜𝑠𝜔𝑡, we get, 𝑥 𝑦 𝑦2 =. 𝑐𝑜𝑠𝛿 + √1 − 2 𝑠𝑖𝑛𝛿 𝑎 𝑏 𝑏 𝑥 𝑦 𝑦2 −. 𝑐𝑜𝑠𝛿 = √1 − 2 𝑠𝑖𝑛𝛿 𝑎 𝑏 𝑏 On squaring, we have, B.Sc IV Sem POLARISATION A D B F G C, H HALLI 2 2 2 𝑥 𝑦 2𝑥𝑦 𝑦 2 + 2 𝑐𝑜𝑠 2 𝛿 −. 𝑐𝑜𝑠𝛿 = (1 − 2 ) 𝑠𝑖𝑛2 𝛿 𝑎 𝑏 𝑎𝑏 𝑏 𝑥2 𝑦2 2 2𝑥𝑦 𝑦2 + 𝑐𝑜𝑠 𝛿 −. 𝑐𝑜𝑠𝛿 = 𝑠𝑖𝑛 𝛿 − 2 𝑠𝑖𝑛2 𝛿 2 𝑎2 𝑏 2 𝑎𝑏 𝑏 𝑥2 𝑦2 2 𝑦2 2 2𝑥𝑦 2 + 2 𝑐𝑜𝑠 𝛿 + − 2 𝑠𝑖𝑛 𝛿 −. 𝑐𝑜𝑠𝛿 = 𝑠𝑖𝑛2 𝛿 𝑎 𝑏 𝑏 𝑎𝑏 𝒙𝟐 𝒚𝟐 𝟐𝒙𝒚 + −. 𝒄𝒐𝒔𝜹 = 𝒔𝒊𝒏𝟐 𝜹 … … … ….. (𝟑) 𝒂𝟐 𝒃𝟐 𝒂𝒃 This is general equation of an ellipse. Special cases: (1) When 𝛿 = 0, equation (3) becomes, 𝑥 2 𝑦 2 2𝑥𝑦 + − =0 𝑎2 𝑏 2 𝑎𝑏 𝑥 𝑦 2 ( − ) =0 𝑎 𝑏 𝑥 𝑦 = 𝑎 𝑏 𝑏 𝑦 =.𝑥 𝑎 This is the equation of a straight line. Therefore, the emergent light will be plane polarized. 𝑥2 𝑦2 (2) When 𝛿 = 𝜋/2, equation (3) becomes, 𝑎2 + 𝑏2 = 1 This represents the equation of symmetrical ellipse. The emergent light in this case will be elliptically polarized provided 𝑎 ≠ 𝑏. (3) When 𝛿 = 𝜋/2, and 𝑎 = 𝑏, equation (3) becomes, 𝑥 2 + 𝑦 2 = 𝑎2 This represents the equation of circle of radius 𝑎. The emergent light will be circularly polarized. Here the vibration of the incident plane polarized on the crystal make an angle of 450 with the direction of the optic axis. Quarter wave plate: It is a uniaxial doubly refracting crystal plate, of suitable thickness, cut with its optic axis parallel to the refracting face and can introduce a phase difference of 𝜋/2 (or path difference of 𝜆/4) between the O and E-rays. When a beam of plane polarized monochromatic light of wavelength 𝜆 is incident normally on a crystal of this type it decomposes into O and E components which travel in the same direction but with different velocities. If ‘t’ be the thickness of the plate then the distance travelled in the crystal is equivalent to distances 𝜇0 𝑡 𝑎𝑛𝑑 𝜇𝑒 𝑡 in air for the O and E rays respectively. Therefore the resultant path difference between the two components will be 𝑃. 𝐷 = 𝑡(𝜇0 − 𝜇𝑒 ) We know that the relation between phase difference and path difference is given by 2𝜋 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =. 𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜆 B.Sc IV Sem POLARISATION A D B F G C, H HALLI 𝜋 To introduce a phase difference of 2 between O and E rays, we have, 𝜋 2𝜋 =. 𝑡(𝜇0 − 𝜇𝑒 ) 2 𝜆 𝝀 𝒕= 𝟒. (𝝁𝟎 − 𝝁𝒆 ) This is the expression for the thickness of the quarter wave plate. Similarly for +ve crystal (Quartz), the expression for thickness of the quarter wave plate is 𝝀 𝒕= 𝟒. (𝝁𝒆 − 𝝁𝟎 ) When plane polarized light is incident on a quarter wave plate, the emergent light is in general elliptically polarized. If plane of polarization of the incident beam makes an angle of 45 0 with the optic axis then the emergent beam is circularly polarized light. Thus quarter wave plate is the simplest device for producing and detecting circularly polarized light. In conjunction with a Nicol prism it is used for analysing all kinds of polarized light. Half wave plate: It is a uniaxial doubly refracting crystal plate, of suitable thickness, cut with its optic axis parallel to the refracting face and can introduce a phase difference of 𝜋 (or path difference of 𝜋/2) between the O and E-rays. When a beam of plane polarized monochromatic light of wavelength 𝜆 is incident normally on a crystal of this type it decomposes into O and E components which travel in the same direction but with different velocities. If ‘t’ be the thickness of the plate then the distance travelled in the crystal is equivalent to distances 𝜇0 𝑡 𝑎𝑛𝑑 𝜇𝑒 𝑡 in air for the O and E rays respectively. Therefore the resultant path difference between the two components will be 𝑃. 𝐷 = 𝑡(𝜇0 − 𝜇𝑒 ) We know that the relation between phase difference and path difference is given by 2𝜋 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =. 𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜆 To introduce a phase difference of 𝜋 between O and E rays, we have, 2𝜋 𝜋=. 𝑡(𝜇0 − 𝜇𝑒 ) 𝜆 𝝀 𝒕= 𝟐 (𝝁𝟎 − 𝝁𝒆 ) This is the expression for the thickness of the quarter wave plate. Similarly for +ve crystal (Quartz), the expression for thickness of the half wave plate is 𝝀 𝒕= 𝟐 (𝝁𝒆 − 𝝁𝟎 ) When a plane polarised light is incident upon a half wave plate, the transmitted light is also plane- polarised. If the plane of polarisation of the incident light makes an angle 𝜃 with the direction of the optic B.Sc IV Sem POLARISATION A D B F G C, H HALLI axis, the plane of polarisation of the emergent beam makes an angle – 𝜃 with the same direction that is the plane has effectively been rotated through an angle 2𝜃. A half wave plate finds its application in the construction of Laurent’s half shade device used in a polarimeter. Production of Plane, Elliptical and Circularly polarized light: Plane polarized light: A beam of monochromatic light is passed through a Nicol prism, while passing through the prism, the beam split up into O-ray and E-ray. The O-ray is totally internally reflected back at the Canada balsam layer, while the E-ray passes through the N-prism. The emergent beam is plane polarized. Circularly polarized light: Circularly polarized light is the resultant of two waves of equal amplitude, vibrating at right angles to each other and having a phase difference of 𝜋/2. A parallel beam of monochromatic light is allowed to fall on a Nicol prism the beam after passing through the N-prism is plane polarized now place another N-prism at some distance from the first one in the crossed position, that is the field of view will be dark as viewed by the eye in this position. A quarter wave plate is introduced between N-prisms and it is rotated until the field of view is dark, then from this position quarter wave plate is rotated through 450. In this case, the vibrations of the plane polarized light falling on the quarter wave plate makes an angle 450 with the direction of the optic axis of the quarter wave plate. The light which is emerging out from this quarter wave plate is circularly polarized. Elliptically polarized light: Elliptically polarized light is the resultant of two waves of unequal amplitude vibrating at right angles to each other and having a phase difference of 𝜋/2. A parallel beam of light is allowed to fall on an N-prism the beam after passing through the N-prism is plane polarized. Now place another N-prism at some distance from the first one in the crossed position so that field of view is dark as viewed by the eye in this position. A quarter wave plate is introduced between the two N-prisms. The plane polarized light from the first N-prism falls normally on the quarter wave plate. The light which is emerging out of QWP is elliptically polarized (the precaution should be taken that light should not incident at an angle of 450 with optic axis). Detection of plane, elliptical and circularly polarized light: Plane polarized light: The beam is allowed to fall on a N-prism. On rotation of the N-prism, if light extinguishes completely twice in each rotation, then the beam is plane polarized. Circularly polarized light: The beam is allowed to fall on an N-prism. The intensity of the beam remains uniform when the N- prism is rotated. The beam in this case is either circularly polarized or unpolarised. To distinguish between the two, the beam is allowed to fall on a quarter wave plate and then on an N- prism. If the beam is circularly polarised after passing through the quarter wave plate, the O-ray and E-rays will undergo a further phase difference of 𝜋/2. The beam after passing through the quarter wave plate becomes B.Sc IV Sem POLARISATION A D B F G C, H HALLI plane polarised. This beam is passed through a rotating N-prism. If the light extinguished completely twice in each rotation then the incident beam is circularly polarised light otherwise it is unpolarised. Elliptically polarised light: When elliptically polarised light is passed through a rotating N-prism, the intensity varies from maximum to minimum. This behaviour is same even for partially polarised light (combination of polarized and unpolarised). To distinguish between the two, the elliptically polarised light is passed through quarter wave plate and then on N-prism. If the beam is elliptically polarised the O-ray and E-ray will undergo a further phase difference of 𝜋/2. The beam after passing through the quarter wave plate becomes plane polarised. If this beam extinguishes completely twice in each rotation, then the incident beam is elliptically polarised otherwise it is partially polarised. Babinet’s compensator: It is an optical device employed in the production and analysis of elliptically polarised light. A quarter wave plate or a half wave plate produces only a fixed path difference between the ordinary and the extra-ordinary rays and can be used only for light of particular wavelengths. For different wavelengths different quarter or half wave plates are to be used. To avoid this difficulty, Babinet designed a compensator by means of which a desired path difference can be introduced. It consists of two wedge shaped sections A & B of quartz, they are mounted in such a way that it forms a rectangular block. The crystal A is fixed and B can slide along the surface of A with the help of micrometer screw (figure), by this the thickness of the optical path can be varied for desired value. Thus Babinets compensator will introduce any desired path difference and it can be used for light of any wavelength. Optical activity: When a plane polarised light is made to pass through a certain substances the plane of polarization of the emergent beam has been rotated through a certain angle. This phenomenon is known as optical activity and the substances which rotate the plane of polarization are said to be optically active. There are two types of optically active substances. Substances which produces clock wise rotation are known as dextrorotatory (right-handed) substances and the substances which rotate the plane of polarization in the anti-clock wise direction are called laevorotatory (left handed) substances. Examples: Aqueous solutions of various kinds of sugars and of tartaric acid, turpentine oil. Quartz etc. Fresnel’s explanation of optical rotation: (Fresnel’s theory) Fresnel’s explanation of optical rotation is based upon the assumption that when plane polarized light is allowed to pass through the substance, it decomposes into two circularly polarized vibrations rotating in opposite direction with the same frequency. In an optically inactive substance, as these two vibrations travels with the same velocity, the plane of emergence lies in the plane of the incidence hence there is no optical rotation (Fig 1). In dextrorotatory B.Sc IV Sem POLARISATION A D B F G C, H HALLI substance, right handed motion travels faster than the left handed (Fig 2) while in a laevorotatory substances left- handed motion travels faster than the right handed (Fig 3). As a result of this the rotation of the plane of polarization is to the right or left according as the right handed or the left handed component is faster and is equal to half the phase difference on emergence between the two circular vibration as shown in the figure. These results may be arrived easily by means of the equations of the vibration. Optically inactive substance: For clockwise circular vibration 𝑥1 = 𝑎𝑐𝑜𝑠𝜔𝑡; 𝑦1 = 𝑎𝑠𝑖𝑛𝜔𝑡 For anti-clock wise circular vibrations 𝑥2 = − 𝑎𝑐𝑜𝑠𝜔𝑡; 𝑦2 = 𝑎𝑠𝑖𝑛𝜔𝑡 Therefore the resultant vibrations are 𝑋 = 𝑥1 + 𝑥2 = 𝑎𝑐𝑜𝑠𝜔𝑡 − 𝑎𝑐𝑜𝑠𝜔𝑡 = 0 𝑎𝑛𝑑 𝑌 = 𝑦1 + 𝑦2 = 𝑎𝑠𝑖𝑛𝜔𝑡 + 𝑎𝑠𝑖𝑛𝜔𝑡 = 2𝑎𝑠𝑖𝑛𝜔𝑡 Thus the resultant vibration has amplitude 2a and is plane polarized. The plane of vibration is along the original direction. That is, along the plane of incidence hence there is no optical rotation (Fig 1). Optically active substance (dextrorotatory): For clockwise circular vibration 𝑥1 = 𝑎𝑐𝑜𝑠𝜔𝑡; 𝑦1 = 𝑎𝑠𝑖𝑛𝜔𝑡 For anti-clock wise circular vibrations 𝑥2 = − 𝑎𝑐𝑜𝑠(𝜔𝑡 + 𝛿); 𝑦2 = 𝑎𝑠𝑖𝑛(𝜔𝑡 + 𝛿) Therefore the resultant vibrations are 𝑋 = 𝑥1 + 𝑥2 = 𝑎𝑐𝑜𝑠𝜔𝑡 − 𝑎𝑐𝑜𝑠(𝜔𝑡 + 𝛿) 𝜔𝑡 + 𝜔𝑡 + 𝛿 𝜔𝑡 − 𝜔𝑡 − 𝛿 = −2𝑎𝑠𝑖𝑛 ( ). 𝑠𝑖𝑛 ( ) 2 2 𝛿 −𝛿 = −2𝑎𝑠𝑖𝑛 (𝜔𝑡 + ). 𝑠𝑖𝑛 ( ) 2 2 𝛿 𝛿 = 2𝑎𝑠𝑖𝑛 (𝜔𝑡 + ). 𝑠𝑖𝑛 ( ) … … ….. (1) 2 2 𝑎𝑛𝑑 𝑌 = 𝑦1 + 𝑦2 = 𝑎𝑠𝑖𝑛𝜔𝑡 + 𝑎𝑠𝑖𝑛(𝜔𝑡 + 𝛿) 𝜔𝑡 + 𝜔𝑡 + 𝛿 𝜔𝑡 − 𝜔𝑡 − 𝛿 = 2𝑎𝑠𝑖𝑛 ( ). 𝑐𝑜𝑠 ( ) 2 2 𝛿 −𝛿 = 2𝑎𝑠𝑖𝑛 (𝜔𝑡 + ). 𝑐𝑜𝑠 ( ) 2 2 𝛿 𝛿 = 2𝑎𝑠𝑖𝑛 (𝜔𝑡 + ). 𝑐𝑜𝑠 ( ) … … … … … …. (2) 2 2 B.Sc IV Sem POLARISATION A D B F G C, H HALLI From equations (1) and (2) it is clear that the plane of polarization of the emergent beam has been rotated in the clock wise direction (right hand side). The angle through which the plane has been rotated can be calculated by dividing equations (1) by (2) 𝛿 𝑋 sin 2 𝛿 tan 𝜃 = = = tan 𝑌 cos 𝛿 2 2 𝜹 ∴ 𝜽= 𝟐 That is the plane is rotated through half the phase difference between two circular vibrations. One can prove the same relation for laevorotatory substance. Specific rotation: Liquid containing an optically active substance rotate the plane of the linearly polarized light. The angle through which the plane polarized light is rotated depends upon, 1. The length of the solution 2. The concentration of the solution 3. The wavelength of light and 4. The temperature. 𝜃 ∝ 𝑙. 𝐶 𝜃 = 𝑡𝑆 𝜆. 𝑙. 𝐶 𝜆 𝜃 𝑡𝑆 = 𝑙. 𝐶 Hence the specific rotation of the solution is defined as “The amount of rotation produced by a solution of unit length having unit concentration at constant temperature and for light of given wavelength”. Laurent’s half shade polarimeter: It consists of two Nicol prisms polariser and analyser. Behind polariser there is half shade device of quartz which covers one half of the field of view while the other half is a glass plate. The glass plate absorbs same amount of light as the quartz plate. Polarimeter tube is a hollow glass tube having a large diameter at its middle portion. When this tube is filled with the solution there should be no air bubble in the path of light. If any air bubble will appear it should be brought at the upper portion of the wide bore of the tube. Light from a monochromatic source is incident on the converging lens. After passing through polariser the beam is plane polarised. One half of the beam passes through the quartz plate and the other half passes through the glass plate. Suppose the B.Sc IV Sem POLARISATION A D B F G C, H HALLI plane of vibration of the plane polarised light incident on the half shade plate is along AB. Here AB makes an angle 𝜃 with YYl. On passing through the quartz plate the beam split up into ordinary and extraordinary components and on emergence a phase difference of 𝜋 is introduced between them. The vibrations of the beam emerging out of quartz will beam along CD whereas the vibrations of the beam emerging out of the glass plate will be along AB. If the analyser has its principal plane along YYl, the amplitude of light incident on the analyser both the halves will be equal. Therefore the field of view will be equally bright. If the analyser is rotated to the right of YYl, then the right half will be brighter as compared to the left half. On the other hand if the analyser is rotated to the left of YYl the left half is brighter as compared to the right half. Therefore to find the specific rotation of an optically active substance, the analyser is set in the position for equal brightness of field of view first without the solution in the tube. The readings of the Vernier are noted. When a tube containing the solution of known concentration is placed, on introduction of the tube the field of view is not equally bright as set earlier. The analyser is rotated in the clockwise direction until the field of view is equally bright again. The new position of the Vernier reading is noted. The difference in the two readings gives angle through which the plane of polarisation of the incident beam has been rotated by the sugar solution. Experiment is repeated for various concentrations, the corresponding angles of rotation are determined. A graph is platted between concentration C and the angle of rotation 𝜃. The graph is a straight 𝜃 line. Then from the relation, 𝑆 = 𝑙 𝐶 the specific rotation of the optically active substance is calculated. Problems 1. Plane polarized light passes through a quartz plate with its optic axis parallel to the faces. Calculate the least thickness of the plate for which the emergent beam (i) will be plane polarized and (ii) will be circularly polarized light. Given 𝜇𝑒 = 1.553, 𝜇0 = 1.544 𝑎𝑛𝑑 𝜆 = 5000 Å. Solution: 𝜆 5000 × 10−10 𝐹𝑜𝑟 𝑝𝑙𝑎𝑛𝑒 𝑝𝑜𝑙𝑎𝑟𝑖𝑠𝑒𝑑 𝑙𝑖𝑔ℎ𝑡; 𝑡 = = = 2.778 × 10−5 𝑚 2(𝜇𝑒 − 𝜇0 ) 2 × (1.553 − 1.544) 𝜆 5000 × 10−10 𝐹𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟𝑙𝑦 𝑝𝑜𝑙𝑎𝑟𝑖𝑠𝑒𝑑 𝑙𝑖𝑔ℎ𝑡; 𝑡 = = = 1.388 × 10−5 𝑚 4(𝜇𝑒 − 𝜇0 ) 4 × (1.553 − 1.544) 2. Calculate the least thickness of a calcite plate which convert plane polarized into circularly polarized light, if the refractive indices of calcite for sodium light are respectively 1.658 and 1.486 for the ordinary ray and the extraordinary ray. Given 𝜆 = 589.3 𝑛𝑚 Solution: 𝜇0 = 1.658, 𝜇𝑒 = 1.486 𝑎𝑛𝑑 𝜆 = 589.3 𝑛𝑚. 𝜆 589.3 × 10−9 𝐹𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟𝑙𝑦 𝑝𝑜𝑙𝑎𝑟𝑖𝑠𝑒𝑑 𝑙𝑖𝑔ℎ𝑡; 𝑡 = = = 8.565 × 10−7 𝑚 4(𝜇0 − 𝜇𝑒 ) 4 × (1.658 − 1.486) 3. A crystal plate produces an optical path difference of 1.5 × 10−7 𝑚 between the ordinary and extraordinary vibrations calculate the thickness of the plate. Given 𝜇𝑒 = 1.554, 𝜇0 = 1.544 Solution: Expression for path difference is given by B.Sc IV Sem POLARISATION A D B F G C, H HALLI 𝑃. 𝐷 = 𝑡(𝜇𝑒 − 𝜇0 ) 𝑃. 𝐷 1.5 × 10−7 𝑡= = = 1.5 × 10−5 𝑚 (𝜇𝑒 − 𝜇0 ) (1.554 − 1.544) 4. Calculate the rotation of the plane of polarisation in a substance of unit thickness for a light of wavelength 5890 × 10−10 𝑚. The difference between the refractive indices for right and left circularly polarized light in the substance is 7.62 × 10−8. Solution: The relation between the phase difference and the path difference is given by 2𝜋 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑒𝑟𝑒𝑛𝑐𝑒 =. 𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜆 2𝜋 2 × 3.14 × 7.62 × 10−8 × 1 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑒𝑟𝑒𝑛𝑐𝑒 = 𝜙 =. 𝑡(𝜇𝐿 − 𝜇𝑅 ) = = 0.8125 𝑟𝑎𝑑 𝜆 5890 × 10−10 180 𝜙 = 0.8125 × = 46.570 𝜋 5. The rotation of the plane of polarisation is 400 in a substance of thickness 2 m. If the difference between the refractive indices for left and right circularly polarized light in the substance is 3.2738 × 10−8 , calculate the wavelength of light used. Solution: 2𝜋 𝑃ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑒𝑟𝑒𝑛𝑐𝑒 =. 𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝜆 2𝜋 𝜙=. 𝑡(𝜇𝐿 − 𝜇𝑅 ) 𝜆 2𝜋 2 × 3.14 × 3.2738 × 10−8 × 2 × 180 𝜆=. 𝑡(𝜇𝐿 − 𝜇𝑅 ) = = 5.8928 × 10−7 𝑚 𝜙 40 × 𝜋 6. Sugar solution of concentration 100 kgm-3 is kept in a polarimeter tube of length 0.22 m. If the specific rotation of sugar is 0.750 kg-1m2, calculate the rotation of the plane of polarization. Solution: Given:- 𝐶 = 100 𝑘𝑔𝑚−3 , 𝑙 = 0.22 𝑚, 𝑆 = 0.750 𝑘𝑔−1 𝑚2 , 𝜃 =? 𝜃 𝑆= 𝑙𝐶 𝜃 = 𝑆. 𝑙. 𝐶 = 0.75 × 0.22 × 100 = 16.50 7. A 0.2 m long polarimeter tube containing a certain solution of concentration 20% produces an optical rotation of 240. Find the specific rotation of the solution. 20 Solution: Given:- 𝑙 = 0.2 𝑚, 𝐶 = 20% = 100 𝑔𝑐𝑚−3 = 0.2 × 10−3 × 106 𝑘𝑔𝑚−3 , 𝜃 = 240 , 𝑆 =? 𝜃 24 𝑆= = = 0.60 𝑙𝐶 0.2 × 0.2 × 103 8. Determine the concentration of a solution of length 0.25 m which produces an optical rotation of 300. The specific rotation of the solution is 0.0209 rad m2 kg-1. Solution: Given:- 𝑙 = 0.25 𝑚, 𝜃 = 300 , 𝑆 = 0.0209 rad m2 kg −1 = 1.1980 , 𝐶 =? 𝜃 𝑆= 𝑙𝐶 B.Sc IV Sem POLARISATION A D B F G C, H HALLI 𝜃 30 𝐶= = = 100.166 𝑘𝑔 𝑚−3 𝑙𝑆 0.25 × 1.1980 9. A 0.2 m long tube containing 48 × 10−6 𝑚3 of sugar solution produces an optical rotation of 110 when placed in a saccharimeter. If the specific rotation of sugar solution is 0.660. Calculate the quantity of sugar contained in the tube in the form of a solution. Solution: Given:- 𝑙 = 0.2 𝑚, 𝜃 = 110 , 𝑉 = 48 × 10−6 𝑚3 , 𝑆 = 0.660 , 𝑀 =? 𝜃 𝑆= 𝑙𝐶 𝜃 11 𝐶= = = 83.33 𝑘𝑔 𝑚−3 𝑙𝑆 0.2 × 0.660 ∴ 𝑀 = 𝐶 × 𝑉 = 83.33 × 48 × 10−6 = 0.003999 𝑘𝑔 𝑀 =4𝑔 10. 0.2 m length of a certain optically active solution causes right handed rotation of 40 0 and 0.3 m of another solution causes left handed rotation of 240. What will be the optical rotation produced by 0.3 m length of the mixture of the above solutions in volume ratio 1:2. It is given that the solution do not react chemically. Solution: As the length of the mixture is 0.3 m and the solutions are in the volume ratio 1:2, we may assume that 0.1 m length of the first solution and 0.2 m length is of the second solution. 0.1 × 40 𝑇ℎ𝑒 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = = 200 𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑𝑒𝑑(+𝑣𝑒) 0.2 0.2 × 24 𝑇ℎ𝑒 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = = 160 𝐿𝑒𝑓𝑡 ℎ𝑎𝑛𝑑𝑒𝑑(−𝑣𝑒) 0.3 ∴ 𝑇𝑜𝑡𝑎𝑙 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = 200 − 160 = 40 ∴ 𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑜𝑝𝑡𝑖𝑐𝑎𝑙 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 40 𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑𝑒𝑑(𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 + 𝑣𝑒) Note: