Wave Optics Notes PDF

AP-Unit II:Wave Optics UNIT II: WAVE OPTICS Electron microscopes can make images of individual atoms, but why will a visible- light microscope never be able to do this? Why are computer chip manufacturers investing billions of dollars in equipment to etch...

AP-Unit II:Wave Optics UNIT II: WAVE OPTICS Electron microscopes can make images of individual atoms, but why will a visible- light microscope never be able to do this? Why are computer chip manufacturers investing billions of dollars in equipment to etch chips with x-rays instead of visible light? The answers to all of these questions have to do with the subject of wave optics. In Physics, physical optics, or wave optics, is the branch of optics which studies interference, diffraction and polarization and other phenomena for which the ray approximation of geometric optics is not valid. Light: Light has electromagnetic nature. It consists of electric and magnetic field vectors, oscillating in mutually perpendicular directions and perpendicular to the direction of wave motion. Since electric field vector has greater influence on matter than magnetic field vector, here emphasis has been given to the electric field vector. Interference: Interference is an important phenomenon of superposition of waves. When two or more coherent waves of light superposed, the pattern of alternate bright and dark band is observed. “The phenomenon of redistribution of light energy due to the superposition of light waves from two or more coherent sources is known as interference”. The principal of superposition of waves states that when two or more waves are incident on the same point, the total displacement at that point is equal to the vector sum of the displacements of the individual waves. If a crest of a wave meets a crest of another wave of the same frequency at the same point, then the magnitude of the displacement is the sum of the individual magnitudes – this is constructive interference. If a crest of one wave meets a trough of another wave then the magnitude of the displacements is equal to the difference in the individual magnitudes – this is known as destructive interference. Resultant wave Wave 1 1 Dr.S Bompilwar  AP-Unit II:Wave Optics Wave 2 Constructive interference Destructive interference --------------------=0=----------------------------- Geometric path and optical Path: The terms "optical path" and "geometrical path" refer to different concepts in the context of light propagation and optics: The main difference between geometrical path and optical path is that the optical path is the distance light travels multiplied by the refractive index of the medium it's traveling through, while the geometrical path is the actual distance light travels Optical Path  Definition: The optical path is a measure of the phase shift of light as it travels through a medium. It takes into account both the distance traveled and the refractive index of the medium.  Formula: The optical path length (OPL) can be calculated using the formula: OPL = μ x L where μ is the refractive index of the medium and L is the geometrical distance traveled by light.  Significance: The optical path is important in phenomena such as interference, diffraction, and the behavior of light in different media. It helps in understanding how light waves interact based on their phase differences. Thin film interference: ( Due to reflected light): An optical medium is called thin film if its thickness is about 1 wavelength of visible region i.e. in the range of 0.5 μm to 10 μm. Eg. A thin sheet of transparent material such as glass, mica, air film enclosed between two transparent plates or soap bubbles. Plane parallel thin film: A transparent thin film of uniform thickness bounded by two parallel surfaces is known as a Plane parallel thin film. Ray diagram: 2 Dr.S Bompilwar  AP-Unit II:Wave Optics Explanation: Let ‘t’ be the thickness of the transparent thin film on which light AB is incident. Let ‘μ’ is the refractive index of the film. ‘i’ is the angle of incidence. Part of the incident ray AB is reflected along BC & transmitted into the film along BF. The ray BF in turn partly reflected back into the film along FD. Part of the reflected ray FD is transmitted at the upper surface and travels along DE. Since the rays BC and DE are coherent, they produce interference pattern depending on their path difference. Draw DH perpendicular to BC. Ray 1 travels along BC in air and ray 2 travels along DE through the thin film. Geometric path difference between ray 1 and ray 2 is given by (BF+FD) - BH Optical path difference is given by, Δ= μ (BF+FD) – 1*(BH) ------------------1. From Δ BFD, BF=FD Also BF= = ∴ BF+FD = ------------------2. In Δ BDH, BH = BD sin i But BD=2*t*tan r ∴ BH = 2*t*tan r* sin i----------------------3. From Snell’s law, sin i= μ sin r 3 Dr.S Bompilwar  AP-Unit II:Wave Optics ∴ BH = 2t*tan r* μ sin r = 2t* * μ sin r = 2t μ * ------------------4 From eq. 1, 2, and 4, we get, Optical path difference Δ= - 2t μ * = = *cos2 Δ= --------------------5 Since the ray 1 is reflected at the boundary of a rarer to denser medium, path change of λ/2 occurs. ∴Δ= - λ/2 This is the required equation of the path difference due to reflected light. 1. Condition for maxima: (Brightness or constructive interference) Maxima occurs if path difference, Δ= m λ ∴ - λ/2= m λ Or = (2m+1) 2. Condition for minima ( Darkness or destructive interference): Minima occurs if Δ= (2m+1) ∴ - λ/2= (2m+1) Or =mλ 4 Dr.S Bompilwar  AP-Unit II:Wave Optics -----------------=0=----------------------- IMP*** Interference in wedge shaped thin film: Expression of Band width/ Fringe width “A wedge is a thin film of varying thickness having a zero thickness at one end and progressively increasing to a particular thickness at the other end”. Two glass slides resting on each other at one edge and separated by a thin spacer at the opposite edge forms the thin wedge of air film as shown in following figure. Fig. interference in wedge shaped film Consider a wedge having a very small angle θ and refractive index. When a monochromatic beam of light in incident on the wedge from above as AB, it gets partly reflected from the top surface as BC and part of the transmitted ray (BF) is reflected back into the medium from the 5 Dr.S Bompilwar  AP-Unit II:Wave Optics bottom surface as FE. Since both the rays BC and BFDE are derived from the same ray AB by division of amplitude, they are coherent and get interfere to produce alternate bright and dark fringes. (In case of white light, coloured fringes are observed.) Optical path difference between the two rays BC and DE is given by, Δ= - λ/2 where λ/2 takes account the gain of half wave due to abrupt jump of п radian in the phase of wave reflected from the bottom boundary of air to glass. 1. Condition for maxima: (Brightness or constructive interference) Maxima occurs if path difference, Δ= m λ ∴ - λ/2= m λ Or = (2m+1) 2. Condition for minima ( Darkness or destructive interference): Minima occurs if Δ= (2m+1) ∴ - λ/2= (2m+1) Or =mλ For normal incidence of light, cos r=1 ∴ = m λ--------------------1 6 Dr.S Bompilwar  AP-Unit II:Wave Optics If dark fringe will occur at A with the thickness t1 Then, 2μ t1= m λ--------------------2 The next dark fringe will occur at C with the thickness t2 2 μ t2= (m+1) λ-----------------3 From eq.2 and 3, 2 μ (t2- t1) = λ But (t2- t1) = BC 2 μ BC= λ ∴ BC= λ/2 μ--------------------4 From triangle ABC, Thus a quarter coating acts as reflecting coating if μF > Coating of Zinc sulfide ( RI= 2.37) reflects 30% of light. With multiple coating, 100% reflectivity can be achieved. Then the coated surface acts as a mirror. Interference filters: It is an optical system that transmits a very narrow range of wavelength and provides a monochromatic beam of light. ----------------------------=0=-------------------------------- NEWTON’S RINGS: 13 Dr.S Bompilwar  AP-Unit II:Wave Optics The phenomenon of Newton's rings, named after Isaac Newton, who first studied them in 1717, is an interference pattern caused by the reflection of light between two surfaces – a spherical surface and an adjacent flat surface. When viewed with monochromatic light, Newton's rings appear as a series of concentric, alternating bright and dark rings centered at the point of contact between the two surfaces. When viewed with white light, it forms a concentric-ring pattern of rainbow colors, because the different wavelengths of light interfere at different thicknesses of the air layer between the surfaces. Newton’s rings are observed when light is reflected from a plano convex lens of a long focal length placed in contact with a plane glass plate as shown below. Fig. Experimental arrangement for newton’s rings Thin air film is formed in between plate and lens. The thickness of the air film varies from zero at the point of contact and gradually increases outwards. When this system is illuminated with monochromatic light, light gets reflected from the lower surface of the lens & upper surface of the plate. The interference due to these reflected rays produces concentric and bright rings. Newton’s rings 14 Dr.S Bompilwar  AP-Unit II:Wave Optics Principle: Fig. Formation of Newtons rings When a ray AB is incident on the system, it gets partially reflected at the bottom curved surface of the lens and part of the transmitted ray is partially reflected from the top surface of the plane glass plate. These two rays are derived from the same incident ray by the division of amplitude and hence are coherent. Second ray which reflects from air to glass boundary suffers phase change of λ/2. The condition for bright band is , = (2m+1) And the condition for dark band is =m For normal incidence, cosr =1 and for air, =1 ∴2t= (2m+1) for bright band-----------1 & 2t= m for dark band------------------2 Determination of radius of rings: 15 Dr.S Bompilwar  AP-Unit II:Wave Optics Let R-Radius of curvature of lens Let dark ring be located at Q t-be the thickness of air film at Q i.e. PQ=t rn – distance of Q from O i.e radius of nth dark ring. From fig. OM=2R; OQ=r and PQ=t=EO By the theorem of intersecting chords, EP*EH=OE*EM rn*rn=t*(2R-t) rn2=2Rt-t2 as 2Rt>>> t2 ,it can be neglected. ∴ rn2=2Rt For dark band, 2t= n ∴ rn2= n Or rn= √ Or diameter of dark ring is given by, Dn=2 √ For n=0, 1 ,2 , 3------------- r0=0, r1=√ , r2=√ r3=√ 16 Dr.S Bompilwar  AP-Unit II:Wave Optics Thus radius or diameter of the dark ring is proportional to the square root of natural number and radius of curvature of lens. Radius of bright rings: Suppose mth bright band is located at Q. Therefore radius of mth bright band is given by, rm2=2Rt, since for bright band, 2t = (2m+1) ∴ rm2=(2m+1) ∴ rm= √ For m = 0, 1, 2, 3--- r0=√ , r1=√ , r2=√ ------------------- Thus radius or diameter of the bright ring is proportional to the square root of odd natural number and radius of curvature of lens. --------------------=0=------------------------ Determination of wavelength of light: Let diameter of nth dark ring is given by, Dn=2 √ ∴Dn2= Let dn+m be the diameter of (n+m)th dark ring ∴D (n+m)2= ∴D(n+m)2 - Dn2 = ∴ = Thus by measuring the diameter of rings and knowing the radius of curvature of lens, wavelength of monochromatic light can be determined. -----------------------=0=----------------------------- Determination of Refractive Index (RI) of liquid: 17 Dr.S Bompilwar  AP-Unit II:Wave Optics If the gap between the lens and the plane glass plate is filled with a liquid, air film is replace by the liquid film. Let µ is the R I of liquid. The condition for the formation of dark ring is given by , 2 µ t cosr = nλ For normal incidence, 2 µ t = nλ ( since cosr=1) ------------------1 Also the radius of n th dark ring is given by, rn2= n = 2Rt (since n = 2t)-----------2 From equation 1&2, (rn2)L= Or (Dn2)L= -------------------------3 Similarly, diameter of (n+p)th dark ring is, (D(n+p)2)L= ----------------4 From equation 3&4 we get; (D(n+p)2)L-(Dn2)L= But for air, (D(n+p)2)air - (Dn2)air= 4p R ∴ = ------------------------=0=------------------------------ Observations of Newton’s ring experiment: 1. Fringes are circular: in Newton’s ring experiment a thin film of air is enclosed between the plano convex lens and glass plate. The thickness of the air film at the point of contact is zero and it increases gradually outward. Therefore locus of points where the air film has same thickness fall on a circle having centre at the point of contact. Hence the fringes are circular. 2. Rings get closer away from the centre: Diameter of dark ring is given by, Dn=2 √ , where R is radius of curvature of lens and n=1,2,3,------ ∴ D1=2 √ ; D2=2 √ ; 18 Dr.S Bompilwar  AP-Unit II:Wave Optics D3=2(1.7) √ ------------------- As the order of ring increases, the diameter of the ring does not increases with the same proportion and hence rings get closer away from the centre and also are not evenly spaced. 3. Central fringe is dark in reflected light: We know that Newton’s rings are produces due to interference between light rays reflected from the top and bottom surface of air film. The optical path difference between interfering rays is given by, Δ= - λ/2 = - λ/2 But at the centre, t=0; ∴ Δ = λ/2 = Thus two interfering waves at the centre are out of phase and hence produce a center dark spot. -----------------------------=0=----------------------------- 4. Plano-convex lens should have larger radius of curvature: In Newton’s ring experiment, the radius of dark ring is proportional to the square root of radius of curvature®. i.e. rn=√ It shows that, greater the radius of curvature of lens, greater the radius of ring which would reduce the errer in the measurement of daimeter of ring which in turn gives the accurate wavelength of light. Hence plano-convex lens should have larger radius of curvature. 5. Fringes are of equal thickness: Occurance of alternate maxima and minima in Newton’s ring experiment is due to the variation of thickness t, of the film. Each maxima or minima is a locus of constant film thickness. Hence the fringes are called as fringes of equal thickness. -----------------------------=0=--------------------------- BEST LUCK 19 Dr.S Bompilwar