Q-1:- Derive the conditions for constructive and destructive interference due to thin films in reflected system. Ans :

Transcription

1 Q-1:- Derive the conditions for constructive and destructive interference due to thin films in reflected system. Ans : In this discussion, following facts are presumed. Whenever a ray suffers reflection at a surface of denser medium from rarer medium (and not from denser to rarer medium) it undergoes a phase change of π 0 or path change of λ/ in its actual geometric path. A distance `λ` traversed by light in a medium of refractive index `&` has its equivalent optical path ` l` Consider a thin film of uniform thickness `t` and refractive index `µ` placed in air. Let AB and CD be the faces of this parallel-sided film. A plane monochromatic wave of wavelength λ traveling is incident at an angle i on the upper surface AB of the film. Part of the incident light is reflected and refracted along different directions as shown in fig. R 1 R i M i r The waves traveling along AR1 and CR originate by partial refection and refraction of the incident wave at the two surfaces of the film. So they are coherent. If the film is thin, the waves traveling along AR1 and CR shall overlap and they will interfere constructively, if their paths differ by an odd multiple λ/ The path difference between the ray AR1 and CR is given by Δ = (AB + BC ) AD (1) From the fig. AB=BC From Δ QRS BM BM Cos r = = BM = t AB BC t Δ = cos r Now AD () AC AM MC...(3)

2 but AM MC AC AM from Δ AMB sin r = AM AB AM= ABsin r AC =ABsinr..(4) from Δ ACD, we have sin i = AD AC AD = AC sin i t AD =.sin r.sin i cos r Therefore the path difference is - t t Δ =.sin r.sin i cos r cos r sin i = sin r sin i sin r t t Δ =. sin r.sin r cos r cos r Δ = t.cos r t cos r cos r Δ = t cosr The reflection at i is in a rarer to rarer medium from the surface of denser medium, therefore a path change of λ/ will occur in the reflected ray QQ The total path change between QQ and SS` is given by. = t cos r (5) For reflected system ; Condition f or brightness The path difference = nλ, n= 0,1,,3,,.. t cos r = n λ t cos r = (n+1) n = 0,1,,3 Condition for darkness :.(6)

3 The path difference = (n 1) t cos r = (n 1) t cos r = n λ (7) n = 1,,3,4, Que: - Explain the formation of colors in thin films of uniform thickness. Ans: Colors in thin films: The condition for the bright fringe is given by tcos r = (n 1) n = 0,1,,3,4 With monochromatic light alternate dark and bright fringes are observed because λ depends on,t, and r. For the dark band condition of interference is given by t cos r = n λ, n = 1,, 3,. From above relation we see different colors if either u, t, or r are changed. If t and r are constants and is varied, we will get various colors because λ will be varied due to the variation in, since white light consists of seven colors, so we will observe colors in the order of violet indigo,.. red etc.because λ α r i) If `t` and μ are constants and `r` is varied i.e. the angle of incidence is Changed so that angle refraction is also changed we observe different colors with white light because λ *r ii) If `u` and `r` are constants and `t` is changed we will get different colors with white light because λ t At a particular point of the film and for a particular position of the eye, the interfering rays of only certain wavelengths will have path difference satisfying the conditions of bright fringe. Hence only such wavelengths (colors) will be present there other neighboring wavelengths will be present with diminished intensity the colors for which the condition of minima is satisfied are absent in the reflected system. Que3: What are Newton s rings? How are they formed? Explain the experimental setup to obtain the Newton rings in the laboratory. Ans. When a Plano-convex lens of large focal length with its convex surface is placed on a plane glass plate, an air film of gradually increasing thickness is formed between the two. It a monochromatic light is allowed to fall normally and the film viewed in the reflected light. Alternate dark and bright rings are seen are called as Newton s rings. Experimental arrangement. L is Plano convex lens of large radius of curvature.

4 L P O Experimental set up Newtons Ring This lens with its convex surface is placed on a plane glass plate P. The point of contact between them is O. Light from an extended source falls and the air film between the lens L and the glass plate P. This light after reflection from the air film interfere and produces the interference pattern in the form of circular rings, which can be seen with a microscope focused on the film. Que3: Explain the formation of Newton s rings Ans: Principle: Newton s rings are formed due to interference between the waves reflected from the top and bottom surfaces of the air film. Formation of Newton s rings : When a ray AB is incident on the system it gets partially reflected at the bottom curved surface of the lens as ray-1 and transmitted ray is partially reflected as ray- from the plane glass plate. The ray-1 and ray- are derived from the same ray AB and therefore they are coherent. Ray-1 undergoes no path change but ray acquire a path change of upon reflection, since it is reflected from air to glass boundary. The path difference between the ray-1 and ray- is given by - tcos ( r )

5 A B or tcosr ( is very small, it can be neglected) For normal incidence and air film. Cos r = 1 and 1.() Path difference = t- For n th maxima we have t- (n 1) t =n..(4) This shows that a maxima or minima of a particular order n will occur for a constant value of t. here, t remains constant along a circular. So the maxima is in form of a circle. Que4: In Newton rings show that radius of dark rings are proportional to the square root of natural numbers. Ans.: Consider a plane-convex lens L placed on a plane glass plate AB. Let R be the radius of curvature of the lens and r be the radius of the n th dark ring corresponding to the constant film thickness t. The condition for the nth dark ring is given by. t = n..(1) and the condition for n th bright ring is given by t= (n+1), n = 1,,3,4,.()

6 By the property of the circle i.e. (R-t) +r =R (As t is very small, t is neglected) r Rt or t = r R..(3) Diameter of dark ring.: by equation (3) equation (1) gives r. n R r n nr If Dn is the diameter of n dark ring we have rn= Dn/ Dn = n 4 Dn 4nR Dn 4R. n th R Dn n (4) This is equation shows that the diameter of n root of natural number Diameter of bright ring. we have t = ((n+1) r (n 1) R R r = (n+1) th dark ring is proportional to the square

7 If Dn be the diameter of n Dn rn = D n R (n 1) 4 th bright ring we have. D n R.(n 1) Dn R. (n 1) Dn ( n 1)..(5) This equation shows that the diameters of bright rings are proportional to the square roots of odd natural numbers. While counting the orders of the dark rings the central spot is not counted and counting starts from 1,,3, n. Que:5 What is diffraction? Distinguish between Fresnel and Fraunhofer diffraction Ans: Diffraction: Diffraction of the light rays round the corners of an obstacle is called as diffraction Fraunhofer Diffraction Fresnel diffraction 1) The source of light and the screen 1) The source of light and the are at Infinite distance from the screen are at finite distance Obstacle from the obstacle ) The incident wave front is plane ) The incident wave front is either Spherical or cylindrical 3)The diffracted wave front is plane 3)The diffracted wave-front is either Spherical or cylindrical 4) The initial phase of secondary wave 4) The initial phase of secondary Lets is the same at all the points in the wavelets is different at different points Plane diffracting device in the plane-diffracting device 5) It is very important in the optical 5) It has no importance in optical Instruments instruments Que:6 Explain the difference between interference and diffraction

8 Interference Diffraction 1) Interference is the result of 1) Diffraction is the result of interaction of light waves which interaction of light energy which comes from the same source comes from different parts of the same wave front ) Interference fringes in a ) Diffraction fringes in a given pattern particular pattern may or may are never of the same width. not of the same width 3) In interference all the bright 3) In diffraction the bright fringes are fringes are of the same or uniform of varying intensity. intensity 4) The points of minimum dark and 4) The points of intensity are not hence these fringes are extremely perfectly dark and hence these contrasting fringes are not quite contrasting. Que7:- Derive an expression for the intensity at a point in the Fraunhofer type of diffraction produced by a single slit. Ans.: A parallel beam of monochromatic light of wavelength illuminates a narrow rectangular slit AB of width d. Every point of the wave front in the plane of the slit AB acts as a secondary source and sends out secondary these waves and focus them on the screen placed in the focal plane of the lens The secondary wavelets traveling normally to the slit are focused at Po by the lens and the secondary wavelets traveling at an angle θ are focused at P1. The path difference between the rays converging at point Po is zero and meet into same phase producing maximum intensity at point Po N P A θ P1 W d O B C θ R L P0 Screen Consider the effect at an angle with the normal Draw a perpendicular AC on BR. Let the path difference between secondary wavelets from A to P1 and B to P1 is given as P.d = BC. (1)

9 But BC = AB Sin BC d sin () and corresponding phase difference is given by Phase difference =. Path difference = d Sin.(3) Let us consider that given slit is divided into N number of parallel infinitesimal slits each of width dl let a is the amplitude of the wave from each slit. d d ' d1 d1... d1...(4) 1 3 n Each point on the slit sends out secondary waves. They meet at point P1. Since the secondary wavelets are originated from the same wave front each wavelet has same amplitude. Therefore the variation in the intensity at point P1 is because of phase difference only. The resultant amplitude at point P1, is obtained by the method of vector addition of amplitudes. This method is explained with the help of phase amplitude diagram as given below. Consider these N wavelets of amplitude, E1 which are meeting at point P1 on the screen. Each one is out of phase by a small angle Φ. When N is large enough the amplitude phase diagram takes the form of smooth arc of the circle of radius R. & the resultant amplitude = AB =Em If A O R Φ/ AB = AM = AO Sin / AM ( Sin / AM AOSin / ) AO AB RSin /...(7) E Φ/ E M Em Em Sin /,( R ) / the phase difference between the first and last wavelets is then we have Em (6) R Em = amplitude of wavelet, R = Radius ArcAB Radius The magnitude of resultant amplitude AB is given by. Sin / Sin E Em Em ( )...(8) / Where Eθ = N E1 = total amplitude due to N wavelets each of amplitude E1 Since intensity is proportional to square of amplitude i.e. I I R E m A E m B Sin / ( ) /

10 Sin I Io( )...(9) Where Io= Em (N.E.) maximum intensity due to N wave-lefts of amplitude E1 1 Principal maxima or central maxima : The resultant amplitude is given by Eθ = Em ( Sin ) E m = (...) Eθ = Em (1-...) Eθ = Em If 0 (central) Phase difference ( ) =, d sinθ 0 d sin *Minima : The intensity will be zero if sin = 0 i.e. I = 0,, 3,... m For m, where n 1,,3,4,... d sin But phase difference (α = / d.sin d sin m ( d sin m m 0 Que8:-: What is plane transmission grating explain how it can be used for determining the wavelength of given monochromatic light Give the theory of plane transmission grating Ans: Diffraction grating : An arrangement consisting of large no. of parallel slits of the same width arranged side by side and separated by equal opaque spaces is called diffraction grating. Grating is prepared by ruling equivalent parallel lines on glass plate using a fine diamond point.

11 *Theory of plane transmission grating. As shown in fig. Consider N parallel slits each of width a Let b be the separation between two successive slits (i.e. width of opaque space) Grating element : The distance between any slit and adjacent opaque space is called a grating element. (d) d = a+b The intensity at point P1, may be considered by applying the theory Fraunhofer diffraction at a single slit. The wavelets proceeding from the points in a slit along the direction are equivalent to a single wave of amplitude A given by Sin E Em ( ) 1 Where d sin If there are (S1, S.Sn) N slits, then we have N diffracted waves coming from middle points of each slit. We want to find out total effect of these N vibrations at an angle at point P1. Path difference between S1, P1, and S P1 = (a+b) Sin..(3) Phasediffe rence ( a b) Sin..(4) Path difference between S1, P1, and S3 P1 = (a+b) sin..(5) Phasedifference [ ( a b) Sin ]..(6) Intensity distribution Principal maxima For principal maxima the path difference between the waves from adjacent slit must be an integral multiple of i.e. (a+b) Sin n,0,1,,...(1) The equation for the phase difference is given by β = (a b) Sin.( n) n β = n () For β n, SinNB becomes indeterminate (or diverges) SinB But by L, Hospital rule, we have Sin d / dbsin NB I = Io ) [lim n ] d / dbsin B

12 Sin NCosN = Io ( ) [ Lim n ] Cos Sin NCosNn = Io ( ) [ ] n 0,1,,3,... Cosn Sin I = Io ( ) N CosNn ( 1 Cosn In this way it we calculate the phase difference for all paths then we can see that the phase difference increases in arithmetic progression. Resultant amplitude of N waves will be given by. sin SinN A = Ao ( )...(7) Sin And the intensity is given by sin SinN I = Io...(8) Sin For single slit we have.d. Sin and for N slits, we have Phase difference = Nβ = N..(a b) Sin ( a b) Sin.(9) equation (9) gives the constant phase difference between the two wavelets sin Io ( ) Sin E m ( ) this factor gives the distribution of intensity Sin NB due to a single slit and gives the distribution of intensity as a combined effect Sin B of all the slits. Thus, the intensity of principal maxima increases with increasing N. Since = n π ( a b) Sin n ( a b) Sin n.(3) These maxima will be on both sides of zero order maxima arranged symmetrically *Minima: When I = O the intensity pattern is of minimum or zero intensity SinN i.e. 0weget Sin Sin N = 0 but Sin 0 N =,, 3,... m

13 N m But ( a b) Sin N ( a b) Sin m N.( a b) Sin m (4) Where m takes all integral values except 0, N, N, 3N, m. N. Because for these values Sin B = O & we get principal maxima Thus we have m = 1,, 3,... ( N 1) This shows that in between any two successive principal maxima there are (N-1) minima. Que9: Explain the Phenomenon of double refraction in calcite crystal. Ans : Birefringence or double refraction : It was observed by E. Bartholimus in 1669, that when an ordinary beam of unpolarised light is incident on a crystal of calcite or quartz two refracted beams emerge in place of usual one beam. This phenomenon is called as birefringence or double refraction. Consider a beam of ordinary unpolarised light, SA incident normally on one face of a calcite crystal as shown in fig. It gives rise to two refracted beams E and O. If the crystal is rotated along the line of sight, O-beam remains stationary and E-beam rotates around the O-beam when the two beams are tested for their state of polarization by means of a tourmaline, are found to be plane polarized in mutually perpendicular planes. As shown in fig.(b) E-ray has vibrations parallel to the shorter diagonal of the crystal and O-ray has vibrations parallel to the longer diagonal of the crystal. The O-ray obeys the usual smells law of refraction and is called as the ordinary ray. The E-ray has different velocity in different directions, so refractive index of E-ray ( is different in different directions and is called as extra ordinary ray. (In calcite uo_ue) Que 10: Explain Huygen s theory of double refraction Ans: The explanation of double refraction was given first by Huygen by applying his theory of secondary wavelets. According to this theory

14 (1) When any wavefront strikes a doubly refracting crystal, every point of the crystal becomes a source of two waverfronts. One for ordinary ray and the other for extra ordinary ray. () The ordinary ray travels with the same velocity (Vo) in all directions and hence its wave front is spherical. (3) The extra-ordinary ray travels with different velocities (VE)in different directions and hence the corresponding wave front is ellipsoid. (4) In case of ordinary ray the ratio of sin i r e sin = e is constant in all the directions. (5) In case of extra-ordinary ray the ratio of varies with the direction of propagation of the E-ray. (6) The sphere and ellipsoid touch each other at points which lie on the optic axis of the crystal, because the velocities (Vo) and (Ve) are same along the optic axis. (7) In certain crystals e.g. calcite and tourmaline the ellipsoid lies outside the sphere (fig. a.) This means that VE VO. These crystals are known as negative crystals. 8) In certain crystals e.g. ice and quartz the sphere lies outside the ellipsoid (fig.) This means that VE VO.These crystals are called positive crystals. Calcite Quartz Negative crystal Positive crystal Que 11:- What do you understand by retardation plates? What are their types? Explain each in brief Ans :- Retardation Plate :- It is defined as a plate cut from a double refracting crystal so as to produce a definite valve of path difference or phase difference between O-ray and E-ray of polarized light. There are two types of plates 1) Quarter wave plate (QWP) ) Half wave plate (HWP) 1) Quarter wave plate :If the thickness of the crystal plate, cut with its faces parallel to the optic axis is such that it introduces a phase difference of πc/ or a path difference of λ/4 between the O-ray and E-ray then it is called a half wave plate. If it is the thickness of the plate `μ` and μ`e are the refractive index of O-ray and E- ray respt. Then its air equivalent thickness is `μot` and `μet`. Then the path difference between the two rays is. μot μet = (μo - μe ) t..(1) ( μ0 μe) for calcite) for quarter waves plate this path difference should be λ/4 t ( μ0 μe) = λ/4

15 t = for calcite (Ve > Vo) 4( ) 0 e and t = λ 4 (μe μo ) for quartz (Vo > Ve ) Half wave plate : If the thickness of the crystal plate, cut with its faces parallel to the optic axis is such that it introduces a phase difference of π ċ or a path difference of λ/ between the O-ray and E-ray then it is called a half wave plate. If `t` is the thickness of the plate, then its air equivalent thickness is `μ0t` and `μet.` The path difference between O and E rays in terms of thick for this plate is path difference = μ0t μet path different = t (μo μe) For half wave plate this path difference should be λ/ t (μo μe) = λ/ t = and t = λ (μo μe) for calcite (VE VO) μo μe λ (μe μo) for quartz (VO>VE) μe>μo

16 Que 1: Explain principle construction and working of He-Ne laser Principle: He-Ne laser works on the principle of four energy level system Construction: The gain medium of the laser is a mixture of helium and neon gases, in approximately a 10:1 ratio, contained at low pressure in a glass envelope. The gas mixture is mostly helium, so that helium atoms can be excited. The excited helium atoms collide with neon atoms, exciting some of them to the state that radiates 63.8 nm. Without helium, the neon atoms would be excited mostly to lower excited states responsible for non-laser lines. A neon laser with no helium can be constructed but it is much more difficult without this means of energy coupling. Therefore, a He-Ne laser that has lost enough of its helium (e.g., due to diffusion through the seals or glass) will lose its laser functionality because the pumping efficiency will be too low. The energy or pump source of the laser is provided by a high voltage electrical discharge passed through the gas between electrodes (anode and cathode) within the tube. A DC current of 3 to 0 ma is typically required for CW operation. The optical cavity of the laser usually consists of two concave mirrors or one plane and one concave mirror, one having very high (typically 99.9%) reflectance and the output coupler mirror allowing approximately 1% transmission. schematic diagram of a helium neon laser Commercial HeNe lasers are relatively small devices, among gas lasers, having cavity lengths usually ranging from 15 cm to 50 cm (but sometimes up to about 1 metre to achieve the highest powers), and optical output power levels ranging from 0.5 to 50 mw. Working: The mechanism producing population inversion and light amplification in a He-Ne laser is due to inelastic collision of energetic electrons with ground state helium atoms in the gas mixture. As shown in the accompanying energy level diagram, these collisions excite helium atoms from the ground state to higher energy excited states, among them the 3 S 1 and 1 S 0 LS or Russell-Saunder coupling, front number tells that an excited electron is n = state) in long-lived metastable states. Because of a fortuitous near coincidence between the energy levels of the two He metastable states, and the 5s and 4s levels of neon, collisions between these helium metastable atoms and ground state neon atoms results in a selective

17 and efficient transfer of excitation energy from the helium to neon. This excitation energy transfer process is given by the reaction equations: He*( 3 S 1 ) + Ne 1 S 0 He( 1 S 0 ) + Ne*4s + ΔE and He*( 1 S) + Ne 1 S 0 + ΔE He( 1 S 0 ) + Ne*5s where (*) represents an excited state, and ΔE is the small energy difference between the energy states of the two atoms, of the order of 0.05 ev or 387 cm 1, which is supplied by kinetic energy. Energy levels in a He-Ne Laser Excitation energy transfer increases the population of the neon 4s and 5s levels manyfold. When the population of these two upper levels exceeds that of the corresponding lower level neon state, 3p 4 to which they are optically connected, population inversion is present. The medium becomes capable of amplifying light in a narrow band at 1.15 μm (corresponding to the 4s to 3p 4 transition) and in a narrow band at 63.8 nm (corresponding to the 5s to 3p 4 transition at 63.8 nm). The 3p 4 level is efficiently emptied by fast radiative decay to the 1s state, eventually reaching the ground state. The remaining step in utilizing optical amplification to create an optical oscillator is to place highly reflecting mirrors at each end of the amplifying medium so that a wave in a particular spatial mode will reflect back upon itself, gaining more power in each pass than is lost due to transmission through the mirrors and diffraction. When these conditions are met for one or more longitudinal modes then radiation in those modes will rapidly build up until gain saturation occurs, resulting in a stable continuous laser beam output through the front (typically 99% reflecting) mirror.

18 Que 13: Derive an expression for the energy levels of a particle inside an infinite potential well Suppose we have a single particle of mass m confined to within a region 0 < x < L with potential energy V = 0 bounded by infinitely high potential barriers, i.e. V = for x < 0 and x > L. The potential experienced by the particle is then: V (x) = 0 0 < x < L = x L; x 0 In the regions for which the potential is infinite, the wave function will be zero, for exactly the same reasons that it was set to zero, that is, there is zero probability of the particle being found in these regions. Thus, we must impose the boundary conditions ψ(0) = ψ(l) = Meanwhile, in the region 0 < x < L, the potential vanishes, so the time independent Schrodinger equation becomes, ħ m d dx d dx ψ(x) = E ψ(x) m ψ(x) + E ψ(x) = ħ To solve this, let k = m ħ E d dx ψ(x) + k ψ(x) = 0 3 General solution of equation is ψ(x) = A sin(kx) + B cos(kx).4 It is now that we impose the boundary conditions, Eq. (1), to give, i. at x = 0: ψ(0) = B = 0 so that the solution is now ψ(x) = A sin(kx)..5 applying the boundary condition ii. at x = L gives ψ(l) = A sin(kl) = 0 6

19 which tells us that either A = 0, in which case ψ(x) = 0, which is not a useful solution (it says that there is no partilce in the well at all!) or else sin(kl) = 0, which gives an equation for k: kl = nπ, n = 0, ±1, ±,..7 We exclude the n = 0 possibility as that would give us, once again ψ(x) = 0, and we exclude the negative values of n as the will merely reproduce the same set of solutions as the positive values. Thus we have kn = nπ L,..8 n = 1,, n where we have introduced a subscript n. This leads to, from equation for k k = m E = ħ (nπ L ) E = ħ m (nπ L )..9 Thus we see that the boundary conditions of equ (1), have the effect of restricting the values of the energy of the particle to those given by Eq. (9). To find associated wave functions we apply the normalization condition to determine A (up to an inessential phase factor) which finally gives L ψ(x) dx = 1 0 L [Asin nπ L x] dx = 1 0 Solving above equation A = L ψ(x) = L sin (nπx L ) 0 < x < L = 0 x < 0, x > L.