ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM Nature OF Light i Dual Nature of Light: Some phenomenon such as interference, diffraction and polarization can be explained on the basis of wave theory of light. Whereas the photoelectric effect and Compton effect supported the corpuscular theory i.e. particle nature of the light. Consequently in the pracess of emission and absorption It is observed that light radiations have particle like properties. On the other hand in the phenomenon of light propagation it is observed that radiations have wave like properties. Thus discussion about the nature of light shows that light possesses dual nature i.e. wave like and particle like properties. ‘e Fron Whenever waves pass through a medium, its particles execute SHM. The path (locus) of all the particles of the medium having the same phase is known as “Wave Front”. Spherical Wave Fro! In case of a point source of light the wave front will be concentric spheres with centre at the source. Such as a wave front is known as “Spherical Wave Front”. Plane Wave Front: Ata very large distance from the source a small portion of a spherical wave front will become plane wave front. = > Ray of Light: / / The direction in which wave moves |s always normal to the wave front. Thus a ray of light means the direction in which a light wave propagates and itis always along the normal to the wave front. LE Bee etd tl Hig aL pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q The principle states that; “Every point on a wave front can be cansidered asa source of secondary spherical wave front. ‘The new position of the wave front after a time “t” can be founded by drawing a plane tangent to the secandary wave”. Explanation: ‘ Figure illustrates two simple examples of fa Huygen's construction. First, consider the plane wave front moving through medium old New t asin figure (a). At t=0, the wave front is Wavefront | “4 Wave front indicated by the plane labeled AA’. According to Huygen's principle, each point on this wave J front is considered as a point source. Using ” _ those points as sources for the wavelets, we tO tet . draw circles of radius “ct", where “C" is the o speed of light and “t” is the time of propagation old New ‘Wave front Wave front from one wave frant to the next. The plane tangent to these wavelets is BB’, which is parallel to AA’. In a similar manner figure (b) shows Huygen’s construction for spherical wave fronts, The superposition of two light rays oscillating with same phase and amplitude is referred as, “Interference” of Light. Types of Interference: 1. Constructive Interference: If crest of one wave interferes the crest of other, then the resultant energy of the superposed wave will be increased to maximurn. The point of intersection of two waves is called Constructive point and the wave phenomenon at this point is known as, “Constructive Interference”. The constructive interference can be seen on a screen inform of Straight or circular bright band. Destructive Interference: If crest of one wave interferes the trough of other, then the resultant energy of the superposed wave will fall to zero. The point of intersection of two waves is thus called Destructive point and the wave phenomenon at this point is known as, "Destructive Interference”. The destructive interference can be seen on a screen in form of straight or circular dark band. 2. LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q Conditions For Interference of Light: The interference of light can anly be seen when following two conditions are accomplished: 1. Phase Coherenci If two wave generators produce exactly similar waves such that if one produces a crest then the other also produces a crest or both praduce trough at the same time then the generators are called Coherent generators. Waves produced by coherent generators are said to have phase coherence. Path Difference: Suppose that two wave pulses (light rays) emitted out of two phase-coherent sources are reaching a reference point “p”.If one wave pulse (ray) travels greater distance to reach point “P” than the other, then path difference is said to be set between the two pulses. Itis dented by “a”. How can the path difference be achieved?? Path difference between the two pulses can be achieved by separating two phase coherent sources by a small distance “d”. If greater be the value of “d”, more effective path difference between the two pulses can be achieved. Suppose that two wave pulses are emitted by two phase coherent sources “S;” and “S2” and finally interfere at point “p”, IFS,P and S,P be the distance travelled by the two wave pulses, then their path difference is given by; N o=5,P -SP By triangle; sq sin = == 251 d ink sind = 5 o = dsin@ Condition For Constructive Interference: Far constructive interference the path difference between two waves having phase coherence must be an integral multiple of their wave length "A". ie. Path difference = ma Where m =0,1,2,3..... Hence for constructive interference path difference between the interference waves must be 0,,24,32. ma. Condition For Destructive Interference: For destructive interference the path difference between the two waves must be an odd multiple of 4/2. ie. LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , Path difference = (m+ 4 A Where m=0,1,2,3.... Hence for destructive interference path difference between the interference waves must be AM SATA A Fai penn + Vee Zz) 2) 2° Young’s Double Slits Experiment The first successful attempt was made by Thomos Young in 1801 for producing the interference of light. His experiments defined below: Experimental Arrangement (Or) How did Young achieve the condition for interference of ligt Thomos Young selected a single light source instead of two sources and the beam of light ‘coming out of a source is further sub-divided into two phase-coherent beams of light. Far this, he took Double Slits arrangement in which each slit behaves like a phase coherent source. The separation between the two slits produces a suitable Path Difference between the two beams emerging out of the slits. These beams of light falls on a screen placed at a distinct distance from the slits and produce interference fringes due to constructive and destructive interference. Each bright fringe will be appeared on the screen due to constructive interference and each dark fringe will represent the destructive interference. The distance between two consecutive bright or two consecutive dark fringes remains constant, is known as “Fringe Spacing”. Derivation For Fringe Spacing: Consider Young's double slits arrangement in two dimensions. Light source Screen In this arrangement, suppose that; The separation between the two slits = The distance between the slits and the screen =L The path difference between two rays = ¢ = dsin®) The distance of point “p” on the screen from the central point = y As, decL therefore, limit (@}- 0 and thus, Sin? = Tané In AOCP; al sl ind = 1 ree sind = Tand = 5 = LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , Multiplying by “a” on both sides, dsino = d® so =dsin8 Positions For Bright Fringes Or Maxima: A bright fringe appears when the constructive interference takes place. In case of constructive interference, ¢ = ma Eat) yeot L yame Where m = 0,1,2,3. ie 2LA 3LA ad fd to Thus, Fringe Spacing = a (distance between two bright fringes} Positions For Dark Fringes Or Minima: A dark fringe appears when the destructive Interference takes place. In case of destructive interference, ¢ = (m +3)a Eati) y=ou 1) .L rtrd soloed Where m = 0,1,2,3, _ la 31a SLA YF 2a 3d 20” Thus, Fringe Spacing = = (distance between two dark fringes) The above analysis clearly explains that the fringe spacing between two consecutive bright or a da dark fringes remain constant i.e|Fringe Spacing =—. This spacing also suggest that, if the values of “L”, “d” and fringe spacing is measured from the experimental arrangement, the wavelength of incident light can be determined, (F.S)xd ie JAZ qe i 0 eter In 1881, Abraham Michelson introduced an optical instrument used for determining wavelength of incident monochromatic light after getting successful interference of light. Due to the fact, it is called Interferometer. Later on, Michelson used the same Instrument to determine the speed of light. LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF q WWW.USMANWEB.COM Construction: (Movable Mirror) Diffuse Light Source My (Fixed Mirror) Compensator Plate Asimple type of Michelson’s interferometer consists of following parts: i. ve Monochromatic Light Source: It isa light source that emits radiations in unit wavelength range. Convex Lens: Adouble-convex lens is used to collimate incident mono-chromatic light rays. For this, fight source is kept at the focal point of convex lens which refract all of incident rays parallel to the principle axis. Thus, convex lens is acted as, “Collimator”. Semi-Silvered Glass Plate: {t isa transparent glass plate in which silver particles are imbedded with equal spacing. When light beam is incident on such a plate, it sub-divides into two parts through reflection and transmission, Moreover; this plate is oriented at an angle of 45° to obtain normal reflection. Compensating Plate: It isa pure transparent glass plate which is used to equal the additional distance travelled by the reflection beam. It is also kept at an angle of 45° to that it remains parallel to the semi-silvered glass plate. Plane Mirrors: Two plane mirrors “Mi” and “Ms” are used to reflect back the reflected beam and transmitted beam, coming out of semi-silvered glass plate towards the reference point “pg”, the mirror “Mb” is kept movable on a micrometer scale. Whereas; the mirror “Mi" remains fixed at the original position. LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , vi. Telescope: ATelescope is focused at the reference point “p” where interference of light beams takes place. By using telescope we may able to count the total number of light flashes appears at point “p” due to interference of light, Working: A monochromatic light is incident on a convex lens which collimates the rays of light and finally the beam falls on a semi-silvered glass plate. This plate sub-divides the beam into two parts through reflection and transmission. The two beams are then reflected and concentrated to reference point “p” by using twa plane mirrors “Mi” and “Mz”. An interference phenomenon. can be seen at reference point “p” by focusing a telescope. Determination of Wavelength of Incident Monochromatic Light: For this, let us assume that movable mirror of Michelson’s interferometer is moved by "x" on. micrometer scale. In this way, the reflected beam of light will travel an additional distance of "2x" as compare to the transmitted beam. Hence, path difference between the two rays of light will become “2x”, Le. a= 2x--——{i) as, reference point “p” lies in constructive region when successful interference of light takes place, therefore, the path difference between the two beams should be “m2”. Le. thus, after measuring the distance travelled by the movable mirror on scale and to count the total number of flashes (fringes) appears at reference point “p” we may determine wavelength of incident monochromatic light. Thin Films A thin layer of transparent medium is often called, “Thin Film”. Eg; the surface of a soap bubble, avery thin layer of air and a layer of kerosene oil an the surface of water can be treated as, “Thin Film". A thin film is commonly used to produce interference of light and also for determining wavelength of incident monochromatic light. In practice, a thin film is further classified into two types: i. Regular Thin Film:A thin film of uniform thickness is referred as, “Regular Thin Film”, In case of interference through regular thin film, path difference between the two LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , vi. Telescope: ATelescope is focused at the reference point “p” where interference of light beams takes place. By using telescope we may able to count the total number of light flashes appears at point “p” due to interference of light, Working: A monochromatic light is incident on a convex lens which collimates the rays of light and finally the beam falls on a semi-silvered glass plate. This plate sub-divides the beam into two parts through reflection and transmission. The two beams are then reflected and concentrated to reference point “p” by using twa plane mirrors “Mi” and “Mz”. An interference phenomenon. can be seen at reference point “p” by focusing a telescope. Determination of Wavelength of Incident Monochromatic Light: For this, let us assume that movable mirror of Michelson’s interferometer is moved by "x" on. micrometer scale. In this way, the reflected beam of light will travel an additional distance of "2x" as compare to the transmitted beam. Hence, path difference between the two rays of light will become “2x”, Le. a= 2x--——{i) as, reference point “p” lies in constructive region when successful interference of light takes place, therefore, the path difference between the two beams should be “m2”. Le. thus, after measuring the distance travelled by the movable mirror on scale and to count the total number of flashes (fringes) appears at reference point “p” we may determine wavelength of incident monochromatic light. Thin Films A thin layer of transparent medium is often called, “Thin Film”. Eg; the surface of a soap bubble, avery thin layer of air and a layer of kerosene oil an the surface of water can be treated as, “Thin Film". A thin film is commonly used to produce interference of light and also for determining wavelength of incident monochromatic light. In practice, a thin film is further classified into two types: i. Regular Thin Film:A thin film of uniform thickness is referred as, “Regular Thin Film”, In case of interference through regular thin film, path difference between the two LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , reflected beams can be changed by varying the angle of incidence of incident beam of light. ii, Irregular Thin Film Athin film of non uniform thickness is referred as, “Irregular Thin Film®.In case of interference through irregular thin film, path difference between the two reflected beams will already be changed due to non uniform thickness of thin film, Interference Through Thin Film Consider a beam of monochromatic light traveling [n air and falling on avery thin film of a refracting medium of refractive index “n" This beam of ight s partially reflected from > ‘the upper surface of the film and partially fff refracted into it, The reflected part “bc” undergoes phase reversal {phase change of 180°), It means that an incident crest after reflection from the film becomes a trough. ‘The refracted part also undergoes reflection from the lower surface of the thin film, this b part does not undergo phase reversal because: t this part of light is traveling in a denser medium ‘ ' (medium of higher refractive index) and is d reflected from the surface of a rare medium (medium of lower refractive index}. Phase reversal of 180° takes place when light waves are reflected from the surface of a medium whose refractive index is greater (denser medium) than the refractive index of the medium (rare medium) in which the light wave was traveling, Hence the refracted part after reflection from the lower surface of the film emerges out as “ef” without phase reversal (crest as crest or trough as trough) parts “be” and “ef” will interfere on reaching the eye according ta their path difference. Since “be” has suffered phase reversal of 180° therefore condition for constructive and destructive interference are interchanged. When light travels from air (rare medium) to another medium (denser medium) of refractive index “n” its wavelength changes (decreases) and is given by; Where “2” Is the wavelength of light in air and “A,” is its wavelength in the medium of refractive index “n”. For Constructive Interference: In this case; 1 c= (n +5] ae @ LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , Where “A,” is the wavelength of light in thin film and in thin film the path difference Is “2t”, where “t” is thickness of thin film, Now; Eat) Where m=0,1,2,3...... For Destructive Interference: In this case; o=mA, Or a 2t=m= The formation of Newton’s rings is an interesting application of interference of light through thin film, In this regards, Newton has used a combination of plane-o-convex lens and plane glass sheet, enclosing air film of varying thickness, to produce interference of light. Mechanism: When a monochromatic light is incident on combination of plane-o-convex lens and plane glass sheet, the irregular thickness of air film provides necessary path difference to the reflected light rays and thus points of maxima and minima will appear at different portions of the thin film resulting concentric circular bright and dark fringes on the screen called, “Newton's Rings”. If white light is used, then each component of wavelength will produce its own set of circular fringes at same place. They all overlap on one anather and thus no clear interference pattern will be appeared on screen except a central dark spat. ‘The Newton's rings are infact, formed due to interference of such light rays that are reflected from the inner and outer surfaces of air film. Since, the particles of air film gets circular shape due to curved surface of plane-o-convex lens therefore, concentric circular fringes will be formed on the screen. LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM Derivation For The Radius Of Nth Newton’s Ring: For this, let us assume that; i. The wavelength of incident monochromatic light = 1 ii. The radius of Nth Newton’s ring = tq, iii, The thickness of air film related to Nth ring =t iv. The refractive index for air film =n =1 v. _ The radius of curvature of curved surface of the plane-o-convex lens = R From the geometry of L ACB; , 1B? = AB? + AC* Ray? t(R-t)? Re =ry?+R? ARE +0? Rt = ty? +t? ----— @ Here, t? « 7,2, therefore, t? = 0 Eq(i) 2Rt = ry" ty = VOR Ty =VR x 2t----—- @ According to this expression, if thickness of the air film related to either bright ring or dark ring isknown, the radius of particular bright ring or dark ring can be determined. For The Radius of Nth Bright Ring: For this, Consider condition of maxima in case of a thin film, 2m=(m #) 2 Forair,n=1, 2t(1) = (u +5), =m 2t { +) Now, substituting the value of "2t” in eqlii) Eqhl)=> ty = fe {m3)a tye fr-ss4) hi y= a (w -3): Here, m=N-1 This expression represents radius of Nth Bright Ring. LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , For the Radius of Nth Dark Ring: For this, Consider condition of minima in case of a thin film, 2im = mA forair,n=1 2t = mA Now, substituting the value of “2t” in eq(ii) Eq(li)=> Ty = VRma. As, m= 0,1,2,3 0.00. for zerath,1%,2™,3%, Therefore, N= 1,2,3,....... for 1%, 29431 Ty = VRNA This expression represents Radius of Nth Dark Ring. Determination of Radius of Curvature of lens (R) and wavelength of light (4). As; ry N—1/2)A This equation shows the radius of curvature of lens. Similarly; = ry ~ (N=1/2)R This equation shows the wavelength of light used. The Central Spot of Newton's Rings Appears Dark, Why??? According to the equation of radius of Nth Bright Ring. ty = (ROW 1/DA n= ¥RG-1/DA v= Gh RA ne It means, 7, >.0 hence, the first bright ring will be situated slightly away from the central spot of Newton's rings. On contrary to this, the path difference between two reflected rays at the central spot remains zero and hence It lies within the destructive region. That's why, central spot of the Newton's rings always appears dark. a For first bright ring, N= 1 LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q Diffraction of Light When a beam of light is disturbed by placing an obstacle in its path, provided the dimension of obstacle is equal to wavelength of incident ight ie. DA, then beam of light appears to be bent around the edge of obstacle, The phenomenan in which a beam of light bends round the corner of an obstacle under appropriate condition is known as, “Diffraction of Light”. This phenomenon was firstly observed by Grimaldi and experimentally verified by Fresnel. There are two basic techniques for getting diffraction pattern on sereen called, “Fresnel-Type Diffraction” and Fraunhoffer-Type Diffraction « I. Fresnel-Type Diffraction: Ifthe distance between (() light souree and slit’s arrangement (i) Slt's arrangement and screen is kept finite, then diffraction pattern obtained on screen is referred as, “Fresnel-Type Diffraction”. Such a diffraction pattem appears on screen due to interfrence of spherical wave fronts. Source Grating Or Screen Apertures ii, Fraunhoffer-Type Diffraction: Ifthe distance between (i) light source and Slit's arrangement (ii) Sli’s arrangement and screen is kept infinite, then diffraction pattern obtained on screen is referred as, “Fraunhoffer-Type Diffraction”. Such a diffraction pattern appears on screen due to interference of plane wave fronts. in laboratory Fraunhoffer diffraction can be produced by using canvex lenses of suitable focal length. A convex lens between source and obstacle makes the rays parallel to each other and the second lens collects the parallel diffracted rays and focuses them to a point on the screen, fi oe || Source Source 5 VW Creel Grating Convex Lens it Grating Convex Lens Infinite latinite Sereen Distance leas LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q Difference Between Fresnel and Fraunhoffer Diffraction: Fresnel Diffraction Fraunhoffer Diffraction In Fresnel diffraction the source of light | In Fraunhoffer diffraction the source of light and the screen where diffraction is and the screen where diffraction is formed are formed are kept at finite distance from | kept at infinite distance fram the diffracting the diffracting obstacle. obstacle. In Fresnel diffraction the wave fronts In Fraunhoffer diffraction the wave fronts falling and leaving the obstacle are falling and leaving the obstacle are plane. spherical, In Fresnel diffraction the corresponding | In Fraunhoffer diffraction the corresponding fays are not parallel. rays are parallel to each other. 4 To get Fresnel diffraction in laboratory _ | To get Fraunhoffer diffraction in laboratory convex lenses are not required. two convex lenses are required. Difference Between Interference And Diffraction of Light: Interference Diffraction 1 Interference is the result of interaction | Diffraction is interaction of light coming from of light coming from two different wave | different parts of the same wave front. fronts originating from the same source. 2. | The fringe spacing may or may not be of | Diffraction fringes are not of same width. the same width, 3 Points of minimum intensity are Points of minimum intensity are not perfectly perfectly dark. dark. 4 All bright bands are of same intensity. | All bright bands are not of the same intensity. in It is an optical instrument used for measuring wavelength of incident monochromatic light after getting successful spectrum on screen. The first grating was designed by Fraunhoffer which contains thin silver wires stretched on 2 metallic frame. Construction: ‘A large number of parallel and equal spaced grooves are set on the surface of a transparent glass sheet by using a Diamond-Cutting pointer whose motion is controlled by a sensitive engine. The untouched portion between any successive grooves will act as Slit to the incident monochromatic light. A spectrum can be seen after getting diffraction of incident light through grating. Due to the fact, grating is also referred as, “Transmission Grating”. Derivation of Formula: Consider a monochromatic light falling normally on a diffraction grating. When light passes through transparent slits, it spreads out. The pattern of spreading cout of light through each slit is identical. Lb Spb wher aS breed Pig aval poet WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q Out of whole pattern we are considering one direction in which light bends at angle 0 with the initial direction, A ‘Screen| Aconvex lens converges each light ray at i of fi ; ens point “P" on the screen. The convex lens i el eel = also makes an angle @ with the grating. d=a+b=Grating element where, a = width of slits and b = width of opaque line for rays 1 and 2; path difference = AD = dsin6 Pp point "P” is maximum if; dsin§ = mA Where; m = 0,41,22,23, nse Positive and negative sign are chosen according to the position of a maximum whether itis above or below the centre of the screen, “m” Is the order of maximum i.e. for 1* order maximum m = 1 and for a 2 order image m=2 etc. By knowing the order of image “m” and the corresponding angular deviation of light @ the wavelength of light can be calculated. Brage’s Law Prof. Von-Laue has generalized that a crystalline structure contains regular and periodic three dimensional arrangement of atoms, ions or molecules, called Lattice. When monochromatic x-rays are incident on atoms of crystal lattice then each atom acts as a source of reflecting radiations of same wavelength. In this way, a crystal structure acts like a series of parallel reflecting planes. The intensity of x-rays beams reflected from two consecutive crystal planes ata certain angle will be maximum when path difference between them remains “}° or integral multiple of “A”. ie, = ATA BAA, we eres ney WA where, n= 1,2,3,4,5. Derivation For Bragg’s Equation: For this, assume a set of twa parallel reflecting planes of a crystal structure that are separated by a small distance of “d”. Let, two parallel x-rays are incident on atoms of these planes at point “B” and point “e" where they get successful reflection, If 8" be the glancing angle of the two x-rays beam, then path differenée between them is LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q given by; o=PE+EQ o =dsin@ + dsin@ nd = 2dsind Where, “A” is the wavelength of x-rays n = 1,2,3,4,5,6.. ss This equation is often called Bragg’s Equation for Diffraction of X-rays through crystals. For first order diffraction, n =1 and thus A = 2dsin@. According to this equation. i, _ If wavelength of incident x-rays and glancing angle for maximum intensity is known to us, the distance between two atomic planes of a crystal structure can be determined. ii — If “d” and “@” are known to us, the wavelength of incident x-rays can be determined, i. Distinguish between iffraction and Interference. Can there be diffraction without interference, and interference without diffraction? Ans: Difference see in notes. Interference can take place without diffraction. Interference, infect is the resultant effect of the superposition of waves coming from two coherent source (i.e. from two different wave fronts originating from the same source). Diffraction cannot take place without interference. It is due to the interference of waves coming from different parts of the same wave front, after the wave fronts have been disturb by an obstacle. li, Describe and explain the interference effect produced by thin film. An observer sees red color at certain position in an oil film, would other observers also see the red color at the same position? Ans: Interference through thin film see in Notes. When an observer sees red color in an oil film, then the film width, angle of incidence and his position are such that for six colors the conditions of destructive interference are satisfied. So he sees red color of white light. Hence, other observers viewing under these conditions will not see the red calor, because of different locations of the eye. iil. - Consider young’s double slit experiment and explain what the following parameters have to do with the light distribution on the screen? Distance between the slit. b) Width of the slit. c) Wavelength of the incident light. Ans: a) Since fringe spacing F.S = 2 Hence F.S 0 ; Thus, the greater the distance between the slits (d),the smaller the fringe spacing. b) The larger the width of the slits, the intensity af the pattern increases, but fringes become mare blurred. LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF q WWW.USMANWEB.COM v vill. c} Since fringe spacing F.S 2% Hence F.S 0 1 Thus, the longer the wavelength of the incident light, the greater the fringe spacing. Give an experiment arrangement for producing Newton's rings, Why are the fringes circular and why is central spot black (dark)? Ans; Experiment arrangement see in notes: Tha Newton’s rings are infact, formed due to interference of such light rays that are reflected fram the inner and outer surfaces of air film, Since, the particles of air film gets circular shape due to curved surface of plane-o-convex lens therefore, concentric circular fringes will be formed on the screen, Central spot is dark reason see in notes. Discuss the statement that a diffraction grating could just as will be called an interference grating. Ans: A diffraction grating consist of many slits. When light from a single source is incident on it, it produces interference pattern on the screen, The fringes are due to the interference of diffracted waves and path difference between them. For a given family of planes in a crystal, can the wavelength of incident x-rays be too large or too small to form a diffraction beam? Ans: The wavelength of incident x-rays can be too small because the distance between two consecutive planes in crystal is too small. Why are x-rays not diffracted by diffraction grating or thin films. ‘Ans: X-rays are electromagnetic waves, They have short wavelength of the order af 10m to 102m, Therefore it is not possible ta produce interference fringes of x-rays by young’s double slit method or by thin film method or by diffraction grating. The reason is that the fringe spacing is given by “Land unless the sts are senarsted by adlstance of the order of 102%m to 10°m. The fringes abtained will be closed together that they cannot be observed, Why the distant flash lights will not produce an interference pattern. Ans: Two light beams which are coherent when they are closer to the source, at large distance they do not remain coherent thus distant flash lights are unable to produce an interference pattern. Why the central point on the screen in young’s double slit arrangement is always bright? Ans: The path difference for interference pattern at centre is zero then interference is constructive and image is bright. LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM mm Ex# 9.1 Exercise based on Wave front, Huygens principle, Interference of light and Young’s Experiment: MCQs i, The lacus of all points in the same phase of vibration is, a) Wave front b) Interference c) Diffraction d) Polarization li, -~ Huygens theory of light says that light consist of. a) Wave front 6) Particles c) Photons d) Dual nature iii. The wave theory of light was proposed by. a) Galileo -b)Huygens c}Kepler -d) Newton iv, © Electromagnetic theory of light was proposed by, a) Faraday b) Maxwell ‘c)Ampere d) De-Broglie v. — Yellow light of a single wavelength can’t be, a) Reflected b) Refracted c) Dispersed d) None see vi. — The path difference in constructive interference must be. a) d=0,d,24,3A b)d=A/2, 3A/2,5A/2 ¢)d=0,1/6, 34/6, SA/6 d)d=0, 34/4, 51/4 vii, . The path difference in destructive interference must be. a) d=0,A,24,3A b)d=A/2, 34/2,5A/2 c)d=0,A/6, 34/6, 54/6 d)d=0, 34/4, 54/4 viii. One condition for interference is that the two sources should be coherent and a) Close together b) Ata far off distance c) Opposite to each other d) Coinciding ix. — Light has a) Wave nature b) Particle nature c) Dual nature d) None of these x. > Light wave are. a) Transverse wave b) Longitudinal waves c) Compressional waves and electromagnetic waves d) None of these xi, . Law f reflection and refraction can also be explained by. a) Particle nature of light b) Wave nature of light c) Quantum nature of light d) None of these xil, Huygens wave theory of light cannot explain. a) Interference b) Diffraction c) Photoelectric ect d) None of these LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM aq xii, Huygens principle is used to determine. a) Speed of light b) Location of wave front c} Wavelensth of light d) None of these xiv, According to Huygens, every point on a wave front may be regarded as a source of, ‘ a) Primary wavelets b) Secondary wavelets c} Base wavelets d) None of these xv. _ According to plank’s theory of light, energy emits in the form of packets or bundles of energy, Is called, 3 a) Quanta b) Photon c) Wavefront d) None of these xvi. To observe interference of light the condition which must be required is that the source must be. a) Monochromatic tH phase coherent ¢} Both of these dj None of these xvil. Coherent sources are thase sources which emit light of. a) Same wavelength b) Constant phase difference c) Both of A and B d} None of these xvii When the crest of one wave coincides with the trough of other wave, it gives to rise to, a) Constructive interference b) Destructive interference c) Polarization d) None of these xix, When the crest of one wave coincides with the crest of other wave, it gives to rise to. + a) Constructive interference b) Destructive interference c} Diffraction d) None of these xx, Young's double silt experiment is used to observe, ni a) Diffraction pattern b) Interference pattern c) Polarization d) None of these xxi. In young’s double slit experiment to obtain condition for phase coherence. a) Twosourcesare used b) Two slitsare used c} One slitsis used d) None of these xxi, In young’s double slit experiment at the center of screen, is formed. a) Dark fringe b) Bright fringe c}No fringe atall dj None of these xxiii, In the young’s double slit experiment at the center of screen is always bright fringe because the path difference is : a) Zero b)Maximum c) Minimum d}None of these wiv. The distance between two consecutive bright or dark fringes is called a) Path difference b) Fringe spacing c) Amplitude content d) None of these xxv. In young's double slit experiment if the distance between the slits and also distance between slit and screen is double then the fringe spacing, x a) Becomes double b) Becomes four times c) Remains same d) None of these LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , xxvi, . The fringe width in young’s double slit experiment can be increased by decreasing a) Width of the slit -b) Slits separation c) Distance between slits and screen d) None of these waxvil. ~ If the slits in young’s double slit experiment are made closer fringe spacing will a) Increase b) Decrease c)Remains same d) None of these xiv. Questions From Past Papers: i, What do you mean by interference of light? Give the conditions of interference of light waves. (1992) ii, ~ Discuss young’s double slit experiment to measure the wavelength of the light. (1992) iii, - Write a short nate on: Wave front af Huygens principle. (1992) iv, - What is interference of light? (1995) v. . Describe young's double silt experiment for demonstrating the phenomenon of interference of light. Derive the expression for fringe spacing. (1995) vi, — Write a short nate on: Interference of light. (1995) i. . IMlustrate the conditions for “Interference of light”. (1997) What is interference of light? Describe young’s double silt experiment for demonstrating the phenomenon of interference of light. Derive the expression for fringe spacing. (1998) ix. - Explain how young’s double-slit experiment proves that interference effect take place in * case of light waves. (2000) x. Derive the expression for fringe spacing in young’s double slit expertment. (2000) xi. What is meant by constructive and destructive interference? Explain young’s double-slit experiment and obtain expressions far the pasition of bright and dark fringes. (2002)Pre Eng xii. Differentiate between constructive and destructive interference. (2002)Pre Med xiii, What are the necessary conditions to observe the interference of light? Differentiate between interference and diffraction patterns. (2003)Pre Med xiv. . Define “Wave front”. State and explain Huygens principle. (2005) xv. Describe young’s double slit experiment and obtain an experiment for the position of bright fringes on the screen. (2005) LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM q xvi, Describe young’s double slit experiment and obtain an experiment for the position of first bright fringe. (2006) yvil, _ Using young’s double slit arrangement obtain an expression for the position of bright fringes. Also calculate the bright fringe spacing. (2008) xviil, Describe young’s double slit experiment and the formula for fringe spacing. (2009) xix, Describe young’s double slit experiment. Derive the relevant expression and the formula for fringe spacing. (2011) xx.’ What are the basic conditions for interference of light? How did Thomus young experimentally confirm the wave nature of light? Derive the expression for fringe spacing, (2013) yxi, Explain the Young’s Double Slits experiment and derive formula for fringe spacing. (2015) xxi. Give the quantitative description of Young's double slit experiment. Derive an expression for fringe spacings. (2017) Numericals From Past Papers: i. Monochromatic light of wavelength 5890A° is used to illuminate two narrow slits Imm apart, im from the screen. What is the separation of fringes obtained on screen in away from the slits? (1980) Ans: 5.89x10“m ii, In ayoung’s double slit experiment the light has @ wavelength of 6x10cm. if the distance between the slits is 0.05cm and the screen is 2m from the slit, what is the distance between two successive bright fringes? (1982) Ans: 2.4x103m iii, Two slits, 0.03cm apart, are illuminated by a monochromatic source. Fringe of width 0,059em are obtained on a screen at a distance of 30cm from the slits, calculate the wavelength of light. (1984) Ans: 5.9x107%m lv. in ayoung’s double slit arrangement the slits are 0.7mm apart and light of wavelength 5890A° is used. What is the separation of the fringes abtained ana screen 140cm away from the slits? (1986) Ans: 0.00178m v. Inayoung’s double slit arrangement the slits are 0.9mm apart and light of wavelength 5890A°is used. What is the separation of the fringes obtained an a screen 180cm away from the slits? (1988) Ans: 1.178x103m vi. Light from a narrow slit passes through two parallel slits 2mm apart. The two successive dark interference fringes ona screen 100m away are 0.2mm apart. Calculate the wavelength of the light used. What would have been fringe separation for light of wavelength 6000A° used? (2990) Ans: 164°, 30mm vil. Two parallel slits are illuminated by the light of two wavelength one of which is dark line of the known wavelength coincides with the fifth bright line of the unknown, wavelength, calculate the unknown wavelength, (1992) Ans: 5.4x107m LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM Ay. viii, Interference fringes were produced by twa slits on a screen 0.8m from them which the light of wavelength 5.8x107m was used. If the separation between the first and the fifth bright fringes is 2.5mm, calculate the separation of the two slits. (1995) Ans: 0.7424x103m ix, Interference fringes were produced by light coming from two slits 0.3mm apart. If five fringes occupied 1.75mm ona screen at 200mm from the slits, find the wavelength of light used, (2000) Ans: 525042 x. Ina double slit experiment the separation of the slit is 1.8mm and the fringe spacing Is 0.30m at distance of 1200mm from the 'slits, find the wavelength of light. © (2002)Pre Eng Ans: 4500A° xi. Interference fringes were produced by two slits 0.25mm apart on a screen 150mm from the slits. If ten fringes occupy 3.75mm what is the wavelength of the light produces fringes? (2008) Ans: 545.83 xii, Ina double slit experiment, eight fringes occupy 2,62mm ona screen 145mm away from the slits. The wavelength of light is 545nm. Find the slit separation. (2012) Ans: 2.11x10¢m xiii, Ina double slit experiment the separation of the slits is 1.9mm and the fringe spacing is 0.31mm at a distance of 1m from the slits. Find the wavelength of the light? (2016) Ans: 0.589x10m Numericals From Book: i. Ina double slit experiment the separation of the slits is 1.9mm and the fringe spacing is 0.3imm at a distance of 1 from the slits. Find the wavelength of the light? Ans: 0.589x10%m ii. -: Interference fringes were produced by two slits 0.25mm apart on a screen 150mm fram the slits. If eight fringes occupy 2.62mm. What is the wavelength of the light producing the fringes? Ans: 5.46x107m Assignment # 9.1 |. What is the Interference of light? Explain its conditions, li, Describe the conditions of constructive and destructive interference. lil, Describe Young's double slits experiment for Interference of light. Derive the expression for fringe spacing and wavelength of light. Iv.” Red light of wavelength 644nm, from a point source, passes through two parallel and narrow slits which are Imm apart. Determine the distance between the central bright fringe and the third dark fringe formed ona screen im away. Ans: 1.61mm LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM x Ex 9.2 Exercise Base on Michelson’s Interferometer: MCQs i. Michelson’s interferometer is usually used to measure. a} Amplitude of light b) Wavelength of light c) Nature of light d) nivig onthe fi, Michelson’s interferometer can also be used to determine the. a) Speed of light b) Amplitude of light c) Direction of light d) None ites iil. In Michelson’s Interferometer a convex lens is used to, a} Collimate incident light b) Disperse incident light c) Taina igen light d) None of these iv. Semi-silvered glass plate Is used to. a) Disperse incident light b) Collimate incident Tht ¢} Divide incident light d) None of these v. Wavelength of incident ve is given by a) =% byd== cg A=Z a) None ot these ™ 2m Questions From Past Papers: i. Write short note on: Michelson’s interferometer. (2000) li. Given the construction and working of Michelson’s interferometer. How it is used to determine the wavelength of amonochromatic light? (2002)Pre Eng iii, Describe the construction and working of the Michelson’s interferometer. (2002)Pre Med iv. Give construction of Michelson’ interferometer. Draw a neat diagram. How it can be used to determine the wavelength of light? (2004) Numericals From Past Papers: i... How many fringes will be pass a reference point if the maveable mirror of the Michelson’s interferometer is moved by 0.08mm? The wavelength of light used is S5800A°. (1998) Ans: 276 ji How much should the moveable mirror of the Michelson’s interferometer be moved in order to observe 400 fringes with reference toa paint? The wavelength of the light used is 5890A°, (2002)Pre Eng Ans: 1.17mm LES Bee tid tl Fig ail pee WWW.USMANWEB.COMUSMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM ” iil, 271 fringes pass a reference point when the movable mirror of Michelson’s interferometer is moved by 0.08mm. find the wavelength of the light used in Angstrom. (2002)Pre Med Ans: 5904A° Numericals From Book: i, How many fringes will be pass a reference point if the moveable mirror of the Michelson's interferometer Is moved by 0.08mm. The wavelength of light used is SB00A°, Ans: 275 Assignment # 9.2 i, _ Give the construction and working of Michelson’s interferometer. How is it used to determine the wavelength of a monochromatic light? ji, When one leg of a Michelson interferometer is lengthened slightly, 150 dark fringes sweep through the field of view. |f the light used has wavelength of 480nm, how far was the mirror in that leg moved? Ans: 0.036mm Ex # 9.3 Exercise based on thin films & Newton’s Rings: MCQs. 1. In thin film interference, the position of constructive and destructive interference are interchanged due to. ; a) Phase coherence b) Diffraction c)Phase reversal d) None of these ii. Asoap film ar oil film on water in day light exhibits beautiful color due to a) Dispersion b) Diffraction c) Interference of light d) None of these iii, .. Newton's rings are formed due to. ‘i a) ‘Diffraction b) Polarization c) Interference d) None of these iv, Allens used in the experiment to get Newton's rings is t a) Plano-convex b) Plano-cancave c) Double convex d) None of these v, _ Newton’s rings are. - a) Rectangular b) Spherical c) Concentric circle d) None of these vi. _ In Newton's rings the center spot is always be a) Bright spot b) Dark spot c) Colored spot d) None thee vil. . We use. in Newton's rings. a) White light -b) Monochromatic light c} Both A and 8 d) None of these vil. In the formation of Newton’s rings the condition of constructive and destructive Interference are reversed due to, a) Phase coherence b} Phase reversal c} tienes d) None of these LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF " WWW.USMANWEB.COM ii, ili, Wy. ve vi. vii. vill. c € B > o = Questions From Past Papers: vi. vii. viii, ix, e xi. Numericlas From Past Pape! tii. What are Newton's rings? They prove an important property of light. What is this property? (1994) Show how Newton's ring can be used to find the radius of curvature of lens? (1994) For the radius of Newton's ring derive an expression of the relation between the radius of curvature of lens and wavelength of light, (1997) ‘What are Newton’s ring? Give experimental arrangement for producing Newton's ring. (1999) Derive an expression for the radius of curvature of the lens used in the arrangement. (1999) How are Newton's ring formed? Derive an expression for the radius of n™ bright Newton's rings? (2001) What are Newton's rings? Derive an expression for the wavelength of light used in Newton's rings, (2003)Pre Med Describe the experimental arrangement for observing Newton's rings. Derive an expression for the radius of n'* bright ring, (2007) What are Newton's rings? Derive the expression for the radius of nth bright ring. (2012) What are Newton's rings? Derive the expression for the radius of nth bright ring. (2014) What are Newton's rings? Derive the expression for the radius of nth bright ring. (2016) if the radius of 14" dark ring is 1mm, and the radius of curvature of the lens is 126mm. Calculate the wavelength of light. (1997) Ans: 5.6689x107m If the radius of the 14% Newton's ring is 1.0mm when the light of wavelength 5,89x10"m is used. What is the radius of curvature of the lower surface of the lens used? (1999) Ans:12.5em Determine the wavelength of a monochromatic light if 14" bright Newton's has radius. of Imm when a plan convex lens of radius of curvature 126mm is used. (2001) Ans: 5878.90" Ifthe diameter of the 10" bright Newton's ring is 0,0025m when the light of wavelength 5893A° Is used, What is the radius of curvature of the Plano-Convex lens? Also calculate the thickness of the air film corresponding to this ring. (2003)Pre Med Ans: 0.279m, 2.8x10%m Ifthe radius of 12" dark Newton’s ring is Imm when the light of wavelength SB90A° is used. What is the radius of curvature of the lower surface of the lens used? {2003)Pre Eng Ans: 0.1414m LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM vi. vii. If the radius of the 14" bright Newton’s ring is 1mm and the radius of curvature of the lens is 125mm. Calculate the wavelength of the light. (2010) Ans: 5925A° If the radius of the 5‘ dark Newton's ring is 3mm when light of wavelength 5.89x107m is used, what will be the radius of curvature of lower surface of the lens used? (2017) Numericals From Book: fit. Newton's rings are formed between a lens and a flat glass surface of wavelength 5.88x107m. if the light passes through the gap at 30° to the vertical and the fifth dark ting is of diameter 9mm. What is the radius of the curvature of the lens? Ans: 23,8m If the radius of the 14% Newton's ring is imm. when light of wavelength 5.89x107m is used. What is the radius of curvature of the lower surface of the lens used? ‘Ans: 125.7mm Assignment # 9.3 ‘What are Newton's rings? Give experimental arrangement for producing Newton's rings. Derive the expressions for; a) The radius of nth bright ring b} The radius of nth dark ring c) Wavelength of light 4) Radius of curvature of Plano-convex lens. Ex #9.4 Exercise based on Diffraction of light: MCQs The phenomenon of bending of light around the edges of an obstacle is called . a) Interference ‘b) Diffraction c) Polarization d) None of these Diffraction effect is ‘ a) More forround edge b) More for sharp edge c) Less for sharp edged) None of these The diffraction is found to be prominent when the wavelength of light is : a) Small as compared with the size of obstacle 'b) Large as compared with the size of obstacle ¢) Equal as compared with the size of obstacle d) None of the these In Fresnel diffraction the source of light and screen are kept at, a) Finite distance b) Infinite distance c) Adjacent to each other d) None of these LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM x vy. In Fraunhofer diffraction the source of light and screen are kept at, a} Finite distance -b) Infinite distance c) Adjacent to each other d} None of these vi. In Fresnel diffraction the incident wave front is, a) Spherical b) Plane c} Parallel d) None of these vil. In Frauhofer diffraction the incident wave front Is, a} Plane b)Spherical c}Circular d) None of these vill, . Aglass plate upon which thousands of equally spaced opaque lines are ruled is called, a} Interferorneter b) Thin film ¢} Diffraction grating d) None of these ix. Atypical diffraction grating has lines per centimeter. b} 10to 100 b) 100 te 500 ¢}5000to 6000 d) None of these x. The distance between centers of two adjacent slits Is called, a} Grating space b) Gratingelement c} Grating distance d) None ofthese xi The spacing between the parallel lines is served as a) Slit b)Fringe c) Source oflight d) None of these i] it tM Lv fv lw [wi [me | x |x B 8 A 8 A A c c B A o Questions From Past Papers: i. Write ashort nate on; Diffraction grating (1993) li. Write a short nate on: Diffraction of light (1995) ii, What are the difference between Fresnel and Fraunhofer diffraction? Explain. Describe a diffraction grating. How it can be used to find the wavelength of sodium light? (1996) iv. Write a short note on: (2000) What is diffraction grating? Give an expression for the determination of the wavelength of a monochromatic source by diffraction grating. (2001) vi. _ Differentiate between the following: Fresnel and Fraunhofer diffraction. (2002)Pre Med vil. Differentiate between interference and diffraction patterns. (2003)Pre Eng vill. Describe diffraction grating. How it can be used to determine wavelength of a monochromatic light? (Derive a relevant mathematical expression). (2003)Pre Eng ix. Whats the difference between Fresnel and Fraunhofer diffraction? Derive Brage’s law for X-ray diffraction. (2004) x. Define diffraction grating? How it is used to determine the wavelength of a monochromatic light? Derive the mathematical expression? (2008) xl. What is diffraction grating? How it Is used to determine the wavelength sodium light? Derive the relevant mathematical expression. (2007) LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM xi, xiii, xiv. xv. xvie xvii. xviii. What Is the difference between Fresnel and Fraunhofer diffraction? Derive Bragg’s law for X-ray diffraction. (2008) ‘What is interference of light? Give the differences of Fresnel’s and Fraunhofer's diffraction. (2010) What is diffraction of light and what is diffraction grating? Derive an expression for the wavelength of light by diffraction grating. (2010) Why are X-rays not diffracted by diffraction grating or thin films? (2011) Differentiate between Fresnel's and Fraunhofer’s diffraction. (2013) Differentiate between Fresnel’s and Fraunhofer’s diffraction. (2014) What are the necessary conditions to observe the interference of light? Differentiate between interference and diffraction, (2015) Numericals From Past Papers: vi. vii, vill. How many lines per cm are there in grating which gives first order spectrum at an angle of 30° from the normal with the light of wavelength 6x10%cm? (1979) Ans: 8333.33 lines/cm How many lines percm are there in grating which gives second order spectrum at an angle of 12° from the normal with the light of wavelength 4160A°? Take sin12° = 0.208. (1981) Ans: 2498.93lines/cm Monochromatic light wavelength of 6x107m falls normally on a diffraction grating having 500 lines per millimeter. Calculate the angular deviation of the secand order spectrum obtained on this side of the normal. (1987) Ans: 36.86° Abeam of monochromatic light is incident normally on a grating which has 300 lines per millimeter. Find the wavelength of the spectral line for which in second order is 20°. (sin20°=0.342) (1989) Ans: 5700.3 A° A diffraction grating produces a deviation of 12° in the second order with the light of wavelength 4160A°, Find the grating element end the number of lines per centimeter of the grating. (1991) Ans: 4,0017x10°,2498.93lines/cm Red light falls normally on a diffraction grating rated 4000lines/cm and the second order image is diffracted 34° from the normal. Compute the wavelength of the red light in angstroms. (1996) Ans: 69904" Agreen light of wavelength 5400A° is diffracted by a diffraction grating having 2000lines/cm. Compute the angular deviation of the third order image. (1998) Ans: 18.9° X-rays of wavelength 1.54A° are diffracted by a crystal whose planes are 2.81A° apart. Find the glancing angle for the first order. (2004) Ans: 15.90° A green light of wavelength $4004’ is diffracted by a diffraction grating having 2000lines/cm. Compute the angular deviation for the 3 order image. (2005) Ans: 18.9° LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF wv WWW.USMANWEB.COM xi, wil. vil. xiv. Ifa diffraction grating produces a 1* order spectrum of light of wavelength 6x107m at an angle of 20° from the normal, calculate the number af lines per millimeter. (2007) Ans: 570lines/mm Aparallel beam of X-ray is diffracted by a crystal. The first order maximum is obtained when the glancing angle of the incident is 6.5°. lf the distance between the atomic planes of the crystal is 2.8A°. Calculate the wavelength of the radiation. {2009} Ans: 0.63394°, Ifa diffraction grating produces a 1" order spectrum of light of wavelength 6x107%m at an angle of 20°from the normal, calculate the number of lines per millimeter. (2014) Ans; 0.570ines/mm ‘Agreen light of wavelength 5400A° is diffracted by a diffraction grating having 2000lines/cm. Compute the angular deviation of the third order image. (2014) Ans: 18.9° Aparallel beam of X-rays is diffracted by a rock salt crystals. The 1* order maximum, being obtained when the glancing angle of incidence is 6 degree and 5 minutes, the distance between the atomic planes of crystal is 2.81x10""m. Calculate the wavelength of radiation, (2015) Ans: 0.5955x10%m A diffraction grating produces 3 order spectrum of the light of wavelength 7x107m at an angle of 30° from the normal. What is its grating element? Calculate the number of lines per mm. (2017) Numericals From Book: Agreen light of wavelength 5400A° is diffracted by a diffraction grating having 2000lines/em. a} Compute the angular deviation of the third order image. b) Isa 10 order image possible? Ans: a) 18.9°, b) impossible Light of a wavelength 6x107m falls normally on a diffraction grating with 4oo lines per mm. At what angle to the normal are the 1%, 2" and 3" order spectra produced? Ans; 13,9°, 28.7°, 46,1° \fa diffraction grating produced a 1 order spectrum of light of wavelength 107m at an angle of 20° from the normal. What is its spacing and also calculate the number of lines per mm? Ans: 1.75x103mm, 5.7x107lines/mm. How far apart are the diffraction plane in a NaCl crystal for which X-rays of wavelength 1.54A° make glancing angle of 15°-54in the 1" order? Ans: 2.81A° A parallel beam of X-rays is diffracted by rocksalt crystal the 1" order maximum being obtained when the glancing angle of incident is 6 degree and 5 minutes. The distance between the atomic planes of the crystal is 2.81x10"m. Calculate the wavelength of the radiation, Ans: 0.5952x10%m LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM vi. vi. vii. viii. wl. Adiffraction grating with 10000lines per centimeter is illuminated by yellow light of wavelength 589mm. At what angles are the 1 and 2" order bright fringes seen? Ans: 0,589, 1.178 Assignment # 9.4 What is diffraction of light? Give differences between Fresnel and Fraunhofer diffraction. Differentiate between interference af light and Diffraction of light. ‘What is the diffraction grating and how it can be used to determine wavelength of a monachromatic light? Red light falls normally ona diffraction grating ruled 4000lines/cm and the second-order image is diffracted 34° from the normal. Compute the wavelength of the light. Ans: 6990A° Green light of wavelength 540nm is diffracted by grating ruled with 2000lines/cm. a) Compute the angular deviation of the third-order image. b) [sa 10"-order image possible? Ans: 18.5°, impossible Green light of wavelength 500nm is incident normally on a grating and the second-order image is diffracted 32° from the normal. How many lines/cm are marked on the grating? Ans: 5300 A narrow beam of yellow light of wavelength 600nm is incident normally on a diffraction grating tuled 2000lines/cm, and images are formed on a screen parallel to the grating and 1m distant. ‘Compute the distance along the screen from the central bright line to the first-order lines. [Hint: dsin@ = mA, tan® =y/L] Ans: Y=12.1em Determine the ratio of wavelengths of two spectral lines if the second-order image of one line concides with the third-order image of the other line, both lines being examined by means of the same grating. [Hint: myAy = mp4) Ans: 3:2 How far apart are the diffraction planes in a NaCl crystal for which X-rays of wavelength 1.54°A make a glancing angle of 15°54’ in the first order? Ans: 2.81°A MCQs From Past Papers (1994) ‘The dispersion of white light into seven different colors when passed length a prism is due to: a) Different intensities b) Different amplitude c) Different wavelength When light enters intoa denser medium its velocity a) Increases b) Decreases c} Remains the same (1995) ‘A monochromatic light beam is entering fram one medium into anather which one of the following properties remalns unchanged? ’a) Amplitude b] Velocity c} Wavelength d) Frequency Color in soap bubbles is due to: a) Polarization of light _b) Interference of light c] Reflection of light (1996) Which of the following phenomenon cannot be explained by the wave theary? a) “Interference b) Diffraction c} Photoelectric effect In Newton's ring the central spot is always: a) Dark b)Bright <) Red (1998) LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM vill wi. wil, vil. xa, wll, ‘The structure of crystal can be studied with the help of, of x-rays: a} Interference b) Diffraction’ Polarization ‘The phenomenon of, proves that light waves are transverse in nature, a) Reflection bj Refraction c) interference d) Polarization (2000) ‘When light (white) pastes through thin film colors are formed due to: 2) Diffraction of ight b) interference of light «} Dispersion of light) Both interference and dispersion of lights Photoelectric effect proves that light consists of: 2) Particle b) Waves Polarization of light due to tourmaline crystals takes place because of: a) Reflection b) Selective absorption c) Refraction ‘When both the paint source and the screen are placed at finite the phenomenon is called: a) Fresnel diffraction b) Fraunhofer diffraction (2001) Diffraction of ightisa special type oft ‘3) Reflection h) Refraction c} Interference d) Polarization in Michelson interferometer, semi-silvered plate is used to obtain: a) Dispersion b) Phase coherence ¢) Monochromatic light ) Unpolarized light Which of the following is not electromagnetic waves? a) X-rays b) Radlowaves ¢} Ultra waves d) a-rays ‘The condition for'interference in thin films is reversed because af: a) Small thickness of film b) Refraction c} Phase reversal Which of the following demonstrate the transverse nature af light wave: 3) Interference b) Polarization c) Diffraction d) Reflection (2002)Pre Med Light possesses: a) Transversemature b) Electromagnetic character c) Dual natured) All of these ‘The evidence af transverse nature of light is provided by: a) Polarization b}Diffraction c} Interference d) Dispersion Inthin flms destructive interference occurs when the path difference is: a) Anodd multiple of half-wavelength b) Onlyzn-even multiple of wavelength )_Aninteger multiple of wavelength d)_ None of the abave The number of lines rules per centimeter Ina diffraction grating is 4000. Its grating element is: a) 25x04 b)25x10%m ¢) 4xt0%m d) 4xt0°m (2002}Pre Eng The transverse nature oflight can be verified by: a) Interference bj Diffraction ¢} Polarization d) Refraction Which of the following equation represents the Brags'slaw? a) mA =2dsin® b) md = 2dsinO/2 c)mA=dsin2e the characteristics property of ight which does not change with the medkum is: a). frequency b} Wavelength ¢} Velocity (2003}Pre Eng The appearance of colorin soap bubbles is due to: LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM voi, xvii, wovili. 20%, xo, ool, xvii, roxxiv, wowil. wonvili. cy xd a} Polarization b) Diffraction c) Reflection d) Interference In thin film interference the position of constructive interference and destructive interference are interchanged due ta: a) Phase difference b} Phasereversal c) Diffraction d) interference ‘The bending of light around the sharp corners of an obstacle is called: a) Interference 'b) Polarization c) Diffraction d) Refraction (2003)Pre Med Electramagnetic waves consist of oscillatory electric field and magnetic field, Both fields are: a) Parallel to each other b) Parallel to the direction of propagation c} Perpendicular to each other d) None of these A monochromatic beam of light is entering from one medium into another, The property which remains unchanged Is: a) Amplitude — b) velocity c) Frequency d} Wavelength ‘The bending of light around the obstacle is called: a) Polarization b) Interference ¢) Diffraction d} Refraction ‘The dispersion of white light after passing through on prism Is due to: 3) Different intensities b) Different amplitudes c) Different temperature 4) Different ‘wavelengths (2004) The characteristic property of light which does not change with the medium is: a) Frequency b) Wavelength c} Velocity In young’s double-slit experiment the fringe spacing is: a) da/L . b) AL/d ch} d/ab Which of the following phenomenon cannot be explained by the wave theory? a) Interference -b) Diffraction c} Photoelectric effect ‘The condition for the interference in a thin film is reversed because of: a) Small thickness. b) Phase reversal c) Refraction (2005) Colors in thin soap films are due to: a) Refraction of ight b} Diffraction offight c} Interference of light d) Scattering of light To replace a bright fringe by the next bright in Michelson’s interferometer, the movable mirror is moved through a distance equal ta; a) A B)A/2 c) A/4 Gh2a Which property of light Is used to determine the concentration of an optically active substance such as sugar? a) Interference 'b} Dispersion c} Diffraction d) Polarization ‘The wave theory of light cannot explain: a) Polarization b) Photoelectric effect c) Interference d) Diffraction (2008) In thin film interference the position of constructive Interference and destructive interference are interchanged due to: 3) Phase coherence b} Interference c} Phase reversal dj Diffraction If 2000 lines/cm are ruled on a grating, its grating element is: a) Sx10*m (b)Sx10%m c) 5x10%m d) 5x107m LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF WWW.USMANWEB.COM , (2007) al, Wave front near the point source is: al Plane b) Cylindrical c} Spherical d) Conical xiii. Chromatic Aberration can be reduced by using a combination of; a} Twaconverging lenses 'b) Two diverging lens ¢) A converging lens and a diverging lens of different materials d) None xiv, The phenomenon of light which proves that light waves are transverse Is: a) Reflection bj Refraction ¢) Interference d] Polarization xiv. . The condition for the interference in a thin film is reversed because of: a) Diffraction b) Phase coherence c) Refraction d) Phase reversal ivi. Diffraction of x-rays can be studied by using: a) Athinfilm bj Crystalline solld c} polarization d) Dispersion (2008) ivi, ‘The transverse nature of light can be verified by: a) Interference b) Diffraction ¢} Polarization } Dispersion xlvil The condition for the constructive and destructive interference are reversed in case of thin fms due to: a) Phase reversal of one part of a wave b) Phase reversal of both parts of a wave ¢) Phase reversal of none d) Change in irequency af waves xilx. Yellow light fram a sodium lamp Is used to farm a Newton's ring, The central spot in Newton's ring will be: a} Yellow b) Bright c} Dark d) Neither bright nor dark (2010) |. Inyoung's double slit experiment the condition for the constructive Interference is that the path difference must be: a) An odd multiple of half-wavelength tb} Onlyan even multiple of wavelength ) Aninteger multiple of wavelength d) None ofthe above (2031) |i, The wave theory of light cannot explain: a) Polarization b) Photoelectric effect c) interference d) Diffraction (2012) Ii, According to Maxwell's theory, light travels in the form of: a) Transverse wave b) Longitudinal wave c| Mechanical wave d) Electromagnetic wave lil, sin@ = 8 Is specifically ess than: a} 15 degree b)10degree ¢) 5 degree d) radian liv. Huygen’s principle is used ta: a}. Determine the speed b) Expressed polarization c) Locate the wavefront d) Find the refractive index (2013) v. Monochromatic yellow ight is unable to show: a) Reflection b) Refraction c} Dispersion d) Interference LES Bee tid tl Fig ail pee WWW.USMANWEB.COM> USIMANWEB ALL CLASS NOTES DOWNLOAD IN PDF , WWW.USMANWEB.COM (2014) Ii. Diffraction of light is special type of: a) Reflection ‘b) Refraction cl interference “d) Polarization Ii. In yaung’s double slit experiment, the fringe spating is: a Soy gt wud (015) Iviil. Polarization of light due to tourmaline crystals takes place because of: 3) Reflection b) Absorption __¢} Refraction _d) Collision fix. Newton's rings illustrate the phenomenon of; a) Polarization b) Diffraction _¢) Interference _¢) Dispersion Ik. _ Inthin film interference, the positions of constructive and destructive interference are interchanged due to’ a) Phasecoherence bj Phase reversal ‘c) Diffraction . d) Interference (2016) bi, The condition of interference in thin film are reversed due ta: a) Diffraction b)Phasecoherence c) Refraction — d) Phase reversal ‘ni, This equation represents Brage’s law: a) mA=2dsin® b)mA=dsind ¢)2mA=dsind 4) 2mA = 3dsind (2017) iuiil, The number of lines per cm of a diffraction grating Is 4000, its grating element is; a) 2.5x10“cm b)25xt0%cm c}4xi0%cm dd) 4x10%cm L ii. iti. iv. v vi. vii._| viii. iT x xi, xii, (3 B A B D Bi A B A xiii. xiv. xviii. xix, xX. xxi, boii, Pxxiil. [xiv C B D A c B | -c A_| A xxv, xxvi, 3x, [>xxl. —fowoeil. fowl, frciv. —_rxev, xvi. D B c D A B c 8 ¢ kxvil. xviii, paxin. xl, .|xlh xliv, | vlv. | xlvi -_ [xlvil, klviif, B D B c c o D B c A xlix. I. li. lik, titi. Wi. | wii] Iviii, li. Ix. 5 c B D c c B B c 8 Ixi, | bai. belli, D-{A A LES Bee tid tl Fig ail pee WWW.USMANWEB.COM