11.Interference and Diffraction (Marks 04/06)

Transcription

1 11.Interference and Diffraction (Marks 04/06) Q. State and explain the principle of superposition of waves :- Principle of Superposition of wave Statement :- When two or more waves traveling through a medium arrive at a point of a medium simultaneously, then each wave produces it s own displacement independent of the other waves. Therefore the resultant displacement at that point is equal to the vector sum of the displacement due to all the waves. Explanation :- Let y 1, y, y 3,.. y n are the displacements produced by the individual waves arriving at a point simultaneously then the resultant displacement at that point is Y = y 1 + y + y y n Q. What is Interference? Give their types Interference :- The phenomenon of enhancement or cancellation of the displacement produced due to the superposition of waves is called as interference. There are two types of interference 1)Constructive interference This phenomenon is called constructive interference. ) Destructive Interference :- When the two light waves travelling through a medium arrive at a point of the medium simultaneously in out of phase i.e. crest of one wave coincides with trough of other or trough of one wave coincides with crest of other, therefore the resultant displacement or resultant intensity at that point is minimum and the point becomes dark. This phenomenon is called destructive interference. Q. State the condition to get constructive and destructive interference of light. OR State the conditions for obtaining the constructive and the destructive interference of light. Condition for Constructive interference :- When the two light waves arrive at a point simultaneously in same phase, there is a constructive interference at that point. The resultant intensity at that point is maximum and the point becomes bright. ) Destructive interference Q. Define the terms Constructive and Destructive interference. 1)Constructive interference :- When the two waves (light wave) travelling through a medium arrive at a point of medium simultaneously in same phase i.e. crest of one wave coincides with crest of another wave or trough of one wave coincides trough of other. Therefore the resultant displacement or resultant intensity at that point is will be maximum. Therefore, the point become bright. Thus the condition for brightness of the point is that the two light waves should be in same phase i.e. the phase difference between them is 0, π, 4π, 6π, 8π.. etc. Therefore, Phase difference = 0, π, 4π, 6π. 1

2 Phasedifference=nπ Where n = 0, 1,, 3,...etc. However, a phase difference of π corresponds to a path differance of λ (wave length) Hence, for the brightness of a point Path difference = 0, λ, λ, 3λ, 4λ... Path difference=nλ where n = 0,1,, 3,...etc. In other words, we can say that a point is bright if the path difference between the two light waves is an integral multiple of wavelength of the light. Condition for destructive Interference :- When the two light waves arrive at a point simultaneously in out of phase, there is a destructive interference at that point and the resultant intensity at that point is minimum and the point becomes dark. In other words, we can say that a point is dark, if path difference between the two light waves is an odd multiple of half of the wavelength of the light. Q. Derive an expression for the amplitude and intensity at any point on the screen in Young s double slit experiment. Ans:- Consider two coherent sources S 1 and S. Suppose two light waves are emitted from these two source reaches to the point P on the screen. Let y 1 and y are the displacements produced due to these waves at point P such that waves y 1 = a 1 sinωt and y = a sin ωt + φ Where, a 1 and a are the amplitude of the two φ =Phase difference between the two light waves reaches at point P Now by the principle of superposition of waves,the resultant displacement (Y) is given by Y = y 1 + y Thus, the condition for a dark point is that the two light waves are in out of phase i.e. the phase difference between them is π, 3π, 5π, 7π, 9π,. Phase difference of π difference of λ Hence, for darkness path difference = (n 1) λ Where, n = 1,,3,4.. etc. Corresponds to a path Pat difference = λ, 3 λ 5 λ = a 1 sinωt + a sin ωt + φ = R sin ωt + φ Where, R is the resultant amplitude at point P which is given by R = a 1 + a + a 1 a cosφ And the phase difference (φ) is given by ϕ = tan 1 a sin ϕ a 1 +a cos ϕ The resultant wave is also a simple harmonic wave of the same frequency.

3 The intensity of a wave is directly proportional to the square of amplitude of the wave, we have I R I a 1 + a + a 1 a cosφ If I 1 and I are the intensities of the two light waves then we have, If I 1 = I =I, then I R = I 1 + I 1 + I 1 I cosφ I R = I + I + I I cosφ = I + I + I cosφ = I(1 + cosφ) = I cos φ = 4I cos φ Now for the destructive interference Phase difference (φ) = π, 3π, 5π, 7π, 9π,. Therefore, I min a 1 + a a 1 a Or I min a 1 a R min = ( a 1 a ) I min = a 1 a I min = I 1 I If a 1 = a = a therefore I 1 = I =I then R min = 0 I min = 0 Ratio of the maximum to the minimum intensities at point P is given by Now for the constructive interference Phase difference (φ) = 0, π, 4π, 6π. Therefore, I max I min = a 1 + a a 1 a = Where, r = a 1 a = amplitude ratio r + 1 r 1 I max a 1 + a + a 1 a Or I max a 1 + a R max = ( a 1 + a ) I max = a 1 + a I max = I 1 + I If a 1 = a = a therefore I 1 = I =I then R max = a I max = 4a = 4I 3 If W 1 and W be the widths of the two slits S 1 and S and I 1 and I be the intensities of the light due to the respective slits then But we know that Therefore we can write W 1 W = I 1 I I 1 = a 1 I a W 1 W = a 1 Q. Describe the Young s Double slit Experiment and state it s importance. a

4 OR Describe the Young s Double slit experiment for obtaining a steady interference pattern. 1)It was the first experimental proof to the wave nature of light. ) From this experiment the wavelength of monochromatic light can be determined. What is the steady interference pattern? *Steady Interference pattern :- The pattern containing alternate bright and dark bands which remains unaltered with respect to time is called as steady interference pattern (Stationary interference pattern) * Bright band / Bright fringe :- The locus of all the bright points in the interference of the light is called bright band/ bright fringe. * Dark Band / Dark fringe :- 1)In Young s double slit experiment, a narrow slit S is illuminated by monochromatic light as shown in the above fig. ) The light from the slit S is allowed to fall on two other slits S 1 and S kept parallel to it and in symmetrical manner. 3) The two slits S 1 and S acts as two coherent sources and the light from them is allowed to fall on the screen. 4) A stationary interference pattern is produced on the screen. 5) The interference pattern consist of alternate bright and dark bands which are parallel to the slit.these bands are equally spaced i. e. the distance between two consecutive bright bands or two consecutive dark bands is constant. Importance of Young s Experiment :- The locus of all the dark points in the interference of the light is called dark band. State the conditions necessary for obtaining steady interference pattern. 1)Two sources of light must be monochromatic. ) Two sources of light must be coherent. 3) Two sources of light must be equally bright. 4) Two sources of light must be narrow. 5) Two sources of light must be close to each other. Q. What is monochromatic source of light? It is a source of light which emmits the light waves of single wavelength. Exa:- Sodium Lamp. Q. Why two sources should be equally bright for obtaining a steady interference pattern. 4

5 The two sources of light should be equally bright means they should emit the light waves of equal amplitude, as the intensity of light is directely proportional to the square of amplitude. When the two sources are emmiting the light waves of different amplitude then they are interfering in out of phase and resultant amplitude is not equal to zero and we get less bright point. Therefore, in this case we get the interference pattern containing alternate bright bands and less bright bands, which is not steady. Thus, two sources should have the light waves of equal amplitude (equal brightness) Q. What are coherent source? How are they produced? Coherent Sources :- Two sources of the light are said to be coherent sources if they are derived from same source of light and they are emmitting the light waves having constant phase difference. They are produced in the following manner. 1)In Young s double Slit experiment,a slit is illuminated by a mono chromatic light and the light from the illuminated slit is made to incident on two parallel slits. These slits acts as coherent sources sending out the light waves having constant phase difference. as a bright for a moment and it will appear as less bright or dark at the other moment. Hence, steady interference pattern is not obtained. If the two sources of light are coherent then will emit the light waves of constant phase difference. Hence, the point of constructive interference appears as bright point and destructive interference appears as dark point.the pattern containing alternate bright and dark bands will be obtained. Q. Give the theory of interference bands and hence obtain an expression for the fringe width. OR Give the theory of interference bands and obtain an expression for the path difference. Hence obtain an expression for the band width. OR Give the theory of interference bands and prove that the distance between two consecutive bright bands is equal to the distance between two consecutive dark bands. Theory of Interference Bands :- ii) Fresnel s biprism experiment. Q. Why two coherent sources of light are necessary for obtaining steady interference pattern. OR Explain in brief the need of coherent sources to obtain the steady interference pattern. If two sources of light are not coherent then the phase difference between the two light waves at a point in a medium changes Continuously. Hence the intensity at a point of the medium does not remains constant. Therefore, a point of constructive interference appears 5 Consider two monochromatic and coherent source S 1 and S which are separated by the small distence d.the steady interference pattern is obtained on the screen placed at a distance D from these two sources (S 1 and S )

6 Let OQ is the perpendicular bisector of the segment S 1 S at the point Q. Consider a point P on the screen at a distance x from the point Q. Join S 1 P and S P. Also draw S 1 M and S N perpendicular to the screen. The path difference between the two light wave reaching the point P is (S P- S 1 P) From fig, In S 1 PM, (S 1 P) = S 1 M + PM (S 1 P) = D + (PQ QM) = D + x d In S PN (S P) = S N + PN = D + (PQ + QN) = D + x + d...(1)...() Subtract equation (1) from equation (), we get (S P) (S 1 P) = (D + x + d = xd + xd (S P) (S 1 P) = xd (D + x d Path difference = xd D For constuctive interference(for bright band) :- According to the conditions for bright point (Constructive interference, the point P is bright it the path difference is equal to nλ) Let x n be the distance of n th bright band from the central bright band. here Path difference = nλ and Path difference = x n d D x n d D x n = = nλ nλd d (3) Similiarly, for (n + 1) th bright band we get, x n+1 = (n+1)λd d (4) We know that, the band width/ fringe width is defined as the distance between two consecutive bright bands/ two consecutive dark bands. Here Band width is the distance between two conseutive x n and x n+1 Band width (X) = ( x n+1) x n = n+1 λd d nλd d = λd d (n + 1 n) (S P S 1 P) (S P + S 1 P) = xd (S P S 1 P) = xd S P + S 1 P S P S 1 P D S P S 1 P = xd D X = λd d (5) This is an expression for the distance between two consecutive bright bands. (band width) For destructive interference (For dark band) :- According to the conditions for dark point (destructive interference) the point P is dark if the path 6

7 difference is equal to (n-1) λ or it is an odd multiple of half of wavelength. Let x n be the distance of n th dark band from the central bright band. Path difference = (n-1) λ and here Path difference = x n d D Fig, (1) Experimental Arrangement. (n-1) λ x n = = xnd D n 1 λd d Simillarly for (n+1) th dark band x n+1 = We know that (n+1 1) λd d (6) (7) Biprism Experment is performed to determine the wavelength of monochromatic light used. In this experiment, Fresnet s biprism is used which has the refracting angle of 179 and other two angles are of 0.5 each. This biprism is used to produce two coherent sources, as shown in the fig () Band width is the distance between two consecutive bright bands/two consecutive dark band. Hence, Band width (X) = (x n+1) xn = (n+1 1) λd d - n 1 λd d = λd d (n+-1-n+1) Fig () Ray diagram = λd d X = λd d (8) From eqn (5) and (8), we can say that the distance between two consecutive bright bands is equal to the distance between two consecutive dark bands. Describe the fresnel s biprism experiment for obtaining the wavelength of monochromatic light. Experimental Arrangement :- Biprism experiment consist of an optical bench of two metre length on which four stands are mounted which can move horizontal along length of the bench The scale is attached on optical bench to measure the distance between the stands. A slit (S), biprism (B), biconvex lens (L) and Eyepeice (E) are mounted on these four stands as shown in the fig (1) Procedure :- 7

8 a)the slit (S) is illuminated by a monochromatic light whose wavelength is to be determined. The slit is made narrow. Adjust the biprism such that it s refracting edge is parallel to the slit so that S 1 and S, the two virtual images of the slit (S) are formed and they act as two coherent sources. Therefore, the interferece pattern containing alternate bright and dark band are observed through the eyepiece. Determination of wavelength of mono chromatic light :- (λ) The difference (x -, x 1 ), ( x 3 X ), (x 4 x 3 )...etc gives the band width. Now by taking the average of these band width, we get the band width X. 3) Measurement of d :- We can determine the distance between two sources by introducing a biconvex lens between the biprism and the eye piece.the distance between the slit and the eyepiece must be greater than four times the focal length of lens used. We know that, band width (X) = λd d λ = Xd D where, X = band width d = distance between the coherent sources S 1 and S, D = distance between the slits (two source) and the screen (eye piece) We can determine the wavelength of monochromatic light by measuring the values of X, d and D Fig 3 (a)magnified Images of the slit Now, move the convex lens towards the slit till we get the magnified images of the slits as shown in fig. 3 (a).measure the distance between them as d 1 1)Measurement of D :- Measure the distance D between the slit (S) and the eye piece (E), with the help of the scale attached on the optical bench. ) Measurement of X :- Adjust the vertical cross wire of the eye piece on the centre of any bright band and take first micrometre reading as x 1 Now, adjust the vertical cross wire to the next successive bright bands and take the corresponding reading, x, x 3, X4...etc. Fig 3 (b) Diminished Images of the slit Similarly, move the convex lens towards the eye piece till the diminished images of the two source are formed. Measure the distance between them as d as shown in fig. 3 (b). By using the principle conjugate focii we can determine equation for d and it is given by, d = d 1 d 8

9 Now by using this values of D, X and d, the equation (1) becomes, = xd D = x d1d D This as an expression for wavelength of monochoromatic light. Q. Derive on expression for measurement of distance between two virtual images of the slit in the biprism experiment. OR What is Interference of the light? With the neat ray diagram describe how the distance between two virtual sources in the biprism experiment is measured. Derive the necessary formula. *Interference of light :- The phenomenon of enhancement or cancellation of displacement produced due to the super position of wave is called interference of the light. Measurement of d :- We can measure the value of d by introducing the biconvex lens in between the brprism and eye piece. Fig 1 (b) Diminished Images of the slit Now, move the biconvex lens away from the slit till we get the diminished images of the slits as shown in fig 1 (b). The distance between these two images is measured as d. According to the principle of conjugate focii, size of image size of object distance of image = distance of object For magnified images as shown in fig. (a) d 1 d = v u 1 For diminished images as shown in fig (b) d 1 d = u v Multiplying equation (1) and eqn () we get, d 1 d d d = v u u v d 1 d d =1 d 1 d = d d = d 1 d Fig 1 (a)magnified Images of the slit Move the biconvex lens (L) towards the slit till we get the magnified images of the slits as shown in fig. 1 (a) The distance between these two images is measured as d 1 Q. What is diffraction? Give their types and explain it. Diffraction : When the light waves are passing close to the edges of an obstacle or narrow slit they suffers little 9

10 deviation from its straight line path some of the light enters into the region of geometrical shadow. This is due to the bending of light waves round the edges of an obstacle. Defination:- The phenomenon of bending of light waves round the edges of an obstacle or slit is called diffraction of light Types of diffraction : 1) Fresnel s diffraction : ) Fraunhoffer s diffraction. 1) Fresnel s diffraction :- The diffraction in which the source and the screen are at finite distance from slit is called fresnel s diffraction. Incident wavefront is not a plane wavefront. )Fraunhofer s Diffraction :- The diffraction in which the source and the screen are at infinite distance from the slit is called as fraunhofer s diffraction. Incident wavefront is a plane wavefront. In fraunnoffer s diffraction the biconvex lens is used and the screen is placed in the focal plane of the biconvex lens. Q. Explain the fraunhoffer s diffraction due to a single slit. Fraunhofer s diffraction due to single slit :- Consider a monochromatic parallel beam of light (plane wavefront) of wavelength λ is falling on a narrow slit of width a as shown in the above fig. A diffraction pattern is formed on the screen which is placed in the focal plane of the biconvex lens. let O is the centre of the slit and P 0 is central point on the screen. At this point P 0 the maximum intensity is observed. Therefore it is called central maxima or principal maximum. Let us now consider a point P on the screen which is obtained due to the diffraction of light the light waves changes their path and reaches the point P on the screen with some path difference. If the angle made by the diffracted ray with OP 0 is θ Draw the perpendicular AC then BC represents the path difference between the light waves which reaches the point P on screen. Now, In ABC, sin θ = BC AB sin θ = BC a BC = a sin θ If the path difference is λ then intensity at point p is minimum and it is called secondary minima (First secondary minima) i. e. a sin θ = λ.for first secondary minima. When path difference (PD) = a sin θ = λ, we get second secondary minima 10

11 When path differance PD = a sin θ = 3 λ, we get third secondary minima and we can say that their images are resolved. This ability of an optical instrument to form distinct images of two objects which are very close to each other is called resolving power of that instrument. When P.D. = a sin θ = n λ, we get n th secondary minima. If the P. D. = 3λ then the intensity at point p is maximum and it is called as secondary maxima (first secondary maxima) i.e. P.D. = a sin θ = 3λ When a sin θ = 5λ When a sin θ = 7λ When a sin θ = (n+1)λ i. e. for first sec. maxima. we get second secondary maxima. we get third secondary maxima we get n th secondary maxima. Fig 1 (b) Airy s Disc An objective (object lens) of a telescope or microscope acts as circular aperture and produces diffraction pattern for a point object, the diffraction pattern for a point object, the diffraction is a bright disc surrounded by alternate bright and dark rings of decreasing intensity as shown in the fig. (1) (b) such diffraction pattern is called Airy s disc. Fig. Intensity distribution due to to a single slit diffraction. Rayleigh s criterian for resolution of two objects:- When the two objects are very close to each other then they appear as one when these objects are seen through the telescope or micro scope then they appear as distinct Suppose O and O are the two point object (Sources) then each source will produce it s diffraction pattern with the centres P 1 and P as shown in the fig. According to Rayleigh, the images of two point objects which are very close to each other, are said to be resolved (Separated) if the central maxima of one falls on the first minima of the other. This is called Rayleigh s criterion for resolution. 11

12 Let α is the angular seperation between the two point objects O and O at the optical system and θ is the angular seperation between the central maxima and first minima in the diffraction pattern due to each source. Hence the two objects are just resolves. (b) Just resolved (α = θ) Case III :- Well Resolved (α > θ) Depending upon the value of α the three cases of resolution are observed as given below, Case I :- Just resolved (α < θ) (c) Well resolved (α > θ) The central maxima due to one source is beyond the first minima of the other. Resultant pattern clearly shows that the two principal maxima are well separated (resolved) as shown in the fig (3) (c), Hence the object are completely resolved. Central maxima due to one source is within the first minima due to the other source. The two central maxima show over lapping and the resultant diffraction pattern appears for a single source as shown in fig (3) (a) Hence, the objects are not resolved Case II :- Just resolved (α = θ) *Resolving Power of Microscope :- Write a short note on resolving power of microscope. OR Define the term resolving power of microscope. Give equation for it. Resolving Power :- The ability of on optical instrument to form the distinct images of the two objects which are very close to each other. Resolving Power of Microscope :- The central maxima due to one source coincides exactly with the first minima of the other source. The resultant deffraction pattern shows two peaks with dip between the two central maxima, as shown in the fig (3) (b) 1 Resolving power of microscope is defined as the reciprocal of the least separation between the two close objects so that they appear just separated when seen through the microscope.

13 = = RP of the microscope = 1 d 1 least seperation between two objects 1 limit of resolution limit of resolution = Conclusions :- λ μsinθ μ sinθ RP of microscope= λ i)resolving power of microscope increases with the increase in RI (μ) of the medium between the objective and object. ii) To increase the resolving power, the oil immersion microscope is used. iii) Resolving power of microscope increases with the use of light of lower (shorter wavelength). Suppose a point object O is illiminated by the light of wavelength λ and seen through the microscope. The rays of light are scattered from the object and they enter into the the objective of the microscope in a cone of semi-vertical angle θ. The least separation between the two objects is given by d = λ μ sin θ Where, μ is te refractive index of the medium between the objective of the microscope and the object. μ sinθ=numerical Aperture of the objective of the microscope The least separation (d) is also called as limit of resolution of microscope. limit of resolution = λ μsinθ Where μ is the refractive index of the medium between the objective of the microscope and the object. The least seperation (d) is also called as limit of resolution of microscope. 13 Define Resolving power of telescope Give equation for it. OR Write a short note on Resolving Power of telescope. Defination :- Resolving power of telescope is defined as the reciprocal of least angular seperation between two distant objects so that they appear just separated when seen through the telescope. Consider λ is the wavelength of the light emitted by the two objects O 1 and O (for exa. two stars) These objects are seen through the telescope having the objective of diameter D. A point source at infinite distance from the objective produces it s diffraction pattern at a focal plane of the objective. The least angular seperation (θ) between the two objects O 1 and O is suppose θ It is given by θ = 1.λ D Thus, RP of telescope is given by RP of telescope = 1 least angular seperation (θ)

14 = 1 θ Path difference = n λ RP of telescope = Conclusions :- D 1. λ 1)The resolving power of telescope increases with the diameter of the objective of a telescope. ) The RP of telescope increases by the light of smaller wavelength (Shorter wavelength) *Distinguish between Interrerance and Diffration. Interference 1)Interference is the result of interaction of light coming from two different wavefronts from two coherent sources. ) Interference fringes may or not be of same width. 3) All bright bands are of equal intensity. 4) The region of minimum intensities are perfectly dark. 5) In interference, the bands are large in number. Diffraction Diffraction is the result of interaction of light coming from different parts of same wavefront. Diffraction fingers are not of same width. The intensity of central maxima is highest and secondary maxima are of decreasing intensity. The region of minimum intensities are not perfectly dark. In diffraction, the band are few in number. ********************************************* Important Formulae:- Type I :- 1)For constructive interference (Bright band) :- Phase difference = O, π, 4π, 6π Phase difference = n π Path difference = O, λ, λ, 3λ.. = n λ =(even) λ Where, n = 0,1,, 3 )Destructive Interference (dark band) :- Phase difference = π, 3π, 5π Phase difference = (n - 1) π Path difference = λ, 3λ, 5λ Path difference = (n 1) λ = (odd) λ 3) When the two light waves of amplitudes a 1 and a differing in phase by φ interfere, the amplitude of the resultant light is given by R = a 1 + a + a 1 a cosφ And resultant intensity is given by I a 1 + a + a 1 a cosφ (a)the Amplitude or intensity of light will be maximum,when φ = 0 R max = (a 1 + a ) And I max (a 1 + a ) (b)the Amplitude or intensity of light will be minimum,when φ = 180 R min = (a 1 a ) And I min (a 1 a ) 14

15 (c)if W 1 and W be the widths of the two slits S 1 and S and I 1 and I be the intensities of the light due to the respective slits then W 1 = I 1 = a 1 W I a (d) Ratio of the maximum to the minimum intensities at point P is given by I max I min = a 1+ a a 1 a = I 1+ I I 1 I Tyep II :- 1)For Young s double slit experiment r Fresnel s biprism experiment :- Band width (or fringe width) X= λd d Where, λ = wavelength of monochromatic light. D = distance between slit and the screen (or eye piece) d = distance between two coherent source (two virtual images of slit) A)Distance of n th bright band from the central bright band is given by :- (X n ) B = nx = nλd d B)Distance of n th dark band from the central bright band is given by :- (X n ) D = n 1 (X n ) D = n 1 X λd d = (n 1)λD d )On same side of central bright band subtraction(-) 3)On opposite side of central bright band addition (+) 4)In the Young s double slit experiment, if transparent thin film or sheet of mica or glass of thickness t is introduced in the path of light from one of the two slits, the central bright band displaces(shifts) in the direction of that slit. If y is distance through which the central bright band moves, then Fringe shift (y) = D d μ 1 t Additional path difference = μ 1 t If the shift is equivalent to n fringes then y = nx n λd d = D d μ 1 t n = μ 1 t λ Thickness of slab ( t) = 1)For secondary maxima :- 15 nλ μ 1 3)Principle of conjugate foci:- Size of object Distance of Object = Size of image Distance of image d d 1 = x y 4)Distance between two coherent source (or two virtual image) d = d 1 d where, d 1 = distance between two magnified images of the slit. d = distance between two diminished images of slit. Type III Diffraction

16 a)path diff n = (n + 1) λ b) Distance from the central maxima = (n + 1) β β = λd d ) For Secondary minima :- a)path diff n = nλ b) Distance from the central maxima = nβ β = λd d 3)Width of secondary maxima/ minima is given by β = λd d 4)Width of central maxima :- Type IV:- β 0 = β = λd d 1)Resolving power of microscope :- If d least separation between the two objects (limit of resolution) d = λ μ sinθ = λ NA Where,(N.A.) = Numerical Aperture = μ sin θ Resolving power of microscope = 1 d R.P. of microscope = μ sin θ λ = NA λ Type V:- 1)Resolving power of telescope if (θ) is least angular seperation between two distance objects θ = 1. λ Resolving power of telescope = 1 θ = D D 1. λ Where D = diameter of the objective of telescope. TYPE - I 1. The optical path difference between two sets of similar waves from two coherent sources arriving at a point on a screen is wavelengths. Is the point bright or dark? If the path difference is 0.08 mm, calculate the wavelength of light. (Ans. Dark, 5145 A.U.). The path difference between two identical waves arriving at a point is 100.5λ. Is the point bright or dark? If the path difference is 44 micrometer, calculate the wavelength of light (Ans. Dark, 4378 A.U. ) 3. The distance of a point of on the screen from two waves starting form coherent sources differ by 93 wavelengths. Is the point bright or dark? if the path difference is mm, calculate the wavelength of light (Ans. Bright, 5000 A.U.) 4. A point in interference is situated at distance 90. and 90.17cm from two coherent sources. Find the nature of illumination of that point if wavelength of light is 5000A.U. (Ans : bright point) 5. The optical path difference at a point in interference is mm. A light of wavelength 5000A.U. is used. Find the nature of illumination of that point. Give the band number at that point. (Ans : Bright point, 15) Where λ = wavelength of illuminated light θ = semi vertical angle of cone of light rays incident on the objective of the microscope. μ = Refrative index of medium between the object and objective The path difference between the waves starting from two coherent sources in the same phase and arriving at a point is 96.5λ. If the path difference is 57.9 microns, calculate the wavelength of light used if the path waves were to start from the two coherent sources in opposite phase, will the point be

17 bright or dark? (Hint : When the waves are starting from the two source in opposite phase. the path difference should be regarded as 96.5λ + λ/ i.e. 97λ or 96λ. to decide whether the point is bright or dark.) (Ans A 0; bright) 7. Find the ratio of the intensities at the two points X and Y on a screen in Young s double slit experiment where waves from the two sources S 1 and S have path differences of zero and λ/4 respectively. (Ans. :1) 8. If the two slits in Young s double slit experiment have widths in the ratio 9:1, deduce the ratio of intensity at the maxima and minima in the interference pattern. (Ans. 4:1) 9. Two coherent sources of intensity ratio 100:1 interfere. Determine the ratio of intensity at the maxima and minima in the interference pattern. (Ans. 3:) 10. The ratio of intensity at the maxima and minima in the interference pattern is 5 :9. What is the ratio of the two slits in Young s interference experiment? (Ans. 16:1) 11. Two sources of intensities I and 4I are used in an interference experiment. Obtain the intensities at points where the waves from two sources superimpose with a phase difference of (i) 0 (ii) π/ (iii)π(ans. (i) 9I (ii)5i (iii) I) Type -II 1. In a biprism experiment, D = 1 meter, d = 0.1 cm and band width = cm. Calculate the wavelength of light. (Ans A.U.) 13. In a biprism experiment, the distance between the two virtual images of the slit is 7 mm and the distance between the slit and eyepiece is 1 meter. If the distance between two consecutive bright bands is 0.1 mm, calculate the wavelength of light used. (Ans A.U.) 14. In a biprism experiment, the distance between slit and eyepiece is 80 cm and the separation between two virtual images of the slit is 0.5 mm. If the slit is illuminated by light of wavelength 6000 A.U., find the distance between second bright band and central bright band. (Ans mm) 15. In the interference pattern obtained with a biprism, the distance between two consecutive bright bands is 0.04 cm. If the eyepiece is at a distance of 1 meter from the slit and the slit is illuminated by monochromatic light of wavelength m, what must be the distance between the two virtual images of the slit? (Ans. 1.8 mm) 16. While determining the distance between the two virtual sources in a biprism experiment two diminished images were obtained, 1.5 mm apart, in one position of the convex lens. In another position of the lens, two magnified images were obtained 6 mm apart. Calculate the distance between the two virtual sources. (Ans. 3mm) 17. In a biprism experiment, the eyepiece is kept at a distance of 1.5 m from the slit. The distance of nd dark band from central bright band was found to be 1. mm. The size of magnified and diminished images of the slit produced by the convex lens were.4 mm and 0.6mm respectively. Calculate the wavelength of light used. (Ans A 0 ) 18. The slit of a biprism is illuminated by monochromatic light of wavelength 5893 A.U. The distance between slit and biprism is 10 cm and that between eyepiece and biprism is 90 cm. If the virtual images of the slit are 5 mm apart, calculate the band width. (Ans mm) 19. In a biprism experiment, the eyepiece was placed at 110 cm from the slit. The distance between to consecutive dark bands was found to be 0.15 mm. A convex lens was then interposed between the biprism and the eyepiece. In one position of the lens, two diminished images seen through the eyepiece were found to be 4 mm apart. In another position of the lens, two magnified images of the slit were seen and the distance between them was 6 mm. Calculate the wavelength of light used. (Ans A.U.) 0. The slit of a biprism is illuminated by light of wavelength 6 x 10-7 m. If the distance between the two virtual sources is 0.48 mm, what would be the fringe width observed from and eyepiece placed at 75 cm from the slit? (Ans mm) 1. In a biprism experiment, the following observations were made; Distance between slit and eyepiece = 1m; Distance between the virtual images of the slit = 3.6 mm; Distance between third and ninth dark bands = mm; calculate the wavelength of light used. (Ans A.U.) 17

18 . In Young s experiment, two slits 4 mm apart, are illuminated by monochromatic light of wavelength 500 A.U. The screen is at 1. m from the slits. Calculate the distance of the central bright band from (i) sixth bright band and (ii) ninth dark band. (Ans. (i) mm, (ii) 1.36 mm) 3. In the biprism experiment, with the distance between the slit and the screen as 1 meter and separation between the two virtual images of the slit as 0.4 cm, an interference pattern is obtained with light of wavelength 5500 A.U. Find the distance between the third and eighth bright bands. (Ans mm) 4. In a biprism experiment, the slit is illuminated by red light of wavelength 6400 A.U. and the cross wire in the eyepiece is adjusted to be at the centre of the third bright band. When blue light is used instead of red light, it is found that the centre of the fourth bright band is at the cross. wire. Find the wavelength of blue light. (Ans A.U.) 5. In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of one meter form the slit. The distance of the 30th bright band from the slit. The distance of the 30th bright band from the centre is measured as 0.6cm. When a convex lens is introduced at a distance between the two virtual images observed is 1. cm. What is the wavelength of light. (Ans A) 6. In a biprism experiment, a convex lens is introduced at a distance of 40cm from the slit. The eyepiece is at a distance of 60cm from the lens. The distance between the two virtual images obtained is 0.35cm. If the wavelength of light used is 5460 A, calculate the band width. (Ans m) 7. In a biprism experiment, the fringes are observed in the focal plane of the eye-piece at a distance of 1m, from the slit. The distance of the tenth bright band form the central bright band from the central bright band is 0.cm. When a convex lens is interposed between the biprism and the eyepiece at a distance between the magnified images of the slit is found to be 0.90 cm. Find the wavelength of light used.(ans A) 8. In a biprism experiment, the distance between two virtual images of the slit is 1.5 mm and distance between slit and focal plane of eyepiece is 1 m. Find the distance between nd and 8th dark fringes on the same side,if the wavelength of light used is 5000 A. (Ans. mm) Calculate the distance between the first bright band and fourth bright band on the same side of the central bright band of the interference pattern produced by coherent sources, separated by 0.5 mm from each other, on the screen placed 1 meter away. (Given : λ = 5000 A.U. ) (Ans. 0.3 cm) 30. In a biprism experiment, the slit is illuminated by monochromatic light of wavelength 6400A.U. If the images of the slit are mm apart and the screen is 1 m away from the slit, calculate (a) fringe width and (b) distance of the fifth dark band from the central bright band (Ans. (a) 0.03 cm (b) cm) 31. In a biprism experiment the distance of the 0th bright band from the centre of the interference pattern is 8 mm Calculate the distance of the 30th bright band. ( Ans 1 mm) 3. In a biprism experiment, fringe width is 0.4 mm when the eyepiece is at a distance of 1 m from the slit. Find the change in the fringe width, if the eyepiece is moved through a distance of 5 cm towards the biprism without changing any other arrangement. (Ans. 0.1mm) 33. The distance between two virtual sources in a biprism experiment is 3. mm. With the eyepiece in a certain position, the fringe width is observed to be 0.16 mm. On moving the Eyepiece away from the slit through 0cm, the fringe width becomes 0.mm. Calculate the wavelength of light used and the initial distance between the slit and the eyepiece. (Ans A.U. 80cm) 34.The fringe separation in biprism experiment is 3.x10-4 m when red light of wavelength 6.4 x10-7 m is used. By how much will this change if blue of wavelength 4 x 10-7 m is used with the same setting? (Ans. 1. x 10-4 m) 35. In a biprism experiment, fringes were obtained with a monochromatic source of light. The eyepiece was kept at a distance of 1 meter from the slit and band width was measured. When another monochromatic source was used use d without disturbing the slit and the biprism, the same band width was obtained when the eyepiece was at 80 cm from the slit. Calculate the ratio of the wavelength of the two sources. (Ans. 4:5) 36. A biprism experiment is performed using yellow light of wavelength 5600 A. The yellow light was then replaced by

19 red light of wavelength 6400 A. Find the value of n for which (n+1) th yellow bright band coincides with the nth red bright band, with the same experimental setting.(ans n = 7) 37. The slit of biprism experiment was illuminated by blue light of wavelength 4800A.U. The interference pattern was observed. The blue light was then replaced by red light. It was observed that the 4th blue bright band and the 3rd red bright band coincide with each other, with the same experimental set up. Calculate the wavelength of red light. (Ans 6400A.U.) 38. In a biprism experiment, the distance between two coherent sources is 0.5 mm and that between slit and eyepiece is 1. m. The slit is illuminated by red light of wavelength 6550 A.U. I f the red light is replaced by green light of wavelength 540 A.U. it is found that n th red bright band coincides with (n + 1) green bright band. Calculate the distance of this band from the central bright band. (Ans. 6.88mm) 39. In a biprism experiment without disturbing other settings, only distance between the slit and micrometer eyepiece is increased by 50% What will be the change in the fringe width if initially it was 0.3mm? (Ans 0.15mm) 40. In young s experiment the fringes are observed by using two lights of wavelengths 6500 A.U. and 500 A.U. without changing experimental setup. Find the least distance from central bright band where the bright bands due to two wavelengths overlap. (D = 1.m and d =mm) (Ans. 1.56mm) 40. The path difference between two light waves at a point in interference is mm. A light of wavelength 6000 A.U. is used, Find the number of bright and dark bands between centre of interference and that point. (Ans. 3 dark and bright bands) 41. In biprism experiment the band width is found to be 0.4 mm when the eyepiece is at distance 1m from slit. Find the change in band width if eyepiece is moved by distance 5cm towards biprism without changing experimental arrangement. (Ans. 0.1 mm ) In biprism experiment a red light of wavelength 6750 A.U. was used to obtain an interference. It is replaced by violet light of wavelength 4050 A.U. without changing experimental setup. It is found that n th red bright band coincides with (n + ) th violet bright band. Find the number of that band. (Ans. 3) 43. In biprism experiment, wavelength of light used is 6000A 0 and n th bright band is obtained at a point P on the screen. Keeping the same setting the source is replaced by a source of green light of wavelength 5000A 0 coincides with the point P. Calculate the value of n. (Ans. n =5) 44. In a biprism experiment, the distance between nd and 10 th dark bands on the same side is 0.1cm. That between slit and biprism is 0 cm and that between biprism and eyepiece is 80cm. If slit images given by the lens in the two positions are 4.5mm and mm apart, find the wavelength of light used. (Ans A) 45. Calculate the distance of the 30 th bright band in biprism experiment, if the distance of 15 th bright band from the centre of the interference pattern is 6 mm. (Ans. 1 mm) 46. A certain fringe width is observed when green light of wavelength 5350 A.U. Where the distance between the slit and screen is 1.8m. What should be the distance between the slit and screen if the red light of wavelength 6400 A.U. is used to get the same fringe width without disturbing distance. between the coherent sources? (Ans m) 47. In a biprism experiment the distance between the slit and the eyepiece is 100cm. When a convex lens is introduced between the biprism and the dye piece, images of the slit obtained in the two positions of the lens are x 10-3 m and 4.5 x 10-3 m apart, The wavelength of light used is 6000A. Find the distance between the fourth bright band on one side and fourth dark band on the other fourth of central bright band. (Ans. 1.5mm) 49. In a biprism experiment, interference fringes are observed at a distance of 60cm from the slits illuminated by a monochromatic ray of light of wavelength 5460 A 0. The distance between the slits is 3mm. Find the change in fringe width if the distance between the slits is (1) increased () decreased by 1 mm. ((1).73 x 10 - mm; () 5.46 x 10 - mm)

20 50. Calculate the percentage change in fringe width if the distance between the slits in a biprism experiment is reduced by 10% and the distance between the slit and screen is increased by 15 %. (8%) 51. In the interference experiment with a biprism, the distance of the slits from the screen is increased by 10% and the separation of the slits is decreased by 0% Find the percentage change in the fringe width. ( Increase 37.5%) 5. In a Young s double slit experiment to obtain interference fringes the light used consists of two wavelength λ 1 and λ It is found that the fourth bright band of λ 1 superimposes on the fourth dark band of λ Find λ 1 /λ. (Ans 7:8) 53. In young s experiment, the slit is illuminated by a light of wavelength 5900 A. If the sources are 1 mm apart and the screen is at a distance of 1 m from the slit, calculate the fringe width. (Ans mm) 54 Calculate the distance between the first bright band and the fourth bright band on the same side of the central bright band of an interference pattern produced by the coherent sources separated by 0.5 mm from each other on the screen placed at 1m from them. (Ans. 3mm) 55. In a Young s double slit experiment to obtain interference fringes the light used consists of two wavelength one of which is 5000 A 0. Find the unknown wavelength if the 5th dark band of 5000 A 0 coincides with (a) the 5th bright band fo the unknown wavelength (b) the fourth bright band of the unknown wavelength other experimental arrangements remaining the same. (Ans 4500A 565A) 56. In a young s experiment the distance between the slits is 1.00 mm and the distance between the slit s and the screen is 1.00 m The slits are illuminated by blue light of wavelength 4800A and the distance of the fifth bright band from the centre is measured By how much will the distance of the fifth bright band from the centre change if the blue light is substituted by orange light of wavelength 6000A? (Ans. 0.6mm) 57. In biprism experiment, the distance between the slit and the eyepiece is 1m and the wavelength of light used is 5600 A. When a convex lens is interposed between of 30cm. form the slit, the distance between the two images obtained is 0.7 mm. Find (i) the band width (ii) the distance of the fourth dark band from the central bright band. (Ans. (i) mm (ii) mm) 58. The following observations were made in a biprism experiment. Distance between the virtual images of the slit =5 mm, distance between the slit and the eyepiece = 75cm, distance between the second dark band and the tenth dark band on the same side of the central bright band = mm. Calculate the wavelength of light used. (Ans A) 59. In a biprism experiment, the fringe width is 0.83 mm, when the eyepiece is at a distance of 1m from the slit. Find the change in the fringe width if the eyepiece is moved towards the slit by 5cm. (Ans. X 1 -X = mm) 60. In a biprism experiment, the slit is illuminated by light of wavelength 5890A. The distance between the slit and the eyepiece is 80cm. The two virtual images of the slit are formed 0.cm apart. Calculate the change in the band width if the eyepiece is displaced by 10cm away form the slit. (Ans mm) 61. In a biprism experiment, the distance between the slit and the focal plane of the eyepiece is 1m, and the wavelength of light used is 6000 a. When a convex lens is interposed between the biprism and the eyepiece, the distance between the images of the two virtual sources given by the lens in the difference positions are 3.6 mm and.5 mm, respectively. Calculate the fringe width. (Ans. 0.mm) 6. Calculate the distance between the second dark band and the fifth bright band on the same side of the central bright band of an interference pattern produced by coherent sources separated by 1. mm, from each other. The screen is placed at one meter form the coherent sources and the wavelength of light used is 6000 A. (Ans mm) 63. In young s experiment, the two slits are illuminated by a monochromatic light of wavelength 500 A. A screen is placed at 1m form the slits. If the two slits are 1.3 mm apart, find (i) The fringe width (ii) The distance between the seventh bright band on one side and the sixth dark band on the other side of teh central bright band. (Ans. (i) 0.4 mm (ii) 5 mm ) 64. Two straight and narrow parallel slits separated by 3 mm are illuminated by a monochromatic light of wavelength

21 A. Interference fringes are obtained at a distance of 60cm from the slits. Find the change in the fringe width if the distance between the slits is (i) increased by 1mm (ii) decreased by 1 mm. (Ans. (i) X 1 - X = mm ) (ii) X 3 - X 1 = mm) 65. In a biprism experiment, the eyepiece is at a distance of 1 m, from the slit. The distance between the two virtual sources is 3mm and the distance between the fifth dark bands on either side of the central bright band is 1.8 mm. Find the wavelength of the monochromatic light used. (Ans A) 66. In a biprism experiment, the distance between the two virtual images of the slit is 0.1 cm and the distance between the slit and teh eyepiece is 1m. If the wavelength of light used. (Ans A) 67. In a biprism experiment the to determine the wavelength of light the distance between the slit and the eye piece is 1m Wavelength of light used is 5600 A When a convex lens is kept at a distance of 30 cm from the slit the distance between two images of 0.7 mm Find (i) band width (ii) the distance of the 4th dark band from the central bright band (Ans 1.867mm,6.533mm) 68. In young s experiment, interference bands are produced on the screen placed at 1.5 m from the two slits separated by a distance of 0.15 mm and illuminated by a light of wavelength 4500 A. Find the change in the fringe width if the screen is brought towards the slit by 50cm. (X 1 - X = 15mm) 69. In young s experiment, the distance between two consecutive bright bands produced on a screen placed at 1.5m from two slits is 6.5mm. What would be the fringe width if the screen is brought towards the slit by 50cm for the same setting? Type -III ( Ans mm) 70. A light of wavelength 6000 A.U. is incident normally on a slit of width 3mm. Find the width of central maxima on screen placed at distance 3m from slit. (Ans. 1.mm) 71. Secondary second order maxima is at distance 1.4mm from centre of diffraction pattern on screen placed at distance 0.8m from slit of width 0.8 mm. Find the wavelength of light if is incident normally on slit. (Ans.5600 A.U.) 7. A screen is placed at distance of 0.5 m from slit and a light of wavelength 6000 A.U. from distant source is incident on slit. The distance between 1 st and 3 rd minima on same side in diffraction pattern is found to be 3mm. Find the width of slit. (Ans. 0.mm) 73. In single slit diffraction experiment, the slit is illuminated by red light of wavelength 6600 A.U. It is replaced by violet light without disturbing experimental set up. It is found that the first minima of red coincide with first maxima of violet. Find the wavelength of violet light. (Ans A.U.) 74. In Fraunhofer diffraction due to narrow slit of width 0. mm is seen on a screen placed at distance m. The distance of first minima on either side of central maxima is 10mm. Find the wavelength of light used. (Ans A.U.) 75. A single slit of width 0.14 mm is illuminated normally by monochromatic light and diffraction bands are observed on screen m away from slit. the centre of second dark band is 1.6cm from middle of central bright band. Calculate the wavelength of light used. (Ans A.U.) 76. Find the angle of diffraction of first bright fringe in Fraunhofer diffraction of slit width 1 x 10-5 cm when the slit is illuminated by monochromatic light of wavelength 6000 A.U. (Ans ) 77. Fraunhofer diffraction pattern of a single slit of width 0.5 cm is formed by a lens of focal length 40cm. Calculate the distance between first dark and next bright fringe from central bright fringe. The wavelength of light used is 4890 A.U. (Ans mm) 78. In a Fraunhoffer diffraction pattern due to a single slit of width 0.04 mm the 3rd dark band occurs in a direction making an angle. of α with the normal to the plane of he slit. If the wavelength of light used is 6000 A 0 find the value of α. Type -IV (Ans. 0 5 ) 79. The semi-vertical angel subtended by two points at the objective of a microscope is 0 0. If the wavelength of 1