NAWAB SHAH ALAM KHAN COLLEGE OF ENGINEERING & TECHNOLOGY 1

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1 1 I-I, PHYSICS QUESTION BANK WITH ANSWERS UNIT I INTERFERENCE IN THIN FILMS 1. Define 1) physical thickness ) apparent thickness 3) optical thickness of a transparent medium. 1) physical thickness: it is actual thickness of a transparent medium. It is also called geometric length ) Apparent thickness: a denser medium appears to be less thick than its actual thickness where as a rarer medium appears to be thicker than its actual thickness. If t is the geometric or actual thickness and μ is the refractive index then its apparent thickness is given by t/μ. 3) Optical thickness: it is defined as the thickness of vacuum which contains the same number of waves as the medium under consideration. If t is the geometric or actual thickness and μ is the refractive index then its optical thickness is given by μt. The concept of optical thickness is used to find the path difference between two waves traveling in different media. Problem 1: A glass slab has thickness 6 cm. Find its apparent & optical thickness (μ = 3/). What is phase difference and path difference? Phase difference: In a wave the phase changes gradually. If two points are separated by a distance λ the phase difference between them is π. If two points are separated by a distance x then the phase difference between them δ is given as Problem : A wave has wave length 6 m. Find the phase difference between two particles separated by 1) m )0 cm. a) Path difference: suppose two rays reach a point taking two different paths. The difference between optical paths of the two rays travelling in different directions is known as the optical path difference. If two rays of light travel distances t1 and t in air or vacuum their path difference = t t1 If a ray travels a distance t1 in air and t in a medium of refractive index μ then the path difference = μt t1 If a ray travels a distance t1 in a medium of refractive index μ1 and t in a medium of refractive index μ then the path difference = μt μ1t1 Problem 3: Three rays R1,R and R3 travel distances 6 cm each, but R1 in air,r in water and R3 in glass. Find the path difference in R1&R,R1&R3 and R&R3 respectively. b) Phase change due to reflection at boundaries of optical interfaces: Light waves may undergo phase change due to reflection at some point in their path. If the waves are reflected at a rarer-to-denser medium boundary, the reflected waves suffer a phase change of π rad or 180 or a path difference of λ/. Therefore, we must add or subtract λ/ in the calculation of true optical path difference whenever a reflection occurs at a denser medium. 3. What are coherent waves? Two light waves having same frequency, same wave length and same phase or constant phase difference are called coherent waves. They may or may not have same amplitude. 4. What are coherent sources? OR Explain the term coherent sources. Describe the general methods to obtain coherent sources. Coherent sources: If two sources produce light of same frequency, same wave length and same phase or constant phase difference they are called coherent sources. The respective waves are called coherent waves. Two independent monochromatic sources cannot be x

2 coherent. Hence coherent sources can be obtained by either 1) division of wave front or ) division of amplitude 5. What is division of wave front (Dec 016) 1 Division of wave front: in this method a narrow slit is used as source. The wave front is spherical. This wave front is subsequently divided into two parts by two narrow slits S1 and S. This method is used in Young double slit experiment. Fresnel s bi prism, Lloyd s mirror, etc are the other examples where the division of wave front is used. 6. What is division of amplitude Division of amplitude: In this method a broad source is used. The wave front is plane. The amplitude of the light beam is divided by partial reflection into two or more beams. Thin films (parallel sided, wedge shaped, Newton s ring, etc), interferometers such as Michelson s interferometer etc utilize this method in producing interference. 7. Discuss why two independent sources of light of the same wavelength cannot produce interference fringes. Two independent sources of light cannot maintain phase difference constant. Hence they cannot produce sustained interference fringes. 8. What is Stokes s law of change of phase on reflection According to Stokes, when a light wave is reflected at the surface of an optically denser medium, it suffers a phase change of π i.e., a path difference of λ/. No such phase change is introduced if the reflection takes place from the surface of rarer medium 9. What is superposition principle? (June 005) When two waves travel simultaneously in a medium the resultant displacement at any point is the algebraic sum of the displacements due to the individual waves. Suppose y1 and y are the displacements of two waves at a point. The resultant displacement due to the principle of superposition is Y = y1 ± y. If y1 = a sin(ωt) and y = a sin(ωt + δ) then Y = A sin(ωt + θ). The amplitude A = a cos (δ/) and the resultant intensity I = 4a cos (δ/) = 4Io cos (δ/) where Io is the intensity of each source. 10. What is meant by interference of light? State the fundamental conditions for the production of interference fringes. (June 017) The phenomenon of redistribution of light energy due to superposition of two coherent waves is called interference. Suppose y1 and y are the displacements of two waves at a point. The resultant displacement due to the principle of superposition is Y = y1 ± y. If y1 = a sin(ωt) and y = a sin(ωt + δ) then Y = A sin(ωt + θ). The amplitude A = a cos (δ/) and the resultant intensity I = 4a cos (δ/) = 4Io cos (δ/) where Io is the intensity due to each source. For interference maximum cos (δ/) = ±1. i.e., the phase difference δ = 0,π, 4π, nπ Or Path difference = nλ For interference minimum cos (δ/) = 0. i.e., the phase difference δ = π, 3π, 5π, (n±1)π Or Path difference = (n±1)λ/ Conditions for observing sustained interference. 1) The waves must be coherent. ) Their amplitudes should be equal 3) If the waves are polarized their planes of polarization should be parallel 4) The sources must be close.

3 3 11. Explain the interference of light due to thin films. (June 006, May 007) With ray diagram discuss the theory of thin films and the condition for constructive and destructive interference in the case of reflected system. (008) Interference in reflected light. Let us consider a thin transparent film of uniform thickness t and refractive index μ. Let GH and G1H1 be the two surfaces of the film as shown in Fig. Suppose a ray AB of monochromatic light be incident on its upper surface GH. This ray is partly reflected along BR and refracted along BC. After one internal reflection at C, it travels along CD and finally emerges out along DR1 in air. DR1 is parallel to BR. To find out the effective path difference between the rays BR and DR1we draw a normal DE on BR and normal BF on CD. Extend DC and BQ to meet at P. In the figure ABN = i, (the angle of incidence) and QBC = r, (the angle of refraction). From the geometry of the figure BDE = i and QPC = r. The optical path difference between the two reflected light rays (BR and DR1) is given by = Path (BC + CD) in film - Path BE in air = (BC + CD) - BE (1) We know that = sini BE/BD BE μ sinr FD/BD FD BE = (FD).() From equations (1) and () = (BC+CD)-(FD) = (BC+CF+FD) - (FD) = (BC + CF) = (PF) (As BC = PC)..(3) From triangle BPF, cos r = PF/BP or PF = BP cos r = t cos r...(4) Substituting the value of PF from equation (4) in equation (3), we have = xt cos r = t cosr...(5) It should be remembered that a ray reflected at a surface backed by a denser medium suffers an abrupt phase change of which is equivalent to a path difference /. Thus the effective path difference between the two reflected rays is ( t cos r / ). We know that maxima occurs when effective Path difference = n. i.e., t cos r / = n or t cos r = (n 1) /...(6) t cos r = (n + 1) / where n = 0,1,,3,4.. t cos r = (n - 1) / where n = 1,,3,4.. If this condition is fulfilled, the film will appear bright in the reflected light. The minima occurs when the effective path difference t cos r / = (n 1) / Or t cos r = n Here n = 0, 1,, 3 etc. When this condition is fulfilled the film will appear dark in the reflected light.

4 4 1. With ray diagram discuss the theory of thin films and the condition for constructive and destructive interference in the case of transmitted system.(may 007, Dec 016) Interference in transmitted light. Let us consider a thin transparent film of uniform thickness t and refractive index μ. Let GH and G1H1 be the two surfaces of the film as shown in Fig. Suppose a ray AB of monochromatic light be incident on its upper surface GH. Due to multiple reflection and refraction at the two surfaces we obtain two transmitted rays CT and ET1. Since these rays have originated from the same point source they are coherent. Hence they have a constant phase difference and are in a position to produce sustained interference when combined. In order to calculate the path difference between the two transmitted rays, we draw normal CQ and EP on DE and CT respectively. We produce ED and CF to meet at I The effective path difference is given by = (CD + DE) - CP....(1) sini CP/CE CP We know that μ sinr QE/CE QE Therefore CP = (QE) () From equations (1) and () = (CD + DQ + QE) (QE) = (CD+DQ) = (QI) = t cos r Here it should be remembered that inside the film, reflection at different points takes place at the surface backed by rarer medium (air) thus no abrupt change of takes place in this case. The interference maxima occur when effective path difference = n i.e. t cos r = n If this condition is fulfilled, the film will appear bright in transmitted light. The minima occur when the effective path difference t cos r = (n 1) / where n = 1,, 3, etc. When this condition is fulfilled, the film will appear dark. Thus the conditions of maxima and minima in transmitted light are just reverse of the conditions for reflected light. 13. Show that with monochromatic light, the interference patterns of reflected and transmitted light are complementary. Thus the effective path difference between the two reflected rays from a thin film = t cos r /. We know that maxima occur when effective Path difference = n. i.e., t cos r / = n or t cos r = (n 1) /...(1) t cos r = (n + 1) / where n = 0,1,,3,4.. t cos r = (n - 1) / where n = 1,,3,4.. If this condition is fulfilled, the film will appear bright in the reflected light. The minima occurs when the effective path difference t cos r / = (n 1) / Or t cos r = n () where n = 0, 1,, 3 etc. When this condition is fulfilled the film will appear dark in the reflected light. The effective path difference between the two transmitted rays from a thin film = t cos r. The maxima occur when effective path difference = n

5 5 i.e. t cos r = n..(3) If this condition is fulfilled, the film will appear bright in transmitted light. The minima occur when the effective path difference t cos r = (n 1) / (4) where n = 1,, 3, etc. When this condition is fulfilled, the film will appear dark. From equations (1) and (4) it is clear that the condition for maximum in reflected light is same that for minimum in transmitted light. Similarly from equations () and (3) it is clear that the condition for minimum in reflected light is same as that for maximum for transmitted light. Thus the conditions of maxima and minima in transmitted light are just reverse of the conditions for reflected light. 14. Explain the colors in a thin film when exposed to sun light. (June 005) Colours in thin films: when a thin film (of soap solution or oil on the surface of water) is exposed to a source of white light beautiful colours are observed. This is due to the phenomenon of interference in thin films. Explanation: In thin films the path difference between the light reflected from the top and the bottom surface is μt cosr λ/. At a particular point of the film and for a particular position of the eye, the interfering rays of only certain wavelengths satisfy the conditions of maxima. As we change the position of the maximum condition is satisfied by different colours. Hence we see multicolours. 15. Why the colours seen in reflected light are not observed in transmitted light. for thin films the path difference for reflected light is t cos r / whereas for transmitted light the path difference is t cos r. The conditions for maxima and minima in transmitted light are inverse that of reflected light. Hence the colours suppressed or absent in reflected light appear as intense colours in transmitted light or colours of reflected and transmitted light are complementary. 16. Why do soap bubbles appear multicoloured when viewed under sun light. In thin films the condition for brightness is given by μt cos r = (n + 1)λ/. Since the bubble is spherical in shape, the light incident will have different values of r at different points on the soap bubble. Hence we get multicolours. 17. Why does a film appear black in reflected light when it is excessively thin? The path difference in thin films is given by μt cos r λ/ If the thickness of the film is extremely small when compared to λ, then μt cos r term can be neglected and the path difference is λ/. Hence destructive interference occurs. So the film appears dark. 18. Why a thick film seen by reflected light shows no colours but appears white? Why a thick film does not show colors when seen in white light. Light waves can maintain phase up to a certain length only. This is called coherence length. For visible light it is of the order of fraction of a millimeter. In thick films or glass plates of thickness millimeter or more interference is not observed as the light waves are no longer coherent. Problem 4: A parallel beam of light (λ = 5890x10-8 cm) is incident on a thin glass plate (μ = 1.5) such that the angle of refraction onto the plate is 60 o. Calculate the smallest thickness of the glass plate which will appear dark by reflection [3.6 x 10-5 cm] Hint: λ t μ cosr

6 6 Problem 5: White light falls normally on a film of soapy water whose thickness is 5 x 10-5 cm and μ = Which wavelength in the visible region will be reflected most strongly? [ 530 x 10-8 cm, for n = ] Hint: The condition of maxima for reflected light is t cos r = (n 1) / Find λ for different values of n Problem 6: White light is incident on a soap film at an angle sin -1 (4/5) and the reflected light on examination by a spectroscope shows dark bands. Two consecutive dark bands correspond to wavelengths 6.1 x10-5 and 6.0 x 10-5 cm. If the refractive index of the film 4/3, calculate the thickness.[ cm] Hint: or μ t cosr nλ (n1)λ λ1 λ μ tcosr λ λ 1 1 λ or n λ λ 1 INTERFERENCE IN WEDGE SHAPED FILM 19. Discuss the formation of interference fringes in a thin wedge shaped film.(may 017) problem: Find the thickness of wedge-shaped air film at a point where fourth bright fringe is situated. Wave length of light is nm A thin wedge of air film can be obtained by two glass slides resting on each other and separated by a thin spacer at one end. The experimental arrangement is shown. When light is incident from above, it gets partially reflected from the top and bottom surfaces of the air film. The two rays BC and FE are coherent as they are derived from the same ray AB through the division of amplitude. For small film thickness the rays interfere producing dark and bright fringes depending on the phase difference. Since the glass plates are thick the interference is mainly due to the air film. At a point of air film of thickness t the optical path difference between the two rays BC and FE is given by = μt cos r ± λ/.(1) the term λ/ is due to reflection at the air glass boundary at the bottom. For interference maximum the path difference = mλ..() or μt cos r = (m ± 1)λ/.(3) For interference minimum the path difference = (m ± 1)λ/..(3) or μt cos r = m λ.(4) for normal incidence cos r = 1. For m th dark fringe suppose the thickness of the air film is t1. Then μt1 = m λ (4) Suppose the next dark fringe occurs at a thickness t. Then μt = (m + 1) λ..(5) From the above equations we get μ(t t1) = λ..(6) But (t t1) = BC μ(bc) = λ BC = λ/μ From the triangle ABC, BC = AB tanθ hence AB tanθ = λ/μ the fringe width AB, β = λ μ tanθ λ μ θ

7 7 Salient features of interference pattern 1. Fringe at the apex is dark.. Fringes are straight and parallel. 3. Fringes are of equal thickness. 4. Fringes are localized. Problem Given that wave length λ = nm number of bright fringe m = 4 thickness of the film t =? Applying μt cos r = (m - 1 )λ/ we get t = 3.5x589.3 nm or thickness t = 1.75x589.3 nm = nm problem 7: Fringes of equal thickness are observed in a thin glass wedge of refractive index 1.5. The fringe spacing is 1 mm and wavelength of light is 5893Å. Calculate the angle of the wedge.[39.96 seconds of an arc] Hint: β λ μ θ Problem 8:If the angle of wedge is 0.5 o and the wavelengths of sodium lines are 5890Å and 5896Å, find the distance from the apex at which the maxima due to two wavelengths first coincide when observed in reflected light [n = 981] Hint: t = (n + 1)λ1/ = (n + 3)λ/ NEWTON S RINGS 0. Discuss the theory of Newton s rings with relevant diagram (b) Explain how Newton s rings are formed in the reflected light. 1. Derive the expressions for the diameters of dark and bright rings.(june 005 ). Prove that in reflected light, (i) Diameters of the dark rings are proportional to the square roots of natural numbers and (ii) diameters of bright rings are proportional to the square roots of odd numbers. (iii) the difference in the squares of the diameters is constant When a plano-convex lens with its convex surface is placed on a plane glass plate, an air film of gradually increasing thickness is formed between the two. The thickness of the film at the point of contact is zero. If monochromatic light is allowed to fall normally, and the film is viewed in reflected light, alternate dark and bright rings concentric around the point of contact between the lens and glass plate are seen. These are called Newton s rings Experimental arrangement. The experimental arrangement of obtaining Newton s rings is shown in figure1. L is a plano-convex lens of large radius of curvature R. This lens with its convex surface is placed on a plane glass plate P. The lens makes contact with the plate at C. Light from an extended monochromatic source such as sodium lamp falls on a glass plate G held at an angle 45 with the vertical. The glass plate G reflects normally a part of the incident light towards the air film enclosed by the lens L and the glass plate P. A part of the incident light is reflected by the curved surface of the lens L and a part is transmitted which is reflected back from the plane surface of the plate. These two reflected rays interfere

8 8 and give rise to an interference pattern in the form of circular rings. These rings are localized in the air film, and can be seen with a microscope focussed on the film. Explanation of the formation of Newton s rings. Newton s rings are formed due to interference between the waves reflected from the top and bottom surfaces of the air film formed between the plate and the lens. The formation of Newton s rings can be explained with the help of Fig.. AB is a monochromatic ray of light which falls on the system. A part is reflected at C (glass-air boundary) which goes out in the form of ray 1 without any phase reversal. The other part is refracted along C D. At point D it is again reflected and goes out in the form of ray with a phase reversal of. The reflected rays 1 and can produce interference fringes as they have been derived from the same ray AB. As the rings are observed in the reflected light, the path difference between them is ( t cos r + /); For air film = 1 and for normal incidence r = 0. Hence in this case, the path difference is ( t + / ). At the point of contact, t = 0, and the path difference is /, which is the condition of minimum intensity. Thus the central spot is dark. For nth maximum, we have t + / = n This expression shows that a maximum of a particular order n will occur for a constant value of t. Since t remains constant along a circle the maximum is in the form of a circle. For different values of t, different maxima will occur. In a similar way, this can be shown that minima are also in the circular form. Theory : Newton s rings by reflected light Now we shall calculate the diameters of dark and bright rings. Let LOL be the lens placed on a glass plate AB. The curved surface LOL is a part of spherical surface (shown dotted in figure) with centre at C. Let R be the radius of curvature and r be the radius of Newton s ring-corresponding to the constant film thickness t. As discussed above t + / = n, or t = (n -1) / for the bright ring where n = 1,, 3, etc. And t = n for dark ring where n = 1,, 3, etc. From the property of the circle NP x NQ = NO x ND Substituting the values r x r = t x (R - t) = Rt t Rt (approximately) r = Rt or t = r /R Thus for bright ring r λ n 1 R n 1λ R r Replacing r by D/ we get the diameter of the n th bright ring as or D = λ R n 1 Or D n 1 Thus the diameters of the bright rings are proportional to the square roots of odd natural numbers as (n - 1) is an odd number. Similarly for a dark ring D n Thus diameters of dark rings are proportional to the square roots of natural numbers. D 4 n 1 λ R

9 9 With the increase in n the fringe width decreases but the area of the ring remains same. (iii) the diameter of the n th bright ring Dn = λ R n 1 Square of the diameter of the n th bright ring Dn = λ Rn 1 Square of the diameter of the (n+1) th bright ring Dn+1 = λ Rn 1 The difference in the diameters of the successive rings Dn+1 - Dn = 4λR since the above is independent of the number n, it is constant. Newton s rings by transmitted light In case of transmitted light, the rings are just opposite to the rings in reflected light. The central ring is bright. The diameter of the n th bright ring D = n λ R n The diameter of the n th dark ring D = λ R n 1 n 1 3. Describe Newton s rings experiment to determine the radius of curvature of a plano - convex lens (June 014, Dec 016) or determination of wavelength of light. The experimental arrangement is shown in figure. The microscope is adjusted such that Newton s rings are clearly seen. The centre of the cross-wires is adjusted at the centre of the fringe pattern. The microscope is moved to the extreme left of the pattern and the cross wire is adjusted tangentially in the middle of a clearly n th bright or dark fringe. The reading of micrometer is noted. The microscope is now moved to the right and the readings of micrometer screw are noted successively at (n -) th, (n - 4) th rings etc. till we are very near to the central dark spot. Again crossing the Central dark spot in the same direction the readings corresponding to... (n -4) th, (n-) th,nth rings are noted on other side. The diameters of different rings are found out. Now a graph is plotted between number of rings n and the square of the corresponding diameter. The graph is m n a straight line. From the graph we find the slope or m n If R is the radius of curvature of the plano convex lens then the wavelength of the D D 4 m n monochromatic light can be found using m nr If we know the wave length of the light we can find the radius of curvature of the lens using Dm Dn R 4 m n 4. Explain what happens when the air in the inter space is replaced by a transparent liquid in Newton s rings method (June 017) 5. Determine the refractive index of transparent liquid by using Newton s ring method. (June 005) When air in the inter-space is replaced by a liquid the diameters of the Newton s rings decrease. When air is present between the lens and a glass plate the diameter of the n th bright ring D n 1λ R when a liquid of refractive index μ is introduced in the space the diameter of the nth bright ring n 1λ R D` D D

10 10 Experiment to determine the refractive index of the given liquid. The experimental arrangement is as shown. The diameters of the m th and n th dark rings are measured as Dm and Dn. Now the given liquid is introduced between the lens and the glass plate. The rings shrink. The new diameters D1m and D1n are measured. The refractive index of the liquid can be calculated using the formula μ D D NEWTON S RINGS WITH WHITE LIGHT 6. What will happen to Newton s rings if (a) white light is used (b) the lens is lifted slowly from the flat plate. When monochromatic light is used, Newton s rings are alternately dark and bright. In case of white light, first few rings will be colored. Other rings cannot be viewed due overlapping of rings of different colors. b) As the lens is lifted slowly the central dark spot becomes bright and the diameters of the rings increase. Finally the rings disappear. 7.Why the centre of Newton s ring is black? How can it be made bright. at the center the lens surface is in contact with the glass plate. Hence the effective path difference for reflected rays is λ/. Therefore the centre of Newton s ring is black. It can be made bright by providing a space of λ/ with the help of dust particles. Problem 9: In a Newton s ring experiment, the diameter of the 5 th ring was 0.3 cm and the diameter of 5 th ring was 0.8 cm. if the radius of curvature of the plano convex lens is 100 cm, find the wavelength of light used. [687x10-5 cm] D D m n Hint: λ 4m nr Problem 10: A Newton s rings arrangement is used with a source emitting two wave lengths λ1 = 6x10-5 cm and 4.5 x 10-5 cm and it is found that nth dark ring due to λ1 coincides with (n + 1) th dark ring for λ. If the radius of curvature radius of curvature of the curved surface is 90 cm, find the diameter of n th dark ring for λ1. [0.54 cm] Hint: nλ1= (n + 1)λ Dn = 4nRλ1 Problem 11:In Newton's ring experiment the diameters of the 4 th and 1 th dark rings are cm and cm respectively. Find the diameter of the 0 th dark ring. [0.906 cm] D0 D1 D D Problem 1:Newton s rings formed by sodium light between a flat glass plate and a convex lens are viewed normally. What will be the order of the dark ring which will have double the diameter of that of 40 th dark ring. [160] Hint: D D m n m n Problem 13: In a Newton s rings experiment the diameter of the 10 th ring changes from 1.40 to 1.7 cm when a liquid is introduced between the lens and the plate. Calculate the refractive index of the liquid. [1.15] Hint: Dn = 4n λ R μ Problem 14: Light containing two wavelengths λ1 and λfalls normally on a plano convex lens of radius of curvature R resting on a glass plate. If the nth dark ring due to λ1 coincides with D D m 1m n 1n

11 11 (n + 1) th dark ring for λ. If the radius of curvature radius of curvature of the curved surface is R, find the radius of n th dark ring for λ1. HINT: the radius of the nth dark ring r = but nλ1 = (n + 1)λ Or n = 1 Hence r = 1 R 1 n 1 R Problem 15: If the diameter of two consecutive Newton s rings in reflected light of wavelength 5890 Ǻ are.0 and.0 cm respectively, what is the radius of curvature of the lens surface in contact with plane glass surface? [Ans cm] Problem 16: In a Newton s rings experiment, the diameter of the 5 th ring was cm and the diameter of the 15th ring was cm. Find the radius of curvature of the plano-convex lens if the wavelength of light used is 5890 x 10-8 cm. [Ans cm] Problem 17: In a Newton s rings experiment the diameter of the 15 th ring was found to be 0.59 cm and that of 5 th ring is cm. if the radius of curvature of the lens is 100 cm find the wavelength of the light (June 005)[Ans. 588 nm] Problem 18: calculate the thickness of air film at 10 th dark ring when viewed in reflected light of wavelength 500 nm. The diameter of the 10 th ring is mm. (004)[.5 µm] Problem 19: In Newton s rings experiment the diameter of the 10 th ring is 0.5 cm. If λ= 5900 Ǻ find the radius of curvature of the lens.[1.059 m](005,011) Problem 0: Newton s rings are formed by light reflected normally from a plano-convex lens and a plane glass plate with a liquid between them. The diameter of the n th ring is.18 mm and that of (n +10) th ring is 4.51 mm. Calculate the refractive index of the liquid, given that the radius of curvature of the lens is 90 cm and the wavelength of light used is 5893 A. [Ans. 1.7] Problem 1: In a Newton s ring experiment the diameter of the 10 th ring changes from 1.40 cm to 1.7 cm when a liquid is introduced between the lens and the plate. Calculate the refractive index of the liquid. [Ans. 1.15] MULTIPLE CHOICE QUESTIONS Wave nature of light is evidenced a) photoelectric effect b) interference c) black body radiation d) nuclear emission 1. Two sources are said to be coherent if their emitted waves have a) same wavelength J b) same amplitude c) constant phase difference d) all the three. When the light wave is reflected at the glass-air interface, the change of phase of the reflected wave is equal to a) 0 b)π/4 c) π/ d)π 3. When the light wave is reflected at the air-glass interface, the change of phase of the reflected wave is equal to. a) 0 b)π/4 c) π/ d)π 4. Two waves having their intensities in the ratio 9:1 produce interference. In the interference pattern the ratio of maximum to minimum intensity is equal to a) :1 b) 9:1 c) 3:1 d) 4:1

12 1 5. Two beams interfere have their amplitudes ratio : 1. Then the intensity ratio of bright and dark fringes is a) :1 b) 1: c) 9:1 d) 4:1 6. When a thin film of oil or soap bubble is illuminated with white light, multiple colours appear. This is due to a) diffraction. b) polarization c) total internal reflection d) interference 7. In Newton s rings interference is due to light rays reflected from a) lower surface of lens and upper surface of glass plate b) lower surface of glass plate and upper surface of lens c) lower surface of lens and lower surface of glass plate d) upper surface of lens and upper surface of glass plate 8. Choose the correct statement a) In Newton s rings experiment, when viewed through reflected light the centre of the rings pattern is bright b) In Newton s rings experiment, when viewed through transmitted light the centre of the rings pattern appears dark c) In Newton s rings experiment, when viewed through reflected light the centre of the rings pattern is dark, d) In Newton s rings experiment, when liquid of higher refractive index is introduced in between the lens and the glass plate, the radius of the rings increases. 1. The resultant displacement due to the presence of both the waves of displacements yl and y is given by y = y1 + y. This is known as principle of.(principle of superposition). Two waves are said to be.. if their waves have same wavelength, same amplitude and constant phase difference. (coherent) 3. If : 1 is the amplitude ratio of the sources, the intensity ratio of bright fringe to dark fringe is (9:1) 4. If the film thickness is extremely small when compared to λ, the film will appear (dark) 5. Newton s ring experiment, the radius of nth dark ring is given by.. where λ is the wavelength of light and R is the radius of curvature of the lens. n R 6. On introducing a liquid between the lens and the glass plate in Newton s rings experiment, the diameter of the rings observed (decreases)

13 DIFFRACTION 1. What is meant by diffraction of light? The phenomenon of bending of light into the geometrical shadow region is called diffraction. This is shown by all kinds of waves, irrespective of their nature.. What are the types of diffraction and give the differences between them? (June 005, June 011) Explain Fresnel and fraunhofer diffraction (May 017) Diffraction phenomenon can be divided into following two general classes: (1) Fresnel s diffraction: In this class of diffraction, source and screen are placed at finite distances from the aperture of obstacle having sharp edges. In this case no lenses are used for making the rays parallel or convergent. The incident wave front are either spherical or cylindrical. () Fraunhofer s diffraction: In this class of diffraction source and the screen or telescope (through which the image is viewed) are placed at infinity or effectively at infinity. In this case the wave front which is incident on the aperture or obstacle is plane. OR Fresnel diffraction 1 In this class of diffraction, source and screen are placed at finite distances from the aperture of obstacle having sharp edges. The incident wavefront are either spherical or cylindrical. Fraunhofer diffraction In this class of diffraction source and the screen or telescope (through which the image is viewed) are placed at infinity or effectively at infinity. In this case the wave front which is incident on the aperture or obstacle is plane. 3 Source used is small in size Big or extended source is used 4 Lenses are not used Lenses are used 3. What is the difference between interference and diffraction? (June 005) 4. Differentiate between interference and diffraction. (May 007, June 011) 5. Explain what is meant by diffraction of light. How diffraction is different from interference? (June 011) DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION Following are the differences between interference and diffraction phenomena: interference diffraction Interference is due to the interaction Diffraction is due to the interaction between two separate wave fronts between the secondary wavelets originating from two coherent sources originating from different points of same In an interference pattern all the maxima are of same intensity In interference all fringes have equal wave front. In a diffraction pattern the intensity decreases on either side of the central maximum In diffraction the central maximum is very

14 14 widths. In the interference pattern the regions of minimum intensity are usually almost perfectly dark wide. The width decreases on either side In the in diffraction pattern the regions of minimum intensity are not perfectly dark. 6. a) Give the theory of Fraunhofer diffraction due to single slit and hence b) obtain the condition for primary and secondary maxima. Using this c) obtain intensity distribution curve (June 006) 7. Obtain the condition for primary maxima in Fraunhofer diffraction due to a single slit and d) derive an expression for width of central maximum (005, 011) briefly explain fraunhofer diffraction at single slit (May 017) a) FRAUNHOFER DIFFRACTION AT SINGLE SLIT suppose AB is a narrow slit width a and perpendicular to the plane of the paper. Let a plane wave front WW of monochromatic light of wave length λ propagating normally to the slit be incident on it. Let the diffracted light be focussed by means of a convex lens on a screen placed in the focal plane of the lens. The secondary wavelets travelling normally to the slit, i.e., along the direction OPo are brought to focus at Po by the lens. Thus Po is a bright central image. This is called zero order maximum. The secondary wavelets travelling at an angle θ with the normal are focussed at a point P1 on the screen. In order to find out intensity at P1, draw a perpendicular AM to the diffracted rays. b) Suppose the slit is divided into two halves AO and OB. Suppose the waves starting from the top of the two halves have a path difference λ/. These two waves interfere destructively producing minimum. For every point in the first half there is a corresponding point in the second half producing waves having a path difference λ/. a sin θ λ or a sinθ = λ (1) The path difference between secondary wavelets from A and B direction θ. The rays travelling in this direction interfere destructively producing minimum called first order minimum. The higher order minima can be obtained in the direction d sinθ = m λ where m =,3,4,.. In addition to the central maximum there are secondary maxima which lie in between the secondary minima on either side. Hence for secondary maxima a sinθ = (m + 1) λ/ It should be noted that secondary maxima do not fall exactly mid-way between two minima, but they are displaced towards the centre of the pattern, of course, the displacement decreases as the order of maximum increases. Thus the diffraction pattern due to single slit consists of a central bright maximum flanked by secondary maxima and minima on both the sides. c) Intensity distribution in diffraction pattern

15 15 Suppose the width of the slit is divided into n equal parts and the amplitude of the wave from each part is a (because width of each part is same). Suppose the phase difference between any two consecutive waves from these parts would be δ 1 n π a = δ = n 1 α (say) λ sinθ Using the method of vector addition of amplitudes, the resultant amplitude Aθ is given by sinα A θ a for large value of n, α/n is very small. α sin n sin α Hence A θ n a α sin α Aθ Ao α The intensity in any direction is given by sinα Iθ Io α where Io is the intensity of the principal maximum at θ = 0. Figure represents the intensity distribution. It is a graph of sinα Io α (along Y-axis) as a function of α or sin θ (along X-axis). It can be seen that most of the light is confined to the central maximum. The intensity of the secondary maxima falls off rapidly Io/, Io/61.. d) Linear width of the principal maximum Linear width of the principal maximum is distance between the first order secondary minima on either side of the central maximum. If x is the distance of the first secondary minimum from the center then the width of the central maximum W = x. If f is the focal length of the lens used to focus the diffraction pattern we have sinθ = x/f But sinθ = λ/a. Therefore x λ f a or f x λ a Hence, the width of the central maximum is given by f W x From the above it is clear that as the slit narrows the width of the central maximum increases as shown. 8. How do you measure the slit width. Measurement of slit width The slit is illuminated with a laser beam and the diffraction pattern is obtained on a screen placed at a distance one meter. The distance between the first minima on either side x is measured. The slit width a can be measured using λ a f λ a x where f is the focal length of the lens or distance of the screen from the slit. Problem 1: A slit is illuminated with light of wave length λ. find the angular width of the central maximum if the width of the slit is 1) 4λ ) 3λ 3) λ 4) λ. Hint: a sin θ = λ in the above if the screen is at a distance m find the width of the central maximum.

16 16 Problem : A slit of width 1.5 mm is illuminated by a light of wavelength 500 nm and diffraction pattern is observed on a screen m away. Calculate the width of the central maximum. [1.33 mm] Dλ Hint: x a Problem 3: A screen is placed m away from a narrow slit. If the first minima lie at 5 mm on either side of the central maximum when light of wavelength 500 nm is used. Find the width of the slit.[0. mm] Hint: a = λd/x. Problem 4: A lens whose focal length is 40 cm forms a Fraunhofer diffraction pattern of a slit 0.3 mm wide. Calculate the distances of the first dark band and of the next bright band from the axis (wavelength of light used is 5890 Å [0.785 mm, mm] f 3 f Hint: x1 and x a a Problem 5: find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 1x10-5 cm when the slit is illuminated by light of wave length 6000Å.[30 o ] Hint: a sinθ = nλ Problem 6: Find the angular width of the central maximum in the Fraunhofer diffraction using a slit of Width 1 μm when the slit is illuminated by light of wavelength 600 nm.(june 006 ) [θ = o ][θ = o ] Hint: a sinθ = λ Problem 7: Calculate the angular separation between the first order minima on either side of central maximum when the slit is 6 x10-4 cm width and light illuminating it has a wavelength 6000 Å 9. Explain with theory the Fraunhofer diffraction due to N slits. (May 003, June 004, June 011) Fraunhofer Diffraction due to N slits (Diffraction grating) Diffraction grating consists of very large number of narrow slits side by side and separated by opaque spaces. The incident light is transmitted through the slits and blocked by opaque spaces. Such a grating is called transmission grating. When light passes through the grating, each one of the slit diffracts the waves. All the diffracted waves reinforce one another producing sharper and intense maxima on the screen. In practice a plane transmission grating is a plane sheet of transparent material on which opaque rulings are made with a diamond point. The spaces between the rulings are equal and transparent and constitute the parallel slits. The rulings are opaque and are of equal width. The combined width of a ruling and a slit is called grating element. Theory of plane transmission grating Let ABC.H represent the section of grating normal to the plane of the paper. Let the width of each slit be a and that of opaque ruling be b. Now (a + b) which is the combined width of a ruling and a slit is called grating element. It is also the distance between two successive slits. Any two points on successive slits separated by a distance (a + b) are called corresponding points. Let a plane wave front be incident normally on the grating. The points in the slits act as secondary sources of light giving rise to secondary waves. These waves spread in all directions on the other side of the grating. These waves are brought to focus on a screen with the help of a lens. The secondary

17 17 waves travelling in the same direction as that of the incident wave are focused at Po. Since all these secondary waves have travelled equal distance to reach Po, they reinforce constructively and hence the point Po is the position of central bright maximum. Now let us consider secondary waves travelling at an angle θ with the direction of incidence and reaching P 1. The intensity at P 1 depends on the path difference between the secondary waves originating from the corresponding points of two adjacent slits. Since the distance between corresponding points is (a + b) the path difference is (a + b) sinθ. The intensity at P 1 will be maximum if (a + b) sinθ = nλ Intensity distribution Suppose the waves from the grating reach a point on the screen. Suppose the phase difference between two waves form the edges of a slit is given by π a = α (say) λ sinθ and suppose the phase difference between two waves form the adjacent slits is given by π ( λ b) sinθ a = β (say) The intensity at any point is due to diffraction as well as interference. It is given as Asinα I α The factor sinnβ sinβ Asinα α while the factor sinnβ sinβ gives the distribution of intensity due to diffraction at single slit gives the distribution of intensity as a combined effect of all slits. Principal maxima: the most intense maxima are called principal maxima. They are obtained for (a + b) sinθ = ± nλ where n = 0,1,,3.. Asinα α The intensity of principal maxima is I N Minima: in between any two principle maxima there will be (N - 1) minima. Secondary maxima: as there are (N - 1) minima there will be (N - ) 9. Calculate the maximum number of orders possible for a plane diffraction grating. (June 011) The principal maxima in a grating satisfy the relation (a + b) sinθ = nλ or The maximum angle of diffraction can be 90 o. Hence the maximum possible order is given by (n) max o a bsin90 a b λ λ n a b sinθ λ 11. Find the missing order in a grating spectrum In a grating spectrum the total intensity is controlled by diffraction at a single slit and interference due to two adjacent slits. If the intensity due to any one cause is zero the total intensity at that point will be zero.

18 18 Suppose at a point we n th principal maximum satisfying the relation (a + b) sinθ = nλ.(1) Suppose at the same point we get m th diffraction minimum satisfying the relation a sinθ = mλ.() If both the conditions (1) and () are satisfied simultaneously then n th diffraction order will be missing. Dividing equation (1) by () we get (a b)sinθ n (a b) n or.(3) a sinθ m a m 1. Explain experimental method of determination of wavelength of Spectral lines of a given source of light using plane transmission grating. 13. Explain with theory how wavelength of spectral line is determined using Plane diffraction grating (M 003, J 004) what is resolving power. What is its formula? The resolving power of a grating is defined as the capacity to form separate diffraction maxima of two wavelengths λ1 and λwhich are very close to each other. It is given by λ/dλ where λ is the average wavelength and dλ is the difference in wavelengths. For a λ grating it is given as n N where N is the total number of lines on the grating. dλ The resolving power of a grating depends on the total number of lines on the grating and the order of the spectrum. Examine if two spectral lines of wavelengths 5890Å and 5896Å can be clearly resolved in the 1) first order ) second order by a diffraction grating cm wide and having 45 lines/cm (May 017) λ Hint: n N [the lines will be clearly resolved in second order] dλ What is dispersive power. What is its formula? MULTIPLE CHOICE QUESTIONS 10. The Penetration of waves into the regions of the geometrical shadow Is a) interference b) diffraction c) polarization (1) dispersion 11. In a Single slit diffraction, the first diffraction minima is observed at an an angle of 30 o, when the light of wavelength 500 nm is used. The Width of the slit is a) 5x10-5 cm b).5x10-5 cm. c) 10 x 10-5 cm d) 1.5x10-5 cm 1. In Fraunhofer diffraction the wavefront undergoing diffraction has to be a) spherical b) cylindrical c) elliptical d) plane 13. In a single slit experiment if the slit width is reduced a) the fringes becomes brighter b) the fringes become narrower c) the fringes become wider d) the colour of the fringes change 14. Instead of red colour source, if blue colour source is used in single slit experi ment a) the diffraction pattern does not change b) the diffraction bands become wider

19 19 c) the diffraction pattern becomes narrower and crowded together d) the diffraction pattern disappears. 15. The diffraction pattern of a single slit consists of a) wider dark band at the center with alternate bright and dark bands on either side. b) narrow bright band at the center with alternate dark and bright bands of equal intensity on either side. c) wider bright band at the center with alternate dark and bright bands of equal intensity on either side. d) wider and brighter band at the center with alternate dark and bright bands of decreasing intensity on either side. 16. A parallel beam of monochormatic light falls normally on a plane diffraction grating having 5000 lines/ cm. A second order Spectral line is diffracted through an angle of 30. The wavelength of light is a) 5 x 10-7 cm b) 5 x 10-6 cm C) 5 x10-5 cm d) 5 x 10-4 cm /... urn lies 5 : Th W d. m e 1 th 0f Slit Is on elthel' Side Of the central maxrmum % < a) 0.04 mm b) 0 x c).4 mm d) 4:1? r a) 1 pm b) 10 In C) 5 pm d) 5 um 1 0. When white light is incident on a diffraction grating, the light diffracted more a) blue b) yellow c) violet (1) red 1. Monochromatic light falling normally on a grating gives rise to diffracted second order beam at angle 30. If the grating has 5000 lines / cm, the wavelength of light is a) 600 nm ' b) 400 nm c) 500 nm d) 650nm. Maximum number of orders possible with a grating is a) independent of grating element b) directly proportional to grating element. c) inversely proportional to grating element. d) directly proportional to wavelength. 3. Polarization of light conclusively proves that a) light waves are longitudinal b) light waves are transverse c) light waves are longitudinal as well as transverse. d) light travels with velocity3x10 8 m/s 4. The phenomenon causing polarization of light is a) interference b) diffraction c) double refraction d) refraction 5. A calcite crystal-is placed over a dot through the calcite one finds a) one dot. b) two dots which are stationary c) one dot rotating about the other. d) two rotating dots. 6. In a doubly refracting crystal, along optic axis a)μo > μe b) μo < μe c) μo = μe d) none Fill in the Blanks 1, With decrease of slit width in single slit, the width of the fringes We

20 0 13 In a grating, the combined Width of a ruling and a slit is called,4 : 14 Points on successive slits separated by a distance equal to the grating element. are called 15. In Fraunhofer diffraction at a plane transmission grating, when white light source is used, the angle of diffraction for violet is,, than that of red. 16. The phenomenon which confirms transverse nature of light is 17. If the intensity of polarized light passing through rotating analyser falls to zero,. then the light is polarized. 18. The angle of incidence of light for which the reflected beam 1s completely plane polarized IS known as 19. The polarizing angle for glass is, 0. When unpolarized light passes through certain crystals, two refracted beams are produced. This phenomenon is called 1In Nicol prism for A 5893A, the refractive index of Canada balsam cement is. In Nicol prism the refractive index for e-ray varies between ' and slate True or False 1Two monochromatic sources of same amplitude and some wavelength can act as coherent sources. _ True / False In Young 5 double slit experiment, if one of the slits ls closed the fringe contrast decreases.. ' ' True/False 3If the path difference between the two interfering waves is integral multiple of i v\, constructive interference occurs.. True/ False 1In Newton s ring experiment with reflected light the point of contact of lens With the glass plate where thickness of air is zero, appears dark. True/ False i. Radius of 4 dark ring in Newton s ruing experiment is 4 EA. True/ False RThinfilmswhenviewedund hit li t, ul et." persion. er W e gh appears m l 3 _ 06 l 7. For diffraction to occur, the size of the obstacle must be comparable with wave length. True / False B. Converging lens is used to focus parallel rays in Fresnel diffraction. True / False 9. The source and screen are at infinite distance from the obstacle, producin