Case I: Thin film of uniform thickness:

Transcription

1 Interference in Thin Films The film of transparent material like a drop of oil spread on the surface of water, show brilliant colours when exposed to an extended source of light. This phenomenon can be explained on interference basis. Here interference takes place between rays reflected from the upper and from the lower surface of the film. Case I: Thin film of uniform thickness: (a) Reflection Pattern: Let us consider a thin film of thickness t, refractive index µ and a ray AB of monochromatic light of λ is falling on it at an angle i. This ray is partly reflected and partly refracted as shown in Figure 4.4. Path difference = (BD + DE) film - (BH) air = [µ (BD +DE)- BH] air = [µ(bd + DG + GE) BH] = µ (BD + DG) As µge = BH BH = EB sin i GE = EB sin r µ GE = EB = sin i µ = sin i / sin r Extend BL and ED to meet at K we have

2 LKD = LNDE = r Triangles BLD and LKD are congruent BD=DK Then path difference = µ(bd + DG) = µ(kd + DG) = µ KG = µ KB cos r = 2 µt cos r Now, here since reflection from denser medium is taking place by one ray so additional path difference of λ/2 is taken into account. Thus the total path difference is T = 2 µt cos r λ/2 (5) Condition for maxima T = (2 µ cos r- λ /2) = n λ 2 µt cos r = (2n + 1) λ /2 n = 0, 1, 2, 3,... Condition for minima T = (2 µt cos r- λ /2} = (2n + 1) λ /2 2 µt cos r = (n + 1) λ n is the integer so (n + 1) is also integer and can be taken as n 2 µt cos r = nλ n = 0, 1, 2 when cos r is kept constant and thickness increases gradually, the path difference becomes λ /2, λ, 3 λ /2, 2 λ, 5 λ /2,... and as a result the film will appear dark (t = 0), bright, dark and so on. On the other hand if t is constant and r is varied we again get a series of maxima and minima.

3 (b) Transmitted Pattern: Here the path difference two rays OEMS and PR will be = µ (DE + EM) DH = 2 µt cos r (9) In this case there will be no additional path difference so the total path difference T = 2 µt cos r Condition for maxima = 2 µt cos r = nλ n = 0, 1, 2, 3,... Condition for minima = 2 µt cos r = (2n + 1) λ/2 n = 0, 1, 2, 3,... (12) We find that the conditions for maxima and minima are found in case of transmitted pattern are opposite to those found in case of reflected pattern. Under the same conditions of the film looks dark in reflected pattern it will look bright in transmitted pattern. Colours of thin films: When white light is incident on a thin film only few wavelengths will satisfy the condition of maxima and therefore corresponding colours will seen in the pattern. For other wavelengths condition of minima is satisfied, and so corresponding colours will be missing in the pattern. The coloration of film vary with t and r. Therefore if one vary either t or r a different set of colours will be observed. Since the condition for maxima and minima are opposite in case of reflected and transmitted pattern. So the colours found in two patterns will be complimentary to each other.

4 Necessity of broad source: When point source is used only a small portion of the film can be seen through eye and as a result the whole interference pattern cannot be seen. But when a broad source is used rays of light are incident at different angles and reflected parallel beam reach the eye and whole beam and complete pattern is visible. Case II: The film is of varying thickness (Wedge shaped thin film): Consider the wedge shaped film as shown in Figure 4.5. Let a ray from S is falling on the film and after deflections produce interference pattern. = [(PF + FE) film + (PK) air ] = µ (PF + FE) (PK) = µ (PN + NF + FE) PK = µ (PN + NF + FE PN) = µ (NF +FE) = µ (NF + FL) = µ (NL) = 2 µ t cos (r +θ) Then total path difference considering refraction from denser medium is taking place T = 2 µt cos (r+θ) λ/2 Condition for maxima 2µcos (r + θ) = (2n + 1) λ/2 n = 0, 1, 2, )

5 Condition for minima 2µcos (r + θ) = nλ n = 0, 1, 2, 3,... (16) Thus the film will appear bright if the thickness t satisfies the condition of maxima t = (2n + 1) λ / 4µcos (r + θ)... (17) and it will appear dark when t = n λ / 2µ tanθ cos (r + θ) (18) Hence, we move along the direction of increasing thickness we observe dark, bright, dark λ... fringes. For t= 0 i.e., at the edge of film = λ/2 so the film will appear dark. Then width of the fringes so observed can be found β = n λ / 2µ tanθ cos (r + θ) In case of normal incidence r = 0 2 µt= (2n + 1) λ/2 (maxima) 2 µt = n λ (minima) β = λ/ 2µθ = fringe width Principle of Superposition According to this principle the resultant displacement of a particle of medium acted upon by two or more waves simultaneously is equal to the algebraic sum of the displacement of the particle due to individual waves. Thus if two waves are

6 y 1 = a 1 sin wt and y 2 = a 2 sin (wt+ δ) which have same frequency but differing in phase and amplitude, then from the principle of superposition the resultant displacement y will be Here we assume and Y = y 1 + y 2 = a 1 sin wt+ a 2 sin(rot+ δ) = (a 1 + a 2 cos δ) sin wt + a 2 cos wt sin δ θa 1 + a 2 cos δ = A cos θ a2 sin 8 = A sin θ, y = A cos θ sin wt + A sin θ cos wt A 2 =a 1 2 +a a 1 a 2 cosδ Tan θ = a 2 sinδ/a 1 +a 2 cosδ which gives the amplitude and phase of resultant wave. The intensity at any point is given by I= A 2 = a l +a a 1 a 2 cosδ Condition for maximum and minimum intensities Intensity will be maximum of points where cos δ = + 1 or δ = 2nπ [n = 0, 1, 2, 3, ]

7 i.e., intensity will be maximum when phase difference is even multiple of nor path difference ( ) is even multiple of λ/2. δ= 2nπ, n = 0, 1, 2,... = 2nπ/2 I max = (al + a 2 ) 2 The intensity will be minimum, when δ = (2n + 1) π n = 0, 1, 2,... = (2n + 1) A/2 I min = (a l a 2 ) 2 Coherent Sources It is found that it is not possible to show interference due to two independent sources of light because a large number of difficulties are involved. Two sources may emit light wave of different amplitude and wavelength and the phase difference between the two may change with time. The fundamental requirement to get will defined interference pattern is that the phase difference between the two waves should be constant. The two sources are said to be coherent if they emit light waves of same frequency nearly the same amplitude and are always in phase with each other. In actual practice it is not possible to have two independent sources which are coherent. But for experimental purpose two virtual sources formed from a single source can act as coherent sources. Methods for producing coherent sources is divided in following two parts have been divide (i) interference of light take place between waves from two sources formed due to single source (division of wavefront) (ii) interference takes place between the waves from the real source and virtual source (division of amplitude).

8 Analytical Theory of Interference Let S1 and S2 are two coherent sources separated by a distance d and made from a monochromatic source 5. Let a screen be placed at a distance from the coherent source as shown in Figure 4.1 point 0. On the screen is equidistance from 51 and As such path difference at 0 will be zero and the intensity at 0 will be maximum. Consider a point Pat a distance y from 0. The wave reaches the point P from S 1 and S 2. From Figure 4.1, PQ = (y- d/2) and PR = (y + d/2) THE PATH Path difference (S 2 p- S 1 P) [ D + (y +d/2) 2/ 2d ) [ D + (y +d/2) ] 2.= yd/ D Corresponding phase difference will be θ = 2π /λ =2π /λ (yd /d) Position of bright fringes:

9 Path difference = even multiple of ( λ/2) Yd/d = 2nλ/2 = nλ Yn =nλd/d which gives position of different maximum at y 0 = 0, y 1 =λd/d,y 2 = 2λd/d,y 3 =3λd/d, y =y 1 y 0 =y 2 y 1 = =yn yn-1 =λd/d Position of dark fringes: Path difference odd multiple of (λ../2) yd /d = (2n + 1) λ/2 y n =(2n +1)λd/2d which gives the position of different minima Y 0 = λd/2d, Y 1 = 3λd/2d, y 3 = 5λd/2d y = Y 1 -Y 0 = Y 2 -Y 1 = Y 3 -Y 2 = = Y n -Y n-1 - = λd/d (9) Fringe width: The distance between any two consecutive bright or dark fringes is called fringe width and is denoted by β. (β) right = ( y)min =λd/d (β) dark = ( Y)max =λd/d (β) = (β) bright = (β) dark =λd/d It is found that

10 (β) r λ: β r d and β r 1/d Angular fringe width e11: It is given by θ β =θ n+1 θ =y n-1/d yn /d =β/d =λ/ d 1. At the centre we get bright fringe. 2. Then alternately dark, bright, dark fringe on both sides of the central maximum. 3. Fringe width of dark and bright fringe are same. 4. Width of fringe depends upon wavelength, separation of sources and source to screen separation. If white light is used each wavelength will give its own fringe pattern upto a short distance from 0 the fringes of different colour would stand side by side but beyond that a general illumination is observed. The central fringe with white light is white as for all colours the central fringe is formed at 0 and they together produce white light. Condition for Sustained Interference 1. The two sources must be coherent. 2. The sources should emit light of same frequency. 3. The amplitudes of two wave must be equal otherwise a clear cut distinction between maximum (a 1 + a 2 ) and minimum (a 1 a 2 ) is not possible. 4. The two waves must propagate along the same line. 5. The two sources must be narrow otherwise a single source may itself be equivalent to a large number of fine sources which will give their own pattern.

11 6. The separation between sources should be small so that wide fringes are obtained. 7. Separation between source and screen should be large to get wide fringes. 8. If the interfering waves are polarised they must be in the same state of polarisation. Effect of Introducing a Thin Transparent Plate in the Path of One Way Let 51 and 52 be two monochromatic coherent sources of wavelength A. Let a thin plate of thickness t and refractive index, µ is introduced in the path of one ray from source S 1 as shown in Figure 4.2. Now, this ray travel partially in the material if c and v are the velocities of light in air and medium respectively then path difference between two rays. (S 2 P-S 1 P) = [(S 2 P) air - {(S 1 P- t) air + (t) plate }]... (1) = [(S 2 P- S l P ) air + (t air - t plate )] FIGURE 4.2 we know that distance t travelled by light through plate is equivalent to a distance µt travelled in air. Path difference = [(S 2 P- S l P) + t µt] air = [S 2 P- S l P (µ 1) t] = [yd/d -(µ-1)t]

12 For maxima Path difference = [ yd/d - (µ-1)t]= nλ y = d [ nλ + (µ - 1) t] or Without plate (i.e. t = 0) the position of maximum is given by y 0 = ndλ/d The plate causes a shift of maximum and minimum by an account. δ = (y- y 0 ) = D/d (µ- 1) t We know that β = λd/d => D/d = β/λ δ = β/λ(µ- 1) t where β is fringe width This shift is +ve if µ > 1 and the pattern will move by an amount δ toward the side on which the plate is placed keeping the nature of pattern same. If the central fringe moves through a distance formely occupied by the nth bright fringe δ = nβ =β/λ(µ- 1) t n=(µ- 1) t/λ The gives the number of bands through which the system has shifted. Two Classes of Interference 1. Division of wavefront: In this class the wavefront originate from source S is divided into two parts by using any optical system (like Young Double slit experiment, Fresnel biprism, Fresnel mirror Lloyd s mirror etc.). The two wavefront after travelling

13 certain distance again brought together to give interference pattern. Here, point sources are required. 2. Division of amplitude: Here the amplitude is divided into two or more parts either by partial reflection and which after travelling through different path reach to a point and produce interference. Here broad source is required. Newton s Rings, Michelson interferometer and Fabry Perot interferometer are the examples of this class. Interference in Thin Films The film of transparent material like a drop of oil spread on the surface of water, show brilliant colours when exposed to an extended source of light. This phenomenon can be explained on interference basis. Here interference takes place between rays reflected from the upper and from the lower surface of the film. Case I: Thin film of uniform thickness: (a) Reflection Pattern: Let us consider a thin film of thickness t, refractive index µ and a ray AB of monochromatic light of λ is falling on it at an angle i. This ray is partly reflected and partly refracted as shown in Figure 4.4. Path difference = (BD + DE) film - (BH) air = [µ (BD +DE)- BH] air = [µ(bd + DG + GE) BH] = µ (BD + DG)

14 As µge = BH BH = EB sin i GE = EB sin r µ GE = EB = sin i µ = sin i / sin r Extend BL and ED to meet at K we have LKD = LNDE = r Triangles BLD and LKD are congruent BD=DK Then path difference = µ(bd + DG) = µ(kd + DG) = µ KG = µ KB cos r = 2 µt cos r Now, here since reflection from denser medium is taking place by one ray so additional path difference of λ/2 is taken into account. Thus the total path difference is T = 2 µt cos r λ/2 (5) Condition for maxima

15 T = (2 µ cos r- λ /2) = n λ 2 µt cos r = (2n + 1) λ /2 n = 0, 1, 2, 3,... Condition for minima T = (2 µt cos r- λ /2} = (2n + 1) λ /2 2 µt cos r = (n + 1) λ n is the integer so (n + 1) is also integer and can be taken as n 2 µt cos r = nλ n = 0, 1, 2 when cos r is kept constant and thickness increases gradually, the path difference becomes λ /2, λ, 3 λ /2, 2 λ, 5 λ /2,... and as a result the film will appear dark (t = 0), bright, dark and so on. On the other hand if t is constant and r is varied we again get a series of maxima and minima. (b) Transmitted Pattern: Here the path difference two rays OEMS and PR will be = µ (DE + EM) DH = 2 µt cos r (9) In this case there will be no additional path difference so the total path difference T = 2 µt cos r Condition for maxima = 2 µt cos r = nλ n = 0, 1, 2, 3,... Condition for minima

16 = 2 µt cos r = (2n + 1) λ/2 n = 0, 1, 2, 3,... (12) We find that the conditions for maxima and minima are found in case of transmitted pattern are opposite to those found in case of reflected pattern. Under the same conditions of the film looks dark in reflected pattern it will look bright in transmitted pattern. Colours of thin films: When white light is incident on a thin film only few wavelengths will satisfy the condition of maxima and therefore corresponding colours will seen in the pattern. For other wavelengths condition of minima is satisfied, and so corresponding colours will be missing in the pattern. The coloration of film vary with t and r. Therefore if one vary either t or r a different set of colours will be observed. Since the condition for maxima and minima are opposite in case of reflected and transmitted pattern. So the colours found in two patterns will be complimentary to each other. Necessity of broad source: When point source is used only a small portion of the film can be seen through eye and as a result the whole interference pattern cannot be seen. But when a broad source is used rays of light are incident at different angles and reflected parallel beam reach the eye and whole beam and complete pattern is visible. Case II: The film is of varying thickness (Wedge shaped thin film): Consider the wedge shaped film as shown in Figure 4.5. Let a ray from S is falling on the film and after deflections produce interference pattern.

17 = [(PF + FE) film + (PK) air ] = µ (PF + FE) (PK) = µ (PN + NF + FE) PK = µ (PN + NF + FE PN) = µ (NF +FE) = µ (NF + FL) = µ (NL) = 2 µ t cos (r +θ) Then total path difference considering refraction from denser medium is taking place T = 2 µt cos (r+θ) λ/2 Condition for maxima 2µcos (r + θ) = (2n + 1) λ/2 n = 0, 1, 2, ) Condition for minima 2µcos (r + θ) = nλ n = 0, 1, 2, 3,... (16) Thus the film will appear bright if the thickness t satisfies the condition of maxima t = (2n + 1) λ / 4µcos (r + θ)... (17) and it will appear dark when t = n λ / 2µ tanθ cos (r + θ) (18) Hence, we move along the direction of increasing thickness we observe dark, bright, dark λ... fringes. For t= 0 i.e., at the edge of film = λ/2 so the film will appear dark. Then width of the fringes so observed can be found β = n λ / 2µ tanθ cos (r + θ)

18 In case of normal incidence r = 0 2 µt= (2n + 1) λ/2 (maxima) 2 µt = n λ (minima) β = λ/ 2µθ = fringe width Newton s Fringes and Haidinger Fringes We have seen that thin film causes a path difference 2µt cos r between interfering rays. Now, when area of the film is small, the rays from various portions of the film reaching the eye have almost the same inclination. So it is the variation in thickness that gives fringes. Each fringe is the locus of all such points where thickness is same. Such fringes are known as Newton s fringes or fringes of equal thickness. On the other hand when film has uniform thickness the path difference changes with r only. Each fringe in this case represents the locus of all points on the film, from which rays are equally inclined to the normal. Such fringes are called Haidinger s fringes or fringes of equal inclination. To get such fringes the source must be an extended one. Newton s rings: It is a special case of interference in a film of variable thickness such as that formed between a plane glass plate and a convex lens in contact with it. When monochromatic light falls over it normally we get a central dark spot surrounded by alternatively bright and dark circular rings. When white light is used the rings would be coloured.

19 Experimental Arrangement: Let S be the extended source of light, rays from which after passing through a lens L falls upon a glass plate Gat 45. After partial reflection these rays fall on a plano convex lens P placed on the glass plate E. The interference occurs between the rays reflected from the two surfaces of the air film and viewed through microscope M as shown Figure 4.6. Theory: The air film formed is of wedge shape so the path difference produced will be = 2 µt cos (r+θ) λ/2 For normal inciden r = 0 so = 2 µt-λ/2 At the point of contact t = 0 So µ=λ/2 The central fringe will be dark. Condition for maxima and minima 2 µt = (2n + 1) λ/2 2 µt = nλ n = 0, 1, 2 we get alternatively bright and dark rings. Diameter of the rings: From the Figure 4.7. DB X DC = AD X DC DB= DC= r, radius of ring under consideration. CE = DA = BF = t, thickness of the film at B. OG = OA = R, radius of curved surface of lens.

20 Then, r 2 = t (2R- t) = 2Rt- t 2 = 2Rt t = r 2 /2R Diameter of bright rings D n : From the condition of maxima 2 µ r 2 n /2R = (2n + 1) λ/2, r 2 n = (2n + 1) λr / 2µ =( D n / 2) 2 D n 2 = 2 (2n + 1) λr / µ = 2 (2n + 1) λr (For air, µ = 1) D n = (2AR) (2n + 1) = K (2n + 1) D n = odd number. Thus, the diameter of bright rings are proportional to the square root of the odd natural numbers. Diameter of dark ring D n : Using the condition for minimum 2µ r n 2 /2R = nλ which gives (D n ) 2 = 4nλR Or D n = (4 λr) n D n = n Thus, the diameter of dark rings are proportional to the square root of natural numbers Measurement of Wavelength of light by Newton s Rings For dark rings we know (D n ) 2 /n = 4λR

21 λ = (D n ) 2 /n /4nR But as the fringes around the centre are not very clear so n cannot be measured correctly. To avoid any mistake one can consider two clear fringes nth and (n + p)th So, (D n ) 2 /n = 4λR (D n+p ) 2 = 4(n+p) λr Then ((D n+p ) 2 - (D n ) 2 ) = 4(n+p) λr Or λ = ((D n+p ) 2 - (D n ) 2 ) / 4pR Thus, measuring the diameters and knowing p and R, A can be measured Measurement of Refractive Index of Liquid by Newton s Rings For this purpose liquid film is formed between the lens and glass plate. We have, as above, which give [(D n+p ) 2 - (D n ) 2 ] liquid = 4pλR/µ [(D n+p ) 2 - (D n ) 2 ] air = 4pλR One can see that rings contract with the introduction of liquid. Numericals Based on Interference of Light Problem 1. A film of thickness 0.6 µm of soap solution having refractive index 1.4 is formed. It is observed at an angle of incidence 4. Determine that wave lengths of visible light, absent in the reflected light. Hint: The condition for a particular wavelength, to be absent from the reflected light is 2 µt cos r = A., 2λ., 3 λ.,

22 Let these wavelengths be λ 1, λ 2 λ 3,... µ= sin i/ sin r => r =sin -1 (sin 4 /1.4) = λ 1 = 2 X 1.4 X 0.6 X 10-6 X cos = A 2 λ 2 = 2µt cos r ==> λ 2 = 7461 A Similarly 3 λ 3 = 2µt cos r ==> λ3 = 4974 A The visible region is 4000 to 7000 A So the absent wavelength will be 4974 A. Problem 2. A biprism is placed at a distance of 5 em in front of a narrow slit, Illuminated b sodium light (λ =5 890 A) and the distance between the virtual sources is found to be 0.05 cm. Find they width of the fringes observed in eyepiece placed at a distance of 75 cm from the biprism. Hint: B = λ D/ d = 5890 x x (75 + 5) x 10-2 / 0.05 X 10-2 = A. Problem 3. When a thin piece of glass 3.4 x 104 cm thick is placed in the path of one of the interfering beams in a biprism arrangements, it is found that the central bright fringe shifts through a distance equal to the width of four fringes. Find the refractive index of the piece of glass. Wavelength of light used is 5.46 x w-5 cm. Hint: µ = 1.64 (µ- 1) t = nλ Problem 4. A surface of refractive index 1.5 is to be coated with a film of magnesium fluoride (µ= 1.38) to minimise the reflection. Calculate the minimum thickness of film for normal incidence of light of wavelength 5000 A. Hint: t = λ/4µ= 5000/4 * 1.38 = 906 A Problem 5. Light of wavelength 6000 A falls normally on a thin wedge shaped film of refractive index 1.4, forming fringes that are 2 mm apart. Find the angle of the wedge.

23 Hint: θ = 1.07 x 10-4 radian β = λ/2θµ => θ = λ/2µβ Problem 6. Interference fringes are produced by monochromatic light falling normally on a wedge shaped film of cellophane of refractive index 1.4. If angle of wedge is 20 seconds of an arc and the distance between successive fringes is 0.25 cm, calculate the wavelength of light. Hint: λ = 2 θ µβ = 2 x 20 xπ x 1.4 x 0.25 / 60 X 60 X 180 λ= 6790 A Problem 7. A planoconvex lens of radius 300 cm is placed on an optically flat glass plate and is illuminated by monochromatic light. The diameter of 811 dark ring in the transmitted system is 0.72 cm. Calculate the wavelength of light used. Hint: r 2 = (2n 1) λ R / 2 λ= 5760 A Problem 8. In the Newton s rings arrangement, the radius of curvature of the curved surface is 50 cm. The radii of the 9 th and 16 th rings are 0.18 cm and cm respectively. Calculate the wavelength. Hint: D 9 = 2 x 0.18 = 0.36 cm D 16 = 2 x = cm R =50 cm, p = 7 Then λ = ((D n+p ) 2 - (D n ) 2 ) / 4pR = 5015 A Problem 9. In a Newton s rings experiment the diameter of the 12th ring changes from 1.50 to 1.35 cm liquid. when a liquid is introduced between the lens and the plate. Calculate the refractive index of the Hint: For liquid medium D/= 4nµλR ( i)

24 and for air medium D 2 2 = 4nλR (ii) Then µ = (D 2 / D 1 ) 2 =1.235 Problem 10. In Newton s rings pattern the diameter of 61 bright ring is 0.5 cm. The wavelength of the light used is 4590 A. Light enters the film with angle of incidence of 39. Calculate the radius of curvature of lens. Hint: 2µt cos r = (2n+1) λ/2 and t = r 2 /2R Then r 2 n = (2n+1)λR/ 2µ cos r d 2 n = 2(2n+1) λr/ µ cos r R = cm.