1© Manhattan Press (H.K.) Ltd. Young’s double slit experiment Young’s double slit experiment 9.10 Interference of light waves Relationship between x,, D and a Relationship between x,, D and a Changes in the interference pattern Changes in the interference pattern Measurement of the wavelength of light Measurement of the wavelength of light Energy distribution Energy distribution

2 © Manhattan Press (H.K.) Ltd. Interference of light waves 9.10 Interference of light waves (SB p. 76) Conditions required for interference of light: 1. Light waves from coherent sources 2. Same or almost the same amplitudes 3. Small separation between two coherent sources 4. Small path difference

3 © Manhattan Press (H.K.) Ltd. Young’s double slit experiment 9.10 Interference of light waves (SB p. 77) By Huygen’s principle, - C sends out circular wavefronts - Points at A and B emit wavelets in phase - This method is called division of wavefront

4 © Manhattan Press (H.K.) Ltd. Young’s double slit experiment 9.10 Interference of light waves (SB p. 77) Superposition of waves from A and B - interference - constructive interference (bright fringes) - destructive interference (dark fringes)

5 © Manhattan Press (H.K.) Ltd. PB – PA = a sin  = n (constructive interference at P) and sin  = x n /D Relationship between x,, D and a 9.10 Interference of light waves (SB p. 78) 1. Bright fringe

6 © Manhattan Press (H.K.) Ltd. a sin  = (n - ) (destructive interference at P) and sin  = x n /D Relationship between x,, D and a 9.10 Interference of light waves (SB p. 79) 2. Dark fringe Go to Example 11 Example 11 Go to Example 12 Example 12

7 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 84) 1. Change in the space between S and double slit Move slit S and light source to double slit - bright fringes are brighter (intensity of light through double slit is greater) - fringe separation x unchanged

8 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 85) 2. Change in a a increases, x decreases Larger aSmaller a

9 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 85) 3. Change in source of light Use white light (different wavelengths) white multi- coloured at edges Blue on edge nearest central maximum Red on outer edge

10 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 85) 4. Change in slits width wider S, A and B - no interference pattern (wide wavefronts from S, A and B are no more coherent)

11 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 85) 5. Change in medium Transparent medium between double slit and screen o – wavelength in vacuum  – refractive index x o – fringe separation in vacuum

12 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 86) 6. One of the slits covered with a film From A to O, optical path length = AO From B to O, optical path length = BO + (  - 1)d = AO + (  - 1)d Optical path difference = [AO - (  - 1)d] – AO = (  - 1)d  0 Central maximum shifts to O’ If (  - 1)d = n (the nth bright fringe at O’) If (  - 1)d = (n - ) (the nth dark fringe at O’)

13 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 87) 7. Slits covered with polaroids Light are plane polarized - No interference pattern (lights are mutually perpendicular) - or visible interference pattern (planes of polarization are parallel)

14 © Manhattan Press (H.K.) Ltd. Changes in the interference pattern 9.10 Interference of light waves (SB p. 87) 8. Changes in source of light Use two separate sources of light - No interference pattern sources not coherent)

15 © Manhattan Press (H.K.) Ltd. Measurement of the wavelength of light 9.10 Interference of light waves (SB p. 87) 1 – o = length of 20 dark fringes wavelength can be calculated Go to Example 13 Example 13 Go to Example 14 Example 14 Go to Example 15 Example 15 Go to Example 16 Example 16

© Manhattan Press (H.K.) Ltd. Energy distribution 9.10 Interference of light waves (SB p. 93) Amplitude at bright fringe = 2 x amplitude of single slit (a) Intensity at bright fringe = 4 x intensity of single slit (4I o ) [since I  a 2 ] Note: Wave energy can neither be created nor destroyed, but they can be redistributed for the constructive and destructive interference.

17 © Manhattan Press (H.K.) Ltd. End

18 © Manhattan Press (H.K.) Ltd. Q: Q: (a) What is meant by interference? (b) State the conditions necessary for the production of an interference pattern. (c) State the conditions necessary to produce an interference pattern where the fringes are of the same separation and of about the same intensity in a Young’s double slit experiment. (d) In a Young’s double slit experiment, light of wavelength 644 nm was used. Initially, the slit separation is 0.20 mm and the distance between the screen and the double slit is 1.00 m. Throughout the experiment, the interference pattern was viewed from the position of the double slit. (i) What is the angular separation between two neighbouring bright fringes formed on the screen? (ii) If the screen is then shifted further away from the double slit, what changes, if any, will occur to the interference pattern? Solution 9.10 Interference of light waves (SB p. 80)

19 © Manhattan Press (H.K.) Ltd. Solution: Solution: (a) Interference is the effect produced by the superposition of waves from two coherent sources moving through the same region. (b) The conditions necessary for the production of an interference pattern are: 1. The waves must be from two coherent sources. 2. The amplitudes of the waves must be the same or almost the same. 3. The two coherent sources must be close to each other Interference of light waves (SB p. 80)

20 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (c) The conditions necessary to produce evenly spaced fringes are: 1. The use of a monochromatic source. 2. The single slit S must be narrow and lie on the perpendicular divider of AB (see figure above). 3. The slit separation a must be small. 4. The distance of the screen from the double slit, D, must be large Interference of light waves (SB p. 80)

21 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (d) (i) Use the equation x = (1) From the figure above, the angular separation between two neighbouring bright fringes, θ= From equation (1): =  θ= = = 3.22 × 10  3 rad 9.10 Interference of light waves (SB p. 81)

22 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (ii) When the screen is further away from the double slit, 1. the angular separation ( θ= ) remains unchanged. 2. the linear separation x = increases since D is larger. 3. the intensity of the bright fringes decreases. Return to Text 9.10 Interference of light waves (SB p. 81)

23 © Manhattan Press (H.K.) Ltd. Q: Q: (a) What is meant by Huygens’ principle? (b) With the aid of a diagram, use Huygens’ principle to explain how the interference pattern is produced in a Young’s double slit experiment. (c) In a Young’s double slit experiment, the slit separation is 0.05 cm and the distance between the double slit and screen is 200 cm. When blue light is used, the distance of the first bright fringe from the centre of the interference pattern is 0.13 cm. (i) Calculate the wavelength of the blue light used in the experiment. (ii) Calculate the distance of the fourth dark fringe from the centre of the interference pattern. Solution 9.10 Interference of light waves (SB p. 81)

24 © Manhattan Press (H.K.) Ltd. Solution: Solution: (a) Huygens’ principle states that every point on a wavelength acts as a point source emitting secondary wavelets in phase with one other. (b) 9.10 Interference of light waves (SB p. 81)

25 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): Spherical wavefronts emerge from the slit S. Since slits A and B are of an equal distance from S, the same wavefront from S arrives at A and B simultaneously. According to Huygens’ Principle, the points on the wavefront at A and B emit wavelets in phase with one other. Hence, the waves from A and B are coherent. At the points where the optical path difference of waves from A and B is nλ ( n = 0, 1, 2...), constructive interference occurs and a bright fringe is obtained. If the optical path difference is ( n – )λ, then destructive interference occurs and a dark fringe is obtained Interference of light waves (SB p. 82)

26 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (c) (i) Slit separation, a = 0.05 cm = 5.0 × 10 –4 m Distance, D = 200 cm = 2.00 m Fringe separation, x = 0.13 cm = 1.3 × 10 –3 m (ii) Using x = the wavelength of the blue light, = = = 3.25 × 10 –7 m 9.10 Interference of light waves (SB p. 82)

27 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): From the figure, distance of the fourth dark fringe from the centre = = 3.5 × (1.3 ×10 –3 ) = 4.55 × 10 –3 m Return to Text 9.10 Interference of light waves (SB p. 82)

28 © Manhattan Press (H.K.) Ltd. Q: Q: A Young’s double slit system was set up at one end of an empty rectangular glass tank, and a sodium lamp with a narrow slit placed outside the tank. Interference fringes were observed on the opposite side of the tank. When the tank was filled with oil, the fringe separation changes by 35%. Explain why this happens and discuss whether the fringe separation increases or decreases. Calculate the refractive index of the oil. Solution 9.10 Interference of light waves (SB p. 88)

29 © Manhattan Press (H.K.) Ltd. Solution: Solution: Fringe separation, x = ……………….. (1) In oil, the wavelength, 1 = where  = refractive index  Fringe separation, x 1 = ……………….. (2) = ……………….. (from (1)) Also x 1 = 65% or 0.65x i.e. fringe separation decreases Hence = 0.65x Refractive index,  = = 1.54 Return to Text 9.10 Interference of light waves (SB p. 88)

30 © Manhattan Press (H.K.) Ltd. Q: Q: In a Young’s double slit experiment, an interference pattern was obtained using monochromatic light of wavelength 500 nm. When a thin film of transparent material with refractive index 1.5 was placed in front of one of the slits, the central maximum was shifted to the original position of the 9th bright fringe. What is the thickness of the film? Solution 9.10 Interference of light waves (SB p. 89)

31 © Manhattan Press (H.K.) Ltd. Solution: Solution: When the central maximum was shifted to O’, the original position of the 9th bright fringe above O, all the other fringes were also shifted upwards. Therefore, at the point O on the axis of the double slit system, we will find the 9 th bright fringe which was originally below O Interference of light waves (SB p. 89)

32 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): If d = thickness of the film, for light originating from A and B and arriving at O, the optical path difference = nλ (since there is constructive interference) [AO +( μ – 1) d] – BO = 9λ ( n = 9) (μ – 1)d = 9λ ( AO = BO) ∴ Thickness of the film (d) = = 9.0 × 10 –6 m Return to Text 9.10 Interference of light waves (SB p. 89)

33 © Manhattan Press (H.K.) Ltd. Q: Q: In a Young’s double slit arrangement, the distance between the centres of the slits is 0.25 mm and light of wavelength 6.0 × 10 –4 mm is used. Calculate the angle θ subtended at the double slit by the neighbouring maxima of the interference pattern (refer to figure below) Interference of light waves (SB p. 89)

34 © Manhattan Press (H.K.) Ltd. Q: Q: Describe and explain what happens to the interference fringes if (a) the slit A is covered with a thin piece of transparent material of high refractive index, (b) the intensity of light from A is reduced to half of that from B, (c) A and B are covered with thin films of polaroid, and one of the films is rotated slowly, (d) the distance between A and B is increased slowly. Solution 9.10 Interference of light waves (SB p. 89)

35 © Manhattan Press (H.K.) Ltd. Solution: Solution: Using the equation x =, and since the angle  is small,  = tan  = = = = 2.4 ×10  3 rad 9.10 Interference of light waves (SB p. 90)

36 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (a) When the slit A is covered with a thin piece of transparent material of high refractive index, the central maximum which was originally at O is shifted upwards. When A is covered with the transparent material, the optical path for light travelling from A to O is AO + (μ – 1)d where μ is the refractive index of the material and d its thickness. Hence the optical path difference between light travelling from A to O and from B to O is optical path difference = [AO + (μ – 1)d] – OB = (μ – 1) d (AO = OB) ≠ 0 Hence O is no more the position of the central maximum. If (μ – 1)d = nλ where n is an integer then the nth bright fringe from central maximum will be formed at O. If (μ – 1)d = (n – )λ then the nth dark fringe from the central maximum is found at O Interference of light waves (SB p. 90)

37 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (b) Initially, the intensity of the light across the interference pattern is as shown in Fig. (a). When the intensity of the light from A is halved, the intensity of the light across the interference pattern is as shown in Fig. (b). Fig. (a) Fig. (b) The intensity of the maxima has decreased but the intensity of the minima has increased. Hence the difference between the intensities of the maxima and minima is reduced. There is no change in the fringe separation Interference of light waves (SB p. 90)

38 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): (c) When the axis of polarization of the polaroids are parallel, the bright fringes can be differentiated clearly for the dark fringes. When one of the polaroids is rotated, the difference in intensities of the maxima and minima decreases until the axes of the polaroids are mutually perpendicular. Then no interference pattern will be seen. (d) Using the equation x =, when the slit separation a is slowly increased the slit separation x decreases. The fringes become closer and closer until the neighbouring bright fringes cannot be resolved by the eye. Return to Text 9.10 Interference of light waves (SB p. 91)

39 © Manhattan Press (H.K.) Ltd. Q: Q: In a Young’s double slit experiment, a source of light which emits light of wavelengths 7 × 10 –7 m (red) and 5.6 × 10 –7 m (green) is used. Draw a sketch to show the interference fringes of the two colours formed. Show that the fringes of the two colours first exactly overlap while moving out from the centre of the pattern when What is the significance of n? What is the colour of the fringe at that point? Solution 9.10 Interference of light waves (SB p. 91)

40 © Manhattan Press (H.K.) Ltd. Solution: Solution: For red light, the fringe separation = (7 × 10 –7 ) For green light, the fringe separation = (5.6 × 10 –7 )  x 1 When a red fringe and a green fringe overlap for the first time, the nth red fringe overlap with the (n + 1)th green fringe; and their distances from the central maximum are equal. i.e. nx 1 = (n + 1) x 2 n(7 × 10 –7 ) = (n + 1)(5.6 × 10 –7 )  n = Interference of light waves (SB p. 91)

41 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): n represents the fourth red fringe which coincides with the fifth green fringe. The colour of the fringe produced is Note that for n = 0 and n = 4, a red and green fringe coincide exactly to produce a yellow fringe Interference of light waves (SB p. 92) Return to Text