Interference of light

1. I Interference of light Problems Theory

2. Interference: The redistribution of light energy due to superposition of two or more waves with same phase or constant phase difference is called interference of light.

3. Theory of Interference: Consider two light waves represented by and …(1) Interference (superposition) of these two waves at a point leads to the resultant displacement given by, Y=Y1+Y2

4. Define,

5. is the phase difference between the first wave and resultant wave. R is the amplitude of the resultant wave and Squaring and adding equations (4) and (5), it can be shown that dividing (5) by (4), we get,

6. Constructive interference: Constructive interference is said to be taken place, when two or more waves interfere at a point in phase leading to the maximum intensity (Brightness) at that point. Destructive interference: Destructive interference is said to be taken place when two or more waves interfere at a point out of phase leading to the minimum intensity (Darkness) at that point.

7. Condition for Constructive interference: For Constructive interference, R must be maximum, i.e., when cos(fi)=1

8. If

9. Condition for Destructive interference: For Destructive interference, R must be minimum, i.e., when cos(fi)=-1

10. If

11. Experimental Setup: Young's double slit experiment

12. A parallel beam of monochromatic light passes through a single slit and advances towards closely separated double slits S1 and S2 as a single wave front. It is divided into two separate wave fronts by S1 and S2. The two wave fronts, which are coherent, advance towards the screen Y. While moving towards Y, the two wave fronts superpose according to the principle of superposition of waves. At all points on the screen where the crest of one wave falls on the crest of the other wave and the trough of one wave falls on the trough of the other wave, constructive interference takes place – giving rise to bright fringes. At all the points on the screen where the crest of one wave falls on the trough of the other wave, destructive interference takes place-giving rise to dark fringes. A pattern consisting of alternate bright and dark fringes of equal width is obtained.

13. Expression for fringe width

14. The path difference between the two waves interfering at P is given by From the right angled triangle S1QP From the right angled triangle S2RP

15. But, hence P is very close to O and S1P= S2P=D To obtain bright fringe, the condition is

16. Or the distance of the nth bright fringe from O is . Similarly the distance of the (n-1)th bright fringe from O is Similarly if we consider the formation of a dark fringe at P, we get the condition for dark fringe as

17. Or the distance of the nth dark fringe from O is . Similarly the distance of the (n-1)th bright fringe from O is Therefore, for dark fringes, fringe width, Therefore, Bright and dark fringes are of equal width

18. Coherent Waves: • Waves having same wavelength and frequency are called coherent waves. Or • Waves are said to be coherent if the phase difference between them is zero or constant at any instant of time.

19. Methods to produce coherent source: Two coherent sources can be obtained in two ways, one, by division of amplitude and the other, by division of wave front. The second technique is used to produce coherent sources in the Young’s double slit experiment. [Example of obtaining coherent sources by division of amplitude method is by using Biprism]

20. Conditions to obtain sustained and distinct interference fringes • The two sources of light must be coherent. • A distance as small as possible must separate the two coherent sources. • The two slits must be very narrow. • The two waves must have the same (or nearly the same) wavelength and period. • The two waves must have the same or nearly the same amplitude.

21. Interference at thin films Maxima or constructive int. takes place when, Minima or destructive int. takes place when, Where, ref. index of the thin film t thickness of the film r angle of incidence n=0,1,2,….

28. Write a short note on Newton’s rings.

29. In the adjacent diagram C is a Plano-convex lens of large radius of curvature placed on the glass plate P. Light from a monochromatic source is rendered parallel by a convex lens L. Another glass plate Q reflects the parallel beam and allows it to fall on the convex lens C. Here some of the light is reflected from the upper surface of the glass plate and some from the lower surface of the lens. Division of amplitude takes place. Interference fringes are produced. They are circular because the system is symmetrical about the center of the lens and hens the name Newton’s rings. Newton’s rings are circular interference fringes formed by interference at thin air film formed between a plane glass plate and a convex lens of large radius of curvature.

30. Radius of mth dark fringe is given by • Radius of mth bright fringe is given by R Radius of curvature of the lens surface and Wavelength of the light

31. Destructive Interference Constructive Interference

32. Formulae:

33. 1. Calculate the %ge change in the fringe width when the distance between the double slit and screen is increased by 20% while the slit separation is increased by 50% in a young’s double slit experiment. Ans: 20%

34. 2. The distance between two coherent sources is 0.3mm and the screen is 0.9 m from the sources. The 15th dark fringe is at a distance 2mm from the central bright fringe, find the a. wavelength of the light used. b. distance of the 10th bright fringe from the central bright fringe. c. distance of the tenth bright fringe from the central bright fringe when the separation between the screen and the slits is increased by 0.1 m. Ans: 44.4 nm 1.33 mm 1.48 mm

35. 3. In a young’s double slit experiment, the slits are at a separation of 1.5 mm. When the slits are illuminated by a monochromatic source and a screen is held at a distance 1.5 m, the width of 10 fringes is found to be 5.89 mm. When the screen is moved towards the slits through a distance 0.5 m, the width of 10 fringes has a value 3.93 mm. Find the wavelength of the light used. Ans: 589 nm

36. 4. Two slits in a young’s experiment are 0.5 mm apart. Interference fringe of width 1.3 mm are observed on a screen placed 1.2 m away from the slits. Calculate the wavelength of the light used. If the entire apparatus is immersed under water of nw=4/3, find the change in the fringe width. Ans: 541.67 nm 0.325 mm

37. 5. In an young’s experiment, interference fringes are obtained on a screen placed 1.5 m away from the two slits which are 1.5mm apart. If the wavelength of the light used is 650 nm, finda) The fringe width andb) Change in the fringe width if the screen is moved away by 0.5 m. Ans: 0.65 mm 0.216 mm

38. 6. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a young’s double slit experiment.a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength of 650 nm.b) What is the least distance from the central maximum where the bright fringe due to both the wavelengths coincide?The distance between the slits is 2mm and distance between the slits and the screen is 1.2m. Ans: 1.17 mm 1.56 mm

39. 7. When two narrow slits separated by a small distance illuminated by a light of wavelength 5x10-7m, interference fringes of width 0.5 mm are formed on a screen. What should be the wavelength of light source to obtain fringe 0.3mm wide if the distance between the screen and the slits is reduced to half the initial value? Ans: 6x10-7 m

40. 8. In a Young’s experiment, the width of the fringes obtained with light of wavelength 6000 A is 2.0mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33? Ans: 1.5 mm

41. 9. If the two slits in Young’s experiment have width in the ratio 4:9, then find the ratio of intensity at the maximum to the intensity at the minima in the interference pattern. Ans: 25:1

42. 10. In young’s experiment, the distance of screen fro the two slits is 0.1m. When a light of wavelength 6000 A is allowed to fall on the slits, the width of the fringe obtained on the screen is 2.0mm. Determinea) distance between the two slits.b) the width of the fringe, if the wavelength of incident light is 4800 A. Ans: 3x10-4 m 1.6x10-3 m

43. 11. In a Young’s double slit experiment, the slits are separated by 0.03 cm and the screen is placed 1.5 m away. The distance between the central fringe and the fourth fringe is 1cm. Determine the wavelength of the light used in the experiment. [Hint: 4=1cm] Ans: 5000 A

44. 12. If the two slits in young’s experiment have widths in the ratio 1:9, deduce the ratio of intensity at the maxima and minima in the interference pattern. Ans: 4:1

45. 13. The distance between two coherent sources is 0.02x10-2 m. If the 6th band is at a distance of 1.2x10-2 m from the central fringe, find the wavelength of source used. The distance of the screen from the wavelength of source used. The distance of the screen from the coherent sources is 1m. Ans: 4000 A

46. 14. The ratio of the intensities at minima to maxima in the interference pattern is 16:25. Determine the ratio of the widths of the two slits in young’s experiment. Ans: 81:1

47. 15. In Young’s experiment, the interference fringes are produced on the screen placed at 1.5 m from the 2 slits separated by a distance of 0.15x10-2 and illuminated by a light of wavelength 5x10-7.a) Find the fringe widthb) Find the change in fringe width when the screen is brought towards the slits by 0.5m. Ans: 5x10-4 m 1.67x10-4 m

48. 16. Find the distance between 15th and 25th fringe in an interference pattern due to two narrow slits separated by 0.25x10-3m and illuminated by light of wavelength 6x10-7m. The distance between the screen and slit is 1m. Ans: 24x10-3 m

49. 17. Two coherent sources of wavelength 5893 A are separated by 0.6mm. Calculate the fringe width formed on a screen 0.5 m away from the sources. Ans: 491x10-7 m

50. That’s it.