INTERFERENCE OF LIGHT

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Domenic George
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1 INTERFERENCE OF LIGHT Physics Without Fear

2 Want to be good in Physics? Remember: PHYSICS IS AN ACTIVE SPORT Physics is like a sport. To be good in a sport, you must practice. Likewise, to be good in Physics you must practice. You can t afford to be passive. Physics demands participation. Physics can be learnt by critical thinking alone. Hard work & regularity have no substitute.

3 CONTENT Principle of Superposition of Waves Interference Young s Double Slit Experiment

4 Principle of Superposition of Waves PRINCIPLE OF SUPERPOSITION E 1 + E 2 E 1 The resultant displacement at a point due to two or more waves passing simultaneously through a medium is the VECTOR SUM of the displacements due to individual waves. Mathematically, y= y 1 + y 2 + y 3 + y 4 +. y= y 1 + y 2 E 2 Constructive Interference E = E 1 + E 2 1 st Wave (E 1 ) 2 nd Wave (E 2 ) Resultant Wave Reference Line

5 Principle of Superposition of Waves PRINCIPLE OF SUPERPOSITION E 1 E 1 - E 2 E 2 Destructive Interference E = E 1 - E 2 The resultant displacement is the VECTOR SUM of the displacements due to individual waves. Mathematically, y= y 1 + y 2 + y 3 + y 4 +. or Algebraically y= y 1 y 2 1 st Wave (E 1 ) 2 nd Wave (E 2 ) Resultant Wave Reference Line

6 INTERFERENCE OF LIGHT The phenomenon of redistribution of energy in the space when two light waves from two coherent sources superpose is called interference of waves. Bright Band Dark Band S 1 S 2 Bright Band Dark Band Bright Band Crest Trough Bright Band Dark Band S1 & S2 emit light of same frequency in same phase

7 INTERFERENCE OF LIGHT Theory of Interference of Waves: Consider light waves from two coherent sources. The waves have the same speed, wavelength, frequency ( and hence time period) nearly equal amplitudes, travelling in the same direction with constant phase difference Φ. y 1 = a sin ωt y 2 = b sin (ωt + Φ) ω is the angular frequency of the waves, a,b are the amplitudes and y 1, y 2 are the instantaneous values of Electric displacement. Applying superposition principle, the magnitude of the resultant displacement of the waves is y = y 1 + y 2 y = a sin ωt + b sin (ωt + Φ) = a sin ωt + b sin ωt cos Φ + b sin Φ cos ωt y = (a + b cos Φ) sin ωt + b sin Φ cos ωt

8 INTERFERENCE OF LIGHT y = (a + b cos Φ) sin ωt + b sin Φ cos ωt Putting a + b cos Φ = A cos θ (1) We get b sin Φ = A sin θ..(2) y = A sin ωt cos θ + A cos ωt sin θ = A sin (ωt + θ).(3) (where y is the resultant displacement, A is the resultant amplitude and θ is the resultant phase difference). Equation (3) implies that the resulting wave is also a harmonic wave of same frequency as that of the superposing waves. Square & add equations (1) & (2). We get A = (a 2 + b 2 + 2ab cos Φ) tan θ = b sin Φ a + b cos Φ

9 INTERFERENCE OF LIGHT We have A = (a 2 + b 2 + 2ab cos Φ) As intensity I is proportional to square of the amplitude of a wave. We have, I α A 2 i.e. I α (a 2 + b 2 + 2ab cos Φ) For constructive interference, I should be maximum. As I depends on Φ We have cos Φ = +1 for maxima. i.e. Φ = 2nπ where n = 0, 1, 2, 3,. Also path difference λ= phase difference of 2 π For maxima; path difference p = (λ / 2 π) x 2nπ p = n λ

10 INTERFERENCE OF LIGHT Condition for Destructive Interference of Waves: We have A = (a 2 + b 2 + 2ab cos Φ) For destructive interference or minima; we have cos Φ = - 1. i.e. Φ = (2n + 1)π where n = 0, 1, 2, 3,. and A min = a b. For minima; path difference is p = (λ / 2 π) x (2n + 1)π p = (2n + 1) λ / 2

11 INTERFERENCE OF LIGHT Intensity ratio: It is the ratio of intensity maxima to Intensity minima. As A = (a 2 + b 2 + 2ab cos Φ) A max = a + b & A min = a - b I max α (a + b) 2 I min α (a - b) 2 I max I min = I max I min = (a + b) 2 (a/b + 1)2 (a - b) 2 = (a/b - 1) 2 (r + 1) 2 (r - 1) 2 Where r = a / b (amplitude ratio of the waves)

12 YOUNG S DOUBLE SLIT EXPERIMENT Monochromatic Source The waves from S 1 and S 2 reach the point P such that path difference p = S 2 P S 1 P S 1 S d O S 2 D d/2 d/2 P y

13 S 2 P 2 = D 2 + {y + (d/2)} 2.(1) YOUNG S DOUBLE SLIT EXPERIMENT S 1 P 2 = D 2 + {y - (d/2)} 2.(2) A B From (1) & (2); we get S 2 P 2 S 1 P 2 = [D 2 + {y + (d/2)} 2 ] [D 2 + {y (d/2)} 2 ] or (S 2 P S 1 P) (S 2 P + S 1 P) = 2 yd or p (2D) = 2 yd Path difference p = yd / D Note: Use Pythagoras theorem in S 2 BP & S 1 AP to get equations (1) & (2).

14 YOUNG S DOUBLE SLIT EXPERIMENT POSITIONS OF BRIGHT FRINGES: For a bright fringe at P, p = yd / D = nλ where n = 0, 1, 2, 3, yd / D = nλ y = n D λ / d For central maxima n = 0, y 0 = 0 For 1 st maxima n = 1, y 1 = D λ / d For 2 nd maxima n = 2, y 2 = 2 D λ / d For n th maxima/bright Fringe n = n, y n = n D λ / d

15 YOUNG S DOUBLE SLIT EXPERIMENT POSITIONS OF DARK FRINGES: For a dark fringe at P, p = yd / D = (2n+1)λ/2 where n = 0, 1, 2, 3, yd / D = (2n+1)λ/2 y = (2n+1) D λ / 2d For 1 st minima n = 0, y 1 = D λ / 2d For 2 nd minima n = 1, y 3 = 3D λ / 2d For 3 rd minima n = 2, y 3 = 5D λ / 2d.. For n th minima/dark Fringe n = n, y n = (2n+1)D λ / 2d

16 FRINGE WIDTH: YOUNG S DOUBLE SLIT EXPERIMENT For Bright Fringes : Width β = y n y n-1 = n D λ / d (n 1) D λ / d = D λ / d For Dark Fringes: Width β B = y n y n-1 = (2n+1) D λ / 2d {2(n-1)+1} D λ / 2d = D λ / d Hence both the dark & the bright fringes are equally spaced on the screen.

17 Intensity Distribution Curve: YOUNG S DOUBLE SLIT EXPERIMENT INTENSITY y 0 y For the two interfering waves of same amplitude a, We have I max α (a+a) 2 i.e. I max α 4a 2 for all the bright fringes. For all dark fringes I min = 0 Hence all the dark fringes have zero intensity.

18 YOUNG S DOUBLE SLIT EXPERIMENT Conditions for sustained interference: The two sources producing interference must be coherent. As otherwise the path difference will change with time & the pattern will not be permanent. Due to persistence of vision & rapid variation in intensity at a point, the pattern will not be visible giving an impression of general illumination.

19 YOUNG S DOUBLE SLIT EXPERIMENT Conditions for sustained interference: The two interfering wave trains must have the same plane of polarisation. WHY? As otherwise, there will be a poor contrast between the maxima & the minima.

20 YOUNG S DOUBLE SLIT EXPERIMENT Conditions for sustained interference: The two sources must be very close to each other and the pattern must be observed at a larger distance to have sufficient width of the fringe. WHY? We have, fringe width (D λ / d) will be small & the pattern again becomes invisible due to ovelapping of the fringes.

21 YOUNG S DOUBLE SLIT EXPERIMENT Conditions for sustained interference: The sources must be monochromatic. Otherwise, the fringes of different colours will overlap. WHY? As fringe width depends on wavelength, different colours will overlap & the fringes will not be sharp.

22 The two waves must be having same/nearly same amplitude for better contrast between bright and dark fringes. WHY? YOUNG S DOUBLE SLIT EXPERIMENT Conditions for sustained interference: For a=10 & b=1 I max = 121 = 3 I min 81 2 For a=10 & b=9 (Approximately) Resulting in Poor Contrast. I max = 371 I min 1 which is large Resulting in Good Contrast.

23 VISIT: for more topics

WAVE OPTICS GENERAL. Fig.1a The electromagnetic spectrum

WAVE OPTICS GENERAL. Fig.1a The electromagnetic spectrum WAVE OPTICS GENERAL - The ray optics cannot explain the results of the two following experimental situations: a) When passing by small openings or illuminating small obstacles, the light bends around borders