INTERFERENCE OF LIGHT Physics Without Fear
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CONTENT Principle of Superposition of Waves Interference Young s Double Slit Experiment
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
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
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
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
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
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
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
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)
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
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).
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
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
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.
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.
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.
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.
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.
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.
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.
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