Let the waves from two coherent sources of light are represented as:
y1 = a sin ω t (1)
y2 = b sin (ω t + Ø) (2)
Where a and b are the respective amplitudes of the two waves and Ø is the constant phase angle by which second wave leads the first wave.
According to superposition principle, the displacement (y) of the resultant wave at time (t) would be given by
y = y1 + y2 = a sin ω t + b sin (ω t + Ø)
= a sin ω t + b sin ω t cos Ø + b cos ω t sin Ø
y = sin ω t (a + b cos Ø) + cos ω t × be sin Ø (3)
Put a + b cos Ø = R cos θ (4)
b sin Ø = R sin θ (5)
y = sin ω t × R cos θ + cos ω t × R sin θ
= R [sin ω t cos θ + cos ω t sin θ]
y = R sin (ωt + θ) (6)
Thus the resultant wave is a harmonic wave of amplitude R.
Squaring (16) and (17) and adding, we get
R2 (cos2θ + sin2θ)
= (a + b cos Ø)2 + (b sin Ø)2
R2 × 1 = a2 + b2 cos2Ø + 2ab cos Ø + b2 sin2Ø
= a2 + b2 (cos2Ø + sin2Ø) + 2ab cos Ø
R = √a2 + b2 + 2ab cos Ø (7)
As result intensity I is directly proportional to the square of the amplitude of the resultant wave
Therefore, I R2
i.e. I (a2 + b2 + 2ab cos Ø) (8)
For constructive interference
I should be maximum for which,
cos Ø = max = 1 therefore, Ø =0, 2∏, 4∏,…….
i.e. Ø = 2n∏
Where n = 0, 1, 2 …
If x is the path difference between the two waves reaching point P, corresponding to phase difference Ø, then
x = /2∏Ø = /2∏ (2n∏) = n
i.e. x = n (9)
Hence condition for constructive interference at a point is that phase difference between the two waves reaching the point should be zero or an even integral multiple of ∏. Equivalently, path difference between the two waves reaching the point should be zero an integral multiple of full wavelength.
For destructive interference,
I should be minimum,
cos Ø = minimum = - 1
Therefore, Ø = ∏, 3∏, 5∏ …
Or, Ø = (2n – 1) ∏, where n = 1, 2…………….
The corresponding path differences between the two waves:
x = /2∏Ø = /2∏ (2n – 1) ∏ = (2n -1) /2
i.e. x = (2n – 1) /2 (10)
Hence, the condition for destructive interference at a point is that phase difference between the two waves reaching the point should be an odd integral multiple of ∏ or path difference between the two waves reaching the point should be an odd integral multiple of half the wavelength.
1. From the above discussion, we find that interference is constructive, when cos Ø = 1
From (10), R = √a2 + b2 + 2ab × 1
i.e. R = √(a + b)2 = a + b = sum of the amplitudes of two waves, which would be maximum.
Therefore, Rmax = (a + b)
As Imax R2max
Therefore, Imax (a + b)2 (11)
Again, interference is destructive, when
cos Ø = -1
From (7), R = √a2 +b2 +2ab (-1)
i.e. R = √(a – b)2 = (a –b) = difference of the amplitudes of the two waves, which would be minimum
Therefore, Rmin = a – b
As Imin R2min
Therefore, Imin (a – b)2
Or, Imin = k (a – b)2 (12)
Imax/Imin = (a + b)2/(a – b)2
When b = a
Rmax = (a + b) = a + a = 2a
Rmin = (a – b) = a – a = 0
Therefore, Imin = 0 i.e. dark bands will be perfectly dark and the contrast between bright and dark interference bands will be the best.
From (10), R2 = a2 + b2 + 2ab cos Ø
Taking, a2 = I1; b2 = I2 and R2 = IR, the resultant intensity as a result of superimposition of two waves, may be put as
IR = I1 + I2 + 2√I1I2cos Ø
2. If w1 and w2 are widths of two slits from which intensities of light I1 and I2 emanate, then
w1/w2 = I1/I2 = a2/b2
ExpertsMind.com - Physics Help, Optical Physics Assignments, Interference Conditions Assignment Help, Interference Conditions Homework Help, Interference Conditions Assignment Tutors, Interference Conditions Solutions, Interference Conditions Answers, Optical Physics Assignment Tutors