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# Interference Conditions Assignment Help

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Optical Physics - Interference Conditions

**Interference Conditions**

Let the waves from two coherent sources of light are represented as:

y_{1}** = a sin ω t (1)**

y_{2} = 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 = y**_{1} + y_{2} = 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

**R**^{2} (cos^{2}θ + sin^{2}θ)

**= (a + b cos Ø)**^{2} + (b sin Ø)^{2}

**R**^{2} × 1 = a^{2} + b^{2} cos^{2}Ø + 2ab cos Ø + b^{2} sin^{2}Ø

**= a**^{2} + b^{2} (cos^{2}Ø + sin^{2}Ø) + 2ab cos Ø

**R = √a**^{2} + b^{2} + 2ab cos Ø (7)

As result intensity I is directly proportional to the square of the amplitude of the resultant wave

Therefore,** I R**^{2}

i.e. **I (a**^{2} + b^{2} + 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.

**Important notes**

**1.** From the above discussion, we find that interference is constructive, when **cos Ø = 1**

From (10), **R = √a**^{2} + b^{2} + 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** I**_{max} **R**^{2}_{max}

Therefore, **I**_{max} (a + b)^{2} (11)

Again, interference is destructive, when

**cos Ø = -1**

From (7), **R = √a**^{2} +b^{2} +2ab (-1)

i.e.** R = √(a – b)**^{2} = (a –b) = difference of the amplitudes of the two waves, which would be minimum

Therefore, **R**_{min} = a – b

As** I**_{min} R^{2}_{min}

Therefore,** I**_{min} (a – b)^{2}

Or,** I**_{min} = k (a – b)^{2} (12)

**I**_{max}/I_{min} = (a + b)^{2}/(a – b)^{2}

When** b = a**

**R**_{max} = (a + b) = a + a = 2a

**R**_{min} = (a – b) = a – a = 0

Therefore,** I**_{min} = 0 i.e. dark bands will be perfectly dark and the contrast between bright and dark interference bands will be the best.

From (10),** R**^{2} = a^{2} + b^{2} + 2ab cos Ø

Taking, **a**^{2} = I_{1}; b^{2} = I_{2} and R_{2} = IR, the resultant intensity as a result of superimposition of two waves, may be put as

**IR = I**_{1} + I_{2} + 2√I_{1}I_{2}cos Ø

**2.** If **w**_{1} and **w**_{2} are widths of two slits from which intensities of light** I**_{1} and **I**_{2} emanate, then

**w**_{1}/w_{2} = I_{1}/I_{2} = a^{2}/b^{2}

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**Interference Conditions**

y

_{1}

**= a sin ω t (1)**

y

y

_{2}= 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 = y**

_{1}+ y_{2}= 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

**R**

^{2}(cos^{2}θ + sin^{2}θ)**= (a + b cos Ø)**

^{2}+ (b sin Ø)^{2}

**R**

^{2}× 1 = a^{2}+ b^{2}cos^{2}Ø + 2ab cos Ø + b^{2}sin^{2}Ø**= a**

^{2}+ b^{2}(cos^{2}Ø + sin^{2}Ø) + 2ab cos Ø**R = √a**

^{2}+ b^{2}+ 2ab cos Ø (7)As result intensity I is directly proportional to the square of the amplitude of the resultant wave

Therefore,

**I R**

^{2}

i.e.

**I (a**

^{2}+ b^{2}+ 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.

**Important notes**

**1.**From the above discussion, we find that interference is constructive, when

**cos Ø = 1**

From (10),

**R = √a**

^{2}+ b^{2}+ 2ab × 1i.e.

**R = √(a + b)**sum of the amplitudes of two waves, which would be maximum.

^{2}= a + b =Therefore, Rmax = (a + b)

As

**I**

_{max}

**R**

^{2}_{max}

Therefore,

**I**

_{max}(a + b)^{2}(11)Again, interference is destructive, when

**cos Ø = -1**

From (7),

**R = √a**

^{2}+b^{2}+2ab (-1)i.e.

**R = √(a – b)**difference of the amplitudes of the two waves, which would be minimum

^{2}= (a –b) =Therefore,

**R**

_{min}= a – bAs

**I**

_{min}R^{2}_{min}

Therefore,

**I**

_{min}(a – b)^{2}

Or,

**I**

_{min}= k (a – b)^{2}(12)**I**

_{max}/I_{min}= (a + b)^{2}/(a – b)^{2}

When

**b = a**

**R**

_{max}= (a + b) = a + a = 2a**R**

_{min}= (a – b) = a – a = 0Therefore,

**I**i.e. dark bands will be perfectly dark and the contrast between bright and dark interference bands will be the best.

_{min}= 0From (10),

**R**

^{2}= a^{2}+ b^{2}+ 2ab cos ØTaking,

**a**, the resultant intensity as a result of superimposition of two waves, may be put as

^{2}= I_{1}; b^{2}= I_{2}and R_{2}= IR**IR = I**

_{1}+ I_{2}+ 2√I_{1}I_{2}cos Ø**2.**If

**w**

_{1}and

**w**

_{2}are widths of two slits from which intensities of light

**I**and

_{1}**I**emanate, then

_{2}**w**

_{1}/w_{2}= I_{1}/I_{2}= a^{2}/b^{2}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