In interference, two individual amplitudes are A_(0)each and the intensity is I_(0)each. Find resultant amplitude and intensity at a point, where (a) phase difference between two waves is 60^(@). (b) path difference between two waves is lambda/3.
Answers
Given :
- Amplitude of waves = A₀
- Intensity = I₀
To find :
- Resultant amplitude and intensity at a point where,
- phase difference between two waves is 60°
- path difference between two waves is λ/3
Solution :
(a) Phase difference(Ф) between two waves is 60° :
- We know that, resultant amplitude is given by :
A = √[(A₁²) + (A₂²) + 2 A₁.A₂.cosФ]
- A = √[(A₀)² + (A₀)² + 2 A₀.A₀.cosФ]
= A₀√[2(1+cosФ)]
= A₀ √[2(2 cos²(Ф/2)]
= A₀.2cos(Ф/2) ....(1)
= A₀ . 2.cos(60/2)
A = A₀ .
- We know that, I = A² = [2A₀cos(Ф/2)]²
- Hence, I = 4A₀²cos²(Ф/2) .....(2)
= 4A₀²[cos(60/2)]²
= 4A₀²(3/4)
I = 3 I₀ ....( A₀² = I₀)
(b) Path difference(Δx) between two waves is λ/3 :
- We know that, Phase difference Ф = (2π/λ) . Δx
- So, Ф = (2π/λ) . (λ/3)
Φ = (2π/3)
- From equation (1),
A = A₀.2cos(π/3)
A = A₀
- From equation (2),
I = 4A₀²cos²(Ф/2)
= 4A₀²cos²(π/3)
I = A₀²
I = I₀
Answer : (a) A = A₀ , I = 3 I₀
(b) A = A₀ , I = I₀
The resultant amplitude is A = (√3) A0 and I = 3I0 when phase difference is 60∘ and A = A0 and I = I0 when path difference is λ / 3
Explanation:
(a) Substituting ϕ = 60∘ in the equations
A = 2A0 (cos) ϕ / 2 -------(1)
and I = 4I0 cos^2 (ϕ / 2) ------(2)
We get, A = (√3)A0 and I = 3I0
(b) Given data:
Δx = λ3
∴ϕ or Δϕ = (2π / λ).Δx
Δϕ = (2π / λ) (λ / 3) = 2π / 3 or 120∘
Now, substituting ϕ = 120∘ in the above two equations we get
A = A0 and I = I0
Hence the resultant amplitude is A = (√3) A0 and I = 3I0 when phase difference is 60∘ and A = A0 and I = I0 when path difference is λ / 3