Conditions for Constructive and Destructive Interference

Intensity of the resulting wave in interference

The amplitude of the resulting wave interference is given by

$ \large A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A _2 cos\phi} $

Resultant Intensity ;

$ \large I = I_1 + I_2 + 2 \sqrt{I_1 I _2 }cos\phi $

where φ is the phase difference between the interfering waves.

Now A = Amax when cos φ = +1

i.e. φ = 2 n π

where n = 0 , 1 , 2 , 3 , ………

⇒ φ = 2n π

$ \large \frac{2 \pi}{\lambda}\times \Delta x = 2 n \pi $

⇒  Δ x = n λ

⇒ $\Delta x = 2 n \times \frac{\lambda}{2} $

Therefore when the optical path difference between the interfering waves is even multiple of λ/2 , the waves interfere constructively. The amplitude, hence the intensity of the resulting wave becomes maximum.

∴ Amax = A1 + A2

Similarly A = Amin when cos π = −1

⇒ φ = (2n – 1)π where n = 1 , 2 , 3 …

⇒ $\large \frac{2\pi}{\lambda} \times \Delta x = (2n-1)\pi $

⇒ Path difference , $\large \Delta x = (2n – 1)\frac{\lambda}{2} $

That is when the optical path difference between the interfering waves is odd-multiple of λ/2 , the waves interfere destructively. The amplitude and hence intensity of the resulting wave becomes minimum

Therefore Amin = (A1 − A2)

$ \large \frac{I_{max}}{I_{min}} = \frac{(A_{max})^2}{(A_{min})^2} = \frac{(A_1 + A_2)^2}{(A_1 – A_2)^2}$

$ \large \sqrt{\frac{I_{max}}{I_{min}}} = \frac{A_1 + A_2 }{A_1 – A_2} $

$ \large \frac{A_1}{A_2} = \frac{\sqrt{I_{max}} + \sqrt{I_{min}}}{\sqrt{I_{max}}-\sqrt{I_{min}}} $

Intensity of the resulting wave

The amplitude of the resulting wave is given by

$ \large A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 cos\phi} $

∴ I  ∝ A2

$ \large I = I_1 + I_2 + 2 \sqrt{I_1 I_2} cos\phi $

Where I1 and I2 are the intensities of the interfering waves.

Illustration :  The intensity of the light coming from one of the slits in a Young’s double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.

Solution: The intensity of the light originating from the first slit is double the intensity from the second slit. The amplitudes of the two interfering waves are in the ratio √2 : 1 , say √2 A and A .

At the point of constructive interference, the resultant amplitude becomes (√2 + 1) A. At the points of destructive interference , this amplitude is (√2 – 1)A.

The ratio of the resultant intensities at the maxima to that at the minima is

$ \large \frac{(\sqrt 2 + 1)^2A^2}{(\sqrt 2 -1)^2A^2} $

Illustration  : Young’s double slit experiment is carried out using microwaves of wavelength λ = 3cm. Distance between the slits is d = 5 cm and the distance between the plane of slits and the screen is D = 100 cm.

(a) Find the number of maximas, and

(b) their positions on the screen.

Solution : (a) The maximum path difference that can be produced = distances between the sources. (5 cm) Thus in this case we will have only 3 maximum. One central maxima and two on its either side for a path difference of λ (3 cm.)

Exercise  : The width of one of the two slits in a Young’s double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width, find the ratio of the maximum to the minimum intensity in the interference pattern.

Also Read :

Phase difference & Path difference
Interference , Constructive & Destructive interference
Calculation of fringe width in YDSE
Angular fringe width
Change in fringe width due to change in medium
Fringe shift due to thin film
Diffraction of Light

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