What happens when two waves of equal amplitude and opposite phase meet?

They undergo complete destructive interference, and the resultant displacement is zero at that point.

Why is the superposition principle important in understanding interference?

Because interference is a direct result of superposition—interference patterns arise from the addition (constructive) or cancellation (destructive) of wave displacements.

What is meant by phase difference in wave superposition?

:It is the difference in the phase angles of two waves at a given point, which determines how the waves combine—constructively or destructively.

Is energy conserved in the superposition of waves?

Yes, energy is conserved overall. However, it is redistributed in the interference pattern, creating regions of high and low intensity.

Give a real-life example of wave superposition.

Noise-canceling headphones use destructive interference to cancel unwanted sounds by producing sound waves that are out of phase with ambient noise.

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Superposition of Waves

The superposition of waves is a fundamental idea in physics that explains what happens when two or more waves overlap. According to the superposition principle, the total effect at any point is simply the sum of the individual wave displacements. Depending on how the waves line up—whether they’re in phase or out of phase—this can result in constructive interference (where the waves add up) or destructive interference (where they cancel each other out). This principle applies to all types of waves, including mechanical waves like sound and electromagnetic waves like light. It's key to understanding many wave phenomena, such as interference and diffraction.

1.0Superposition Principle

When two or more waves overlap at a point in space, the resultant displacement at that point is equal to the algebraic sum of the displacements due to each individual wave.

$y = y_{1} \pm y_{2} \pm y_{3} + \dots$

2.0Types of Superposition

1.Constructive Superposition

Constructive superposition occurs when two waves meet in phase (i.e., their crests and troughs align), resulting in a larger amplitude. The displacements of the individual waves add up, producing a wave with maximum intensity.

2.Destructive Superposition

Destructive superposition happens when two waves meet out of phase (i.e., the crest of one wave aligns with the trough of another), causing their displacements to partially or completely cancel out, leading to a reduced or zero amplitude.

Relationship between Phase Difference and Path Difference

$Phase difference = \frac{2 π}{wavelength} \times path difference ϕ = \frac{2 π}{λ} \times Δ x$

Related Video:

3.0Superposition of Two Sinusoidal Waves

$Resultant wave at P from S_{1} an d S_{2} is given as y_{1} = a sin ω t y_{2} = b sin (ω t + ϕ) Applying principle of superposition at point P we get resultant wave Y = y_{1} + y_{2} = a sin ω t + b sin (ω t + ϕ) = a sin ω t + b [sin ω t cos ϕ + cos ω t sin ϕ] = sin ω t (a + b cos ϕ) + b cos ω t sin ϕ a + b cos ϕ = R cos θ \dots\dots . (2) b sin ϕ = R sin θ \dots\dots . (3) Y = R sin ω t cos θ + cos ω t sin θ = R sin (ω t + θ) Resultant wave is a simple harmonic wave having amplitude R and phase difference θ$

Amplitude of Resultant Wave

$Squaring and adding equations (2) and (3) (a + b cos ϕ)^{2} + b^{2} sin^{2} ϕ = R^{2} cos^{2} θ + R^{2} sin^{2} θ (a + b cos ϕ)^{2} + b^{2} sin^{2} ϕ = R^{2} (cos^{2} θ + sin^{2} θ) R^{2} = a^{2} + b^{2} + 2 ab cos ϕ R = a^{2} + b^{2} + 2 ab cos ϕ (I n t e n s i t y \propto (am pl i t u d e)^{2}) \Rightarrow I_{a} \propto a^{2}, I_{b} \propto b^{2} I_{R} = K R^{2} = K (a^{2} + b^{2} + 2 ab cos ϕ) I_{R} = I_{a} + I_{b} + 2 I_{a} I_{b} cos ϕ (This is the Resultant Intensity Equation) tan θ = \frac{b s i n ϕ}{a + b c o s ϕ}$

4.0Condition for Constructive and Destructive Interference

1.Constructive Interference

It occurs when the amplitude and the resultant wave is maximum.

$I_{R} = I_{max} if cos ϕ = maximum = + 1 ϕ = 0, 2 π, 4 π \Rightarrow ϕ = n (2 π) where n = 0, 1, 2, 3, \dots Δ x = \frac{λ}{2 π} \times 2 nπ = nλ I_{max} = (a + b)^{2} I_{max} = (I_{a} + I_{b})^{2} R_{max} = (a + b)$

2.Destructive Interference

$It occurs when the amplitude and the resultant wave is minimum. I_{R} = I_{min} if cos ϕ = minimum = + 1 ϕ = 0, 2 π, 4 π \Rightarrow ϕ = n (2 π) where n = 0, 1, 2, 3, \dots Δ x = \frac{λ}{2 π} \times 2 nπ = nλ I_{min} = (a + b)^{2} I_{min} = (I_{a} + I_{b})^{2} R_{min} = (a + b)$

5.0Ratio of Maximum And Minimum Intensity of Light

$\frac{I _{max}}{I _{min}} = (\frac{I _{a} + I _{b}}{I _{a} - I _{b}})^{2} = (\frac{a + b}{a - b})^{2} = (\frac{\frac{a}{b} + 1}{\frac{a}{b} - 1})^{2} = (\frac{r + 1}{r - 1})^{2} (since \frac{a}{b} = r = Amplitude Ratio)$

6.0Relation Between Intensity and Width of Source

$L e t w_{a} an d w_{b} be the width of narrow sources s_{1} an d s_{2} I_{a} \propto w_{a} \Rightarrow I_{a} \propto a^{2} I_{b} \propto w_{b} \Rightarrow I_{b} \propto b^{2} \frac{a ^{2}}{b ^{2}} = \frac{w _{a}}{w _{b}}$

Illustration-1: $S_{1} an d S_{2}$ are two sources of light which produce individual disturbance at point P given by $E_{1} = 3 sin ω t, E_{2} = 4 cos ω t$ . Assuming $E_{1} an d E_{2}$ to be along the same line, find the resultant after their superposition.

Solution:

$E = 3 sin ω t + 4 sin (ω t + \frac{π}{2}) A^{2} = 3^{2} + 4^{2} = 2 (3) (4) cos \frac{π}{2} = 5^{2} tan ϕ_{0} = \frac{4 s i n \frac{π}{2}}{3 + 4 c o s \frac{π}{2}} = \frac{4}{3} = 5 3^{\circ} E = 5 sin (ω t + 5 3^{\circ})$

Illustration-2:

$S_{1} an d S_{2}$ are two coherent sources of frequency 'f ' each.

[Given: $θ_{1} = θ_{2} = 0^{\circ} an d v_{sound} = 330 m/s .$ ]

(i) So that constructive interference at 'p'

(ii) So that destructive interference at 'p'

two sources of light which produce individual disturbance at point P

Solution:

$For constructive interference: k Δ x = 2 nπ \frac{2 π}{λ} \times 2 = 2 nπ λ = \frac{2}{n}, V = λ f \Rightarrow V = \frac{2}{n} f f = \frac{330}{2} \times n For destructive interference: k Δ x = (2 n + 1) π \frac{2 π}{λ} \times 2 = (2 n + 1) π \frac{1}{λ} = \frac{( 2 n + 1 )}{4}, f = \frac{V}{λ} f = \frac{330 \times ( 2 n + 1 )}{4}$

Illustration-3:Two waves are given by the equations

$y_{1} = 4 sin ω t an d y_{2} = 3 sin (ω t + \frac{π}{3})$ When these two waves interfere at a point, what is the approximate amplitude of the resulting wave?

Solution:

$A = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ A = (4)^{2} + (3)^{2} + 2 \cdot 4 \cdot 3 \cdot cos \frac{π}{3} = 37 \approx 6$

Illustration-4:Two waves with intensities 9I 9I and 4I 4I interfere at a point. If the path difference between them is11λ 11 \lambda , what will be the resultant intensity at that point?

Solution: Path difference, $Δ x = \frac{λ}{2 π} \times ϕ \Rightarrow \frac{2 π}{λ} \times 11 λ = 22 π,$ ,i.e. At the same point, the superposition of the waves results in constructive interference.

$I_{R} = (I_{1} + I_{2})^{2} = (9 I + 4 I)^{2} = 25 I$

Illustration-5:In interference if $\frac{I _{max}}{I _{min}} = \frac{144}{81}$ .Determine the ratio of amplitudes of the two waves involved in the interference.

Solution:

$\frac{a _{1}}{a _{2}} = (\frac{I _{max}}{I _{min}} + 1 / \frac{I _{max}}{I _{min}} - 1) = (\frac{\frac{144}{81} + 1}{\frac{144}{81} - 1}) = (\frac{\frac{12}{9} + 1}{\frac{12}{9} - 1}) = \frac{7}{1}$

Illustration-6:Two interfering waves with intensities x and y arrive at a point with a time delay of $\frac{3 T}{2}$ . Determine the resultant intensity at that point.

Solution:

Time difference $T.D. Δ x = \frac{T}{2 π} \times ϕ \Rightarrow \frac{3 T}{2} = \frac{T}{2 π} \times ϕ \Rightarrow ϕ = 3 π$ . This is the condition of constructive interference.

So resultant intensity,

$I_{R} = (I_{1} + I_{2})^{2} = (x + y)^{2} = x + y + 2 x y$

Illustration-7: If n identical waves, each having intensity $I_{0}$ , interfere at a point, what will be the maximum resultant intensity in the following cases?
(1) When the waves are coherent
(2) When the waves are incoherent

Solution:

$In case of interference of two waves: I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ In case of coherent interference ϕ is unchanging with time and so I will be maximum when cos ϕ = max = 1 (I_{m a x})_{co} = I_{1} + I_{2} + 2 I_{1} I_{2} = (I_{1} + I_{2})^{2} So for n n identical waves each of intensity I_{0} (I_{m a x})_{co} = (I_{0} + I_{0} + \dots)^{2} = (n I_{0})^{2} = n^{2} I_{0} In case of incoherent interference at a given point, ϕ varies randomly with time, so (cos ϕ)_{av} = 0 an d h e n ce (I_{R})_{inco} = I_{1} + I_{2} So, in case of n identical waves (I_{R})_{inco} = I_{0} + I_{0} + \dots = n I_{0}$

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Superposition of Waves

The superposition of waves is a fundamental idea in physics that explains what happens when two or more waves overlap. According to the superposition principle, the total effect at any point is simply the sum of the individual wave displacements. Depending on how the waves line up—whether they’re in phase or out of phase—this can result in constructive interference (where the waves add up) or destructive interference (where they cancel each other out). This principle applies to all types of waves, including mechanical waves like sound and electromagnetic waves like light. It's key to understanding many wave phenomena, such as interference and diffraction.

1.0Superposition Principle

When two or more waves overlap at a point in space, the resultant displacement at that point is equal to the algebraic sum of the displacements due to each individual wave.

$y = y_{1} \pm y_{2} \pm y_{3} + \dots$

2.0Types of Superposition

1.Constructive Superposition

Constructive superposition occurs when two waves meet in phase (i.e., their crests and troughs align), resulting in a larger amplitude. The displacements of the individual waves add up, producing a wave with maximum intensity.

2.Destructive Superposition

Destructive superposition happens when two waves meet out of phase (i.e., the crest of one wave aligns with the trough of another), causing their displacements to partially or completely cancel out, leading to a reduced or zero amplitude.

Relationship between Phase Difference and Path Difference

$Phase difference = \frac{2 π}{wavelength} \times path difference ϕ = \frac{2 π}{λ} \times Δ x$

Related Video:

3.0Superposition of Two Sinusoidal Waves

$Resultant wave at P from S_{1} an d S_{2} is given as y_{1} = a sin ω t y_{2} = b sin (ω t + ϕ) Applying principle of superposition at point P we get resultant wave Y = y_{1} + y_{2} = a sin ω t + b sin (ω t + ϕ) = a sin ω t + b [sin ω t cos ϕ + cos ω t sin ϕ] = sin ω t (a + b cos ϕ) + b cos ω t sin ϕ a + b cos ϕ = R cos θ \dots\dots . (2) b sin ϕ = R sin θ \dots\dots . (3) Y = R sin ω t cos θ + cos ω t sin θ = R sin (ω t + θ) Resultant wave is a simple harmonic wave having amplitude R and phase difference θ$

Amplitude of Resultant Wave

$Squaring and adding equations (2) and (3) (a + b cos ϕ)^{2} + b^{2} sin^{2} ϕ = R^{2} cos^{2} θ + R^{2} sin^{2} θ (a + b cos ϕ)^{2} + b^{2} sin^{2} ϕ = R^{2} (cos^{2} θ + sin^{2} θ) R^{2} = a^{2} + b^{2} + 2 ab cos ϕ R = a^{2} + b^{2} + 2 ab cos ϕ (I n t e n s i t y \propto (am pl i t u d e)^{2}) \Rightarrow I_{a} \propto a^{2}, I_{b} \propto b^{2} I_{R} = K R^{2} = K (a^{2} + b^{2} + 2 ab cos ϕ) I_{R} = I_{a} + I_{b} + 2 I_{a} I_{b} cos ϕ (This is the Resultant Intensity Equation) tan θ = \frac{b s i n ϕ}{a + b c o s ϕ}$

4.0Condition for Constructive and Destructive Interference

1.Constructive Interference

It occurs when the amplitude and the resultant wave is maximum.

$I_{R} = I_{max} if cos ϕ = maximum = + 1 ϕ = 0, 2 π, 4 π \Rightarrow ϕ = n (2 π) where n = 0, 1, 2, 3, \dots Δ x = \frac{λ}{2 π} \times 2 nπ = nλ I_{max} = (a + b)^{2} I_{max} = (I_{a} + I_{b})^{2} R_{max} = (a + b)$

2.Destructive Interference

$It occurs when the amplitude and the resultant wave is minimum. I_{R} = I_{min} if cos ϕ = minimum = + 1 ϕ = 0, 2 π, 4 π \Rightarrow ϕ = n (2 π) where n = 0, 1, 2, 3, \dots Δ x = \frac{λ}{2 π} \times 2 nπ = nλ I_{min} = (a + b)^{2} I_{min} = (I_{a} + I_{b})^{2} R_{min} = (a + b)$

5.0Ratio of Maximum And Minimum Intensity of Light

$\frac{I _{max}}{I _{min}} = (\frac{I _{a} + I _{b}}{I _{a} - I _{b}})^{2} = (\frac{a + b}{a - b})^{2} = (\frac{\frac{a}{b} + 1}{\frac{a}{b} - 1})^{2} = (\frac{r + 1}{r - 1})^{2} (since \frac{a}{b} = r = Amplitude Ratio)$

6.0Relation Between Intensity and Width of Source

$L e t w_{a} an d w_{b} be the width of narrow sources s_{1} an d s_{2} I_{a} \propto w_{a} \Rightarrow I_{a} \propto a^{2} I_{b} \propto w_{b} \Rightarrow I_{b} \propto b^{2} \frac{a ^{2}}{b ^{2}} = \frac{w _{a}}{w _{b}}$

Illustration-1: $S_{1} an d S_{2}$ are two sources of light which produce individual disturbance at point P given by $E_{1} = 3 sin ω t, E_{2} = 4 cos ω t$ . Assuming $E_{1} an d E_{2}$ to be along the same line, find the resultant after their superposition.

Solution:

$E = 3 sin ω t + 4 sin (ω t + \frac{π}{2}) A^{2} = 3^{2} + 4^{2} = 2 (3) (4) cos \frac{π}{2} = 5^{2} tan ϕ_{0} = \frac{4 s i n \frac{π}{2}}{3 + 4 c o s \frac{π}{2}} = \frac{4}{3} = 5 3^{\circ} E = 5 sin (ω t + 5 3^{\circ})$

Illustration-2:

$S_{1} an d S_{2}$ are two coherent sources of frequency 'f ' each.

[Given: $θ_{1} = θ_{2} = 0^{\circ} an d v_{sound} = 330 m/s .$ ]

(i) So that constructive interference at 'p'

(ii) So that destructive interference at 'p'

Solution:

$For constructive interference: k Δ x = 2 nπ \frac{2 π}{λ} \times 2 = 2 nπ λ = \frac{2}{n}, V = λ f \Rightarrow V = \frac{2}{n} f f = \frac{330}{2} \times n For destructive interference: k Δ x = (2 n + 1) π \frac{2 π}{λ} \times 2 = (2 n + 1) π \frac{1}{λ} = \frac{( 2 n + 1 )}{4}, f = \frac{V}{λ} f = \frac{330 \times ( 2 n + 1 )}{4}$

Illustration-3:Two waves are given by the equations

$y_{1} = 4 sin ω t an d y_{2} = 3 sin (ω t + \frac{π}{3})$ When these two waves interfere at a point, what is the approximate amplitude of the resulting wave?

Solution:

$A = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ A = (4)^{2} + (3)^{2} + 2 \cdot 4 \cdot 3 \cdot cos \frac{π}{3} = 37 \approx 6$

Illustration-4:Two waves with intensities 9I 9I and 4I 4I interfere at a point. If the path difference between them is11λ 11 \lambda , what will be the resultant intensity at that point?

Solution: Path difference, $Δ x = \frac{λ}{2 π} \times ϕ \Rightarrow \frac{2 π}{λ} \times 11 λ = 22 π,$ ,i.e. At the same point, the superposition of the waves results in constructive interference.

$I_{R} = (I_{1} + I_{2})^{2} = (9 I + 4 I)^{2} = 25 I$

Illustration-5:In interference if $\frac{I _{max}}{I _{min}} = \frac{144}{81}$ .Determine the ratio of amplitudes of the two waves involved in the interference.

Solution:

$\frac{a _{1}}{a _{2}} = (\frac{I _{max}}{I _{min}} + 1 / \frac{I _{max}}{I _{min}} - 1) = (\frac{\frac{144}{81} + 1}{\frac{144}{81} - 1}) = (\frac{\frac{12}{9} + 1}{\frac{12}{9} - 1}) = \frac{7}{1}$

Illustration-6:Two interfering waves with intensities x and y arrive at a point with a time delay of $\frac{3 T}{2}$ . Determine the resultant intensity at that point.

Solution:

Time difference $T.D. Δ x = \frac{T}{2 π} \times ϕ \Rightarrow \frac{3 T}{2} = \frac{T}{2 π} \times ϕ \Rightarrow ϕ = 3 π$ . This is the condition of constructive interference.

So resultant intensity,

$I_{R} = (I_{1} + I_{2})^{2} = (x + y)^{2} = x + y + 2 x y$

Illustration-7: If n identical waves, each having intensity $I_{0}$ , interfere at a point, what will be the maximum resultant intensity in the following cases?
(1) When the waves are coherent
(2) When the waves are incoherent

Solution:

$In case of interference of two waves: I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ In case of coherent interference ϕ is unchanging with time and so I will be maximum when cos ϕ = max = 1 (I_{m a x})_{co} = I_{1} + I_{2} + 2 I_{1} I_{2} = (I_{1} + I_{2})^{2} So for n n identical waves each of intensity I_{0} (I_{m a x})_{co} = (I_{0} + I_{0} + \dots)^{2} = (n I_{0})^{2} = n^{2} I_{0} In case of incoherent interference at a given point, ϕ varies randomly with time, so (cos ϕ)_{av} = 0 an d h e n ce (I_{R})_{inco} = I_{1} + I_{2} So, in case of n identical waves (I_{R})_{inco} = I_{0} + I_{0} + \dots = n I_{0}$

Frequently Asked Questions

What happens when two waves of equal amplitude and opposite phase meet?

Why is the superposition principle important in understanding interference?

What is meant by phase difference in wave superposition?

Is energy conserved in the superposition of waves?

Give a real-life example of wave superposition.

Join ALLEN!

Superposition of Waves

1.0Superposition Principle

2.0Types of Superposition

Relationship between Phase Difference and Path Difference

3.0Superposition of Two Sinusoidal Waves

4.0Condition for Constructive and Destructive Interference

5.0Ratio of Maximum And Minimum Intensity of Light

6.0Relation Between Intensity and Width of Source

Table of Contents

Frequently Asked Questions

What happens when two waves of equal amplitude and opposite phase meet?

Why is the superposition principle important in understanding interference?

What is meant by phase difference in wave superposition?

Is energy conserved in the superposition of waves?

Give a real-life example of wave superposition.

Join ALLEN!

Superposition of Waves

1.0Superposition Principle

2.0Types of Superposition

Relationship between Phase Difference and Path Difference

3.0Superposition of Two Sinusoidal Waves

4.0Condition for Constructive and Destructive Interference

5.0Ratio of Maximum And Minimum Intensity of Light

6.0Relation Between Intensity and Width of Source

Table of Contents