Interference Conditions for Light Waves

1. The Principle of Superposition: A Visual Start

When two or more waves overlap in space, the resulting disturbance at any point is the algebraic sum of the individual disturbances. This is the principle of superposition. Let's visualize what this means.

As you can see, the outcome depends entirely on how the waves align:

• Constructive Interference: If the crest of one wave meets the crest of another, they reinforce each other, creating a wave with a larger amplitude.
• Destructive Interference: If the crest of one wave meets the trough of another, they cancel each other out, resulting in a wave with a smaller amplitude, or even zero.

Before we dive into the mathematics, let's build a strong conceptual intuition for this. The following video from Khan Academy brilliantly explains interference using clear animations.

The video introduces a critical concept: path difference ($\Delta x$). If two waves start in phase from two different sources but travel different distances to reach a point, they may arrive out of phase. The difference in the distance they travel is the path difference.

2. The Mathematical Derivation of Interference Conditions

Now that we have the conceptual picture, let's satisfy your goal for a "vigorous" mathematical treatment. We will derive these conditions from first principles.

Consider two coherent light waves, represented by the displacements $y_1$ and $y_2$, arriving at a point. Since the sources are coherent, they have the same angular frequency $\omega$ and a constant phase difference $\varphi$. We can write their equations as:
$y_1=a_1\sin (\omega t)$
$y_2=a_2\sin (\omega t+\varphi )$
Here, $a_1$ and $a_2$ are the amplitudes of the individual waves.

According to the superposition principle, the resultant displacement $y$ is:
$y=y_1+y_2=a_1\sin (\omega t)+a_2\sin (\omega t+\varphi )$

Our goal is to find the amplitude of this new resultant wave, $y$. To do that, we need to combine these two sine functions.

The following video provides a clear, step-by-step derivation. Follow along as the presenter combines the wave equations and derives the formula for the resultant amplitude.

After applying the trigonometric identity for $\sin (A+B)$ and some algebraic rearrangement, we arrive at the equation for the resultant wave, which has the form $y=R\sin (\omega t+\theta )$, where $R$ is the resultant amplitude. The derivation shows that the square of the resultant amplitude is:

$R^2=a_1^2+a_2^2+2a_1{a}_2\cos \varphi$

This is a monumentally important formula for interference. It shows that the resultant amplitude $R$ depends directly on the phase difference $\varphi$ between the waves.

Condition for Constructive Interference (Maximum Amplitude)

The amplitude $R$ will be maximum when $\cos \varphi$ has its maximum value, which is +1.
$\cos \varphi =+1$
This occurs when the phase difference $\varphi$ is an integer multiple of $2\pi$:
$\varphi =2n\pi ,\text{where }n=0,1,2,\dots$
Substituting $\cos \varphi =1$ into the amplitude formula gives the maximum possible amplitude:
$R_{\text{max}}^2=a_1^2+a_2^2+2a_1{a}_2=(a_1+a_2{)}^2$
$R_{\text{max}}=a_1+a_2$

Condition for Destructive Interference (Minimum Amplitude)

The amplitude $R$ will be minimum when $\cos \varphi$ has its minimum value, which is -1.
$\cos \varphi =-1$
This occurs when the phase difference $\varphi$ is an odd integer multiple of $\pi$:
$\varphi =(2n+1)\pi ,\text{where }n=0,1,2,\dots$
Substituting $\cos \varphi =-1$ gives the minimum amplitude:
$R_{\text{min}}^2=a_1^2+a_2^2-2a_1{a}_2=(a_1-a_2{)}^2$
$R_{\text{min}}=|a_1-a_2|$
The absolute value is used because amplitude cannot be negative.

You can watch the derivations for these specific conditions in the next segment of the same video.

3. From Phase Difference to Path Difference

In experiments, it's often easier to measure lengths than phases. So, we must convert our phase difference conditions into path difference ($\Delta x$) conditions.

The relationship between phase difference ($\varphi$) and path difference ($\Delta x$) is based on the fact that a path difference of one full wavelength ($\lambda$) corresponds to a phase difference of $2\pi$ radians.

$\frac{\varphi }{2\pi }=\frac{\Delta x}{\lambda }⟹\varphi =\frac{2\pi }{\lambda }\Delta x$

Let's use this to translate our conditions:

1. For Constructive Interference:
   $\varphi =2n\pi$
   $\frac{2\pi }{\lambda }\Delta x=2n\pi$
   $\Delta x=n\lambda ,(n=0,1,2,\dots )$
   This confirms our conceptual finding: constructive interference occurs when the path difference is an integer multiple of the wavelength.

2. For Destructive Interference:
   $\varphi =(2n+1)\pi$
   $\frac{2\pi }{\lambda }\Delta x=(2n+1)\pi$
   $\Delta x=(n+\frac{1}{2})\lambda ,(n=0,1,2,\dots )$
   This also confirms our conceptual finding: destructive interference occurs when the path difference is a half-integer multiple of the wavelength.

The final part of the video walks through this exact conversion.

4. A Note on Intensity

For many NEET problems, you'll work with intensity ($I$) rather than amplitude ($A$). The key relationship is that intensity is proportional to the square of the amplitude ($I\propto A^2$).

Using this, our resultant amplitude equation $R^2=a_1^2+a_2^2+2a_1{a}_2\cos \varphi$ can be rewritten for intensity:
$I_{\text{res}}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
From this, we get the maximum and minimum intensities:

• Maximum Intensity (Constructive): $I_{\text{max}}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$
• Minimum Intensity (Destructive): $I_{\text{min}}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

This ratio $I_{\text{max}}/I_{\text{min}}$ is a very common subject of numerical problems.

Conclusion

Today, we built the mathematical engine that drives wave interference. By applying the principle of superposition to two coherent waves, we derived the all-important conditions for when they will reinforce or cancel each other.

Key Takeaways:

• The resultant amplitude $R$ of two interfering waves depends on their individual amplitudes ($a_1,a_2$) and their phase difference ($\varphi$): $R^2=a_1^2+a_2^2+2a_1{a}_2\cos \varphi$.
• The resultant intensity $I_{\text{res}}$ follows a similar formula: $I_{\text{res}}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$.
• Constructive Interference (bright fringe) occurs when:
  • Phase difference $\varphi =2n\pi$
  • Path difference $\Delta x=n\lambda$
• Destructive Interference (dark fringe) occurs when:
  • Phase difference $\varphi =(2n+1)\pi$
  • Path difference $\Delta x=(n+\frac{1}{2})\lambda$

Preview of the Next Lesson:
We are now perfectly equipped to analyze the most famous demonstration of light's wave nature: Young's Double-Slit Experiment (YDSE). In the next lesson, we will apply today's formulas to the specific geometry of the double-slit setup to derive expressions for the location, spacing, and width of the interference fringes seen on a screen.