Intensity in Interference: Why I = I₁ + I₂ + 2√(I₁I₂) cos φ is Your Best Friend
Master the most important formula in wave optics that unlocks interference patterns, fringe visibility, and intensity distribution.
Why This Formula is a Game-Changer
The interference intensity formula $I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos\phi$ is not just another equation—it's the fundamental bridge between wave nature and observable interference patterns. Understanding this formula means understanding interference itself.
🎯 JEE Relevance
This formula appears in 2-3 questions per JEE Physics paper, often as the key to solving complex interference problems involving Young's double slit, thin films, and other wave phenomena.
🚀 Quick Navigation
1. Deriving the Master Formula
Starting from Wave Superposition
Step 1: Wave Equations
Consider two coherent waves with same frequency ω:
where $\phi$ is the phase difference between the waves.
Step 2: Resultant Wave
By principle of superposition:
$$y = y_1 + y_2 = a_1 \sin(\omega t) + a_2 \sin(\omega t + \phi)$$
Using trigonometric identity:
$$y = R \sin(\omega t + \theta)$$
where the resultant amplitude $R$ is:
$$R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos\phi}$$
Step 3: Intensity Relation
Intensity $I$ is proportional to square of amplitude:
$$I \propto R^2 = a_1^2 + a_2^2 + 2a_1a_2 \cos\phi$$
Since $I_1 \propto a_1^2$ and $I_2 \propto a_2^2$, we get:
$$I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos\phi$$
💡 Key Insight
The term $2\sqrt{I_1I_2} \cos\phi$ is the interference term. It's what makes interference patterns possible and contains all the information about constructive and destructive interference.
2. Special Case: Equal Intensities (I₁ = I₂ = I₀)
Simplified Formula
When both waves have equal intensity $I_1 = I_2 = I_0$, the formula simplifies beautifully:
$$I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos\phi$$
$$I = 2I_0 + 2I_0 \cos\phi$$
$$I = 2I_0 (1 + \cos\phi)$$
Using identity $1 + \cos\phi = 2\cos^2(\frac{\phi}{2})$:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
Extreme Cases Analysis
| Condition | Phase Difference | Intensity | Interpretation |
|---|---|---|---|
| Constructive | $\phi = 2n\pi$ | $I_{max} = 4I_0$ | Bright fringe |
| Destructive | $\phi = (2n+1)\pi$ | $I_{min} = 0$ | Dark fringe |
| General | Any $\phi$ | $I = 4I_0 \cos^2(\frac{\phi}{2})$ | Intermediate intensity |
🎯 JEE Application: Young's Double Slit
In Young's experiment with identical slits, $\phi = \frac{2\pi}{\lambda} \Delta x$, where $\Delta x$ is path difference. This gives the familiar interference pattern with:
- Bright fringes when $\Delta x = n\lambda$
- Dark fringes when $\Delta x = (n+\frac{1}{2})\lambda$
3. General Case: Unequal Intensities
Complete Analysis
Using the general formula $I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos\phi$:
Maximum Intensity: When $\cos\phi = +1$
$$I_{max} = I_1 + I_2 + 2\sqrt{I_1I_2} = (\sqrt{I_1} + \sqrt{I_2})^2$$
Minimum Intensity: When $\cos\phi = -1$
$$I_{min} = I_1 + I_2 - 2\sqrt{I_1I_2} = (\sqrt{I_1} - \sqrt{I_2})^2$$
Fringe Visibility (Contrast)
Definition
Fringe visibility measures contrast between bright and dark fringes:
$$V = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}$$
Substituting our expressions:
$$V = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}$$
$$V = \frac{4\sqrt{I_1I_2}}{2(I_1 + I_2)} = \frac{2\sqrt{I_1I_2}}{I_1 + I_2}$$
$$V = \frac{2\sqrt{I_1I_2}}{I_1 + I_2}$$
Special Cases:
- Equal intensities ($I_1 = I_2$): $V = 1$ (Perfect visibility)
- Very unequal intensities ($I_1 \gg I_2$): $V \approx 0$ (No visible pattern)
- General case: $0 < V < 1$
4. Coherent vs. Incoherent Superposition
The Fundamental Difference
| Aspect | Coherent Sources | Incoherent Sources |
|---|---|---|
| Phase Relationship | Constant phase difference | Random phase difference |
| Interference Term | $2\sqrt{I_1I_2} \cos\phi$ (fixed) | $\langle 2\sqrt{I_1I_2} \cos\phi \rangle = 0$ |
| Resultant Intensity | $I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos\phi$ | $I = I_1 + I_2$ |
| Pattern Formation | Clear interference pattern | No interference pattern |
| Examples | Laser light, same source divided | Two separate bulbs, sunlight |
💡 Why Cosφ Averages to Zero for Incoherent Sources
For incoherent sources, the phase difference $\phi$ changes randomly with time. Therefore:
$$\langle \cos\phi \rangle = \frac{1}{2\pi} \int_0^{2\pi} \cos\phi d\phi = 0$$
This makes the interference term vanish, leaving simple addition of intensities.
5. JEE Practice Problems
Apply Your Understanding
Problem 1: Two coherent sources have intensities I and 4I. Find the ratio of maximum to minimum intensity in the interference pattern.
Problem 2: In Young's double slit experiment, the intensity at a point where path difference is λ/6 is I. If I₀ is the intensity of each wave, find I/I₀.
Problem 3: Two waves with intensities I and 9I interfere. Calculate the fringe visibility.
Problem 4: Prove that for two incoherent sources, the resultant intensity is simply I₁ + I₂.
📋 Formula Quick Reference
General Formulas
- • Resultant Intensity: $I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos\phi$
- • Max Intensity: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$
- • Min Intensity: $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$
- • Fringe Visibility: $V = \frac{2\sqrt{I_1I_2}}{I_1 + I_2}$
Equal Intensity Case (I₁ = I₂ = I₀)
- • General: $I = 4I_0 \cos^2(\frac{\phi}{2})$
- • Bright Fringe: $I_{max} = 4I_0$
- • Dark Fringe: $I_{min} = 0$
- • Visibility: $V = 1$
🎯 JEE Exam Strategy
Recognize when to use the general formula vs. equal intensity simplification.
Remember $\phi = \frac{2\pi}{\lambda} \Delta x$ for path-related problems.
Always check if $I_{max} \geq I_{min}$ and $0 \leq V \leq 1$.
Clearly write the formula substitution for partial credit.
Mastered Interference Intensity?
Continue your wave optics journey with diffraction and polarization concepts