The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern

Q: Why is the maximum intensity $4I_0$ and not $2I_0$? Shouldn't energies just add up?

Energy is indeed conserved, but it is redistributed! In an interference pattern, the energy from the dark fringes (where intensity is 0) is shifted to the bright fringes. The *average* intensity across the whole screen is still $2I_0$, but the peaks reach $4I_0$.

Q: What exactly does 'coherent sources' mean?

Coherent sources emit light waves of the exact same frequency and maintain a constant phase difference over time. This is a strict requirement to produce a stable, observable interference pattern. If the phase difference fluctuated randomly, the pattern would wash out.

Q: What if the two sources had different intensities?

If $I_1 \neq I_2$, the minimum intensity would not be zero. The dark fringes would not be perfectly dark, and the contrast (or visibility) of the interference pattern would decrease.

Q: How does the formula $I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ come from the general intensity equation?

The general equation is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. For a maximum, $\cos \phi = 1$. This gives $I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2}$, which is the algebraic expansion of the perfect square $(\sqrt{I_1} + \sqrt{I_2})^2$.

Q: Can we use this formula for incoherent sources like two separate light bulbs?

No. For incoherent sources, the phase difference $\phi$ changes randomly and rapidly millions of times per second. The average value of $\cos \phi$ becomes zero. So, the resultant intensity is simply the sum of individual intensities: $I = I_1 + I_2$, everywhere on the screen, with no fringes.

Question

The light waves from two coherent sources have same intensity $I_1 = I_2 = I_0$. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima ?

Options:

(A) $I_0$

(B) $2I_0$

(C) $5I_0$

(D) $4I_0$

Sigma Percentile · Accepted Answer

### The Magic of Interference: When 1 + 1 = 4

Imagine you are standing by a calm pond, and you drop two pebbles into the water at the exact same time. The ripples spread out, and where they meet, something fascinating happens. In some places, the waves build on each other to create a giant ripple, and in other places, they perfectly cancel each other out, leaving the water completely flat. This beautiful phenomenon is called **interference**, and it happens with light waves just as it does with water waves.

In our problem, we are dealing with two **coherent sources** of light. Coherent means these sources are perfectly in sync—they have the same frequency and maintain a constant phase difference. When their light waves overlap on a screen, they create a pattern of bright and dark bands known as fringes.

### The Master Equation of Intensity

When two light waves interfere, their energies don't just simply add up like $1 + 1 = 2$. Instead, the resultant intensity $I$ at any point on the screen is governed by a master equation:

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

Here, $I_1$ and $I_2$ are the intensities of the individual waves, and $\phi$ is the phase difference between them at that specific point on the screen. The term $2\sqrt{I_1 I_2} \cos \phi$ is the "interference term," and it's responsible for all the magic.

### Constructive Interference: The Brightest Spots

To find the maximum intensity (the brightest spots on the screen), we need the interference term to be as large as possible. This happens when $\cos \phi = 1$, which corresponds to **constructive interference** (crests meeting crests).

Substituting $\cos \phi = 1$ into our master equation, we get:

$$I_{	ext{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

If you look closely, this is a perfect square! We can rewrite it elegantly as:

$$I_{	ext{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$$

### Solving the Problem

The question gives us a beautiful symmetry: both sources have the exact same intensity, so $I_1 = I_2 = I_0$. Let's substitute this into our simplified formula for maximum intensity:

$$I_{	ext{max}} = (\sqrt{I_0} + \sqrt{I_0})^2$$

Adding the terms inside the bracket gives us $2\sqrt{I_0}$. Now, we just need to square this entire term:

$$I_{	ext{max}} = (2\sqrt{I_0})^2 = 4I_0$$

And there we have it! The maximum intensity is **$4I_0$**.

### The Conservation of Energy Paradox

You might be wondering: "Wait a minute! If I have two sources of intensity $I_0$, shouldn't the total intensity be $2I_0$? How can we get $4I_0$? Is energy not conserved?"

This is a brilliant question. Energy is absolutely conserved! The trick is that interference doesn't create new energy; it merely **redistributes** it. While the bright spots (maxima) get a massive boost up to $4I_0$, the dark spots (minima) drop all the way down to zero.

Let's verify this with the formula for minimum intensity (where $\cos \phi = -1$):

$$I_{	ext{min}} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{I_0} - \sqrt{I_0})^2 = 0$$

The energy that "went missing" from the dark fringes has been perfectly relocated to the bright fringes. If you were to average the intensity across the entire screen, it would still perfectly equal $2I_0$. Nature's accounting is always flawless!

Animated Solution for Physics - Optics: The light waves from two coherent sources have same intensity $I_{1} = I_{2} = I_{0}$ . In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima ?

Select Answer:

Visualized Solution

Interference Pattern

Maximum Intensity Formula

Substituting Values

Simplifying the Expression

Final Calculation

Checking the Minima

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Magic of Interference

The Master Equation of Intensity

Constructive Interference

Solving the Problem

The Conservation of Energy Paradox

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