In a Young's double slit experiment, the two slits act as coherent sources of waves of equal amplitu

Q: Why is the phase difference zero at the middle point?

At the exact middle point of the screen, the distance from both slits is equal. Since the path difference is zero, the phase difference is also zero.

Q: What exactly are incoherent sources?

Incoherent sources emit light waves that do not maintain a constant phase difference over time. The phase relationship changes randomly and rapidly.

Q: Why does the interference term $2\sqrt{I_1 I_2}\cos\phi$ disappear for incoherent sources?

Because the phase difference $\phi$ fluctuates randomly, the cosine of $\phi$ takes on all values between -1 and 1. Over any macroscopic time interval, its average value is exactly zero.

Q: Does the intensity of incoherent sources depend on the position on the screen?

No, since the interference term averages to zero everywhere, the resultant intensity is simply the sum of individual intensities ($I_1 + I_2$) uniformly across the entire screen.

Q: If the slits had different widths, would the ratio still be 2?

Not necessarily. If the intensities were $I_a$ and $I_b$, $I_{coherent} = (\sqrt{I_a} + \sqrt{I_b})^2$ and $I_{incoherent} = I_a + I_b$. The ratio would depend on the exact values of $I_a$ and $I_b$.

Question

In a Young's double slit experiment, the two slits act as coherent sources of waves of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I_1$ and in the second case is $I_2$, then the ratio $\frac{I_1}{I_2}$ is

Options:

(A) 4

(B) 2

(C) 1

(D) 0.5

Sigma Percentile · Accepted Answer

This problem beautifully contrasts the behavior of coherent and incoherent light sources in a Young's Double Slit Experiment (YDSE) setup. It tests our fundamental understanding of how wave intensities combine under different phase relationships.

### Analyzing the Setup

We are presented with two distinct scenarios using the same physical arrangement. In the first case, the two slits act as **coherent sources**. This is the standard assumption in YDSE, meaning the light waves emitted from the slits maintain a constant phase difference over time.

In the second case, the slits act as **incoherent sources**. This means the phase difference between the waves emitted from the two slits fluctuates randomly and rapidly. We are asked to compare the resultant intensity at the exact middle point of the screen for both cases.

Let the intensity of the light from each individual slit be $I_0$. Since the amplitude $A$ is the same for both slits, their individual intensities are equal ($I_a = I_b = I_0$).

### The Master Equation for Interference

When two waves interfere, the resultant intensity $I$ at any point is given by the general superposition formula:

$$I = I_a + I_b + 2\sqrt{I_a I_b}\cos\phi$$

Here, $\phi$ is the phase difference between the two waves arriving at that specific point. The term $2\sqrt{I_a I_b}\cos\phi$ is the crucial **interference term** that dictates whether the waves construct or destruct.

### Case 1: Coherent Sources ($I_1$)

For coherent sources, the phase difference $\phi$ is well-defined and constant. We are looking at the middle point of the screen. At this central point, the distance traveled by the light from both slits is exactly the same.

Because the path difference $\Delta x$ is zero, the phase difference $\phi$ is also zero ($\phi = 0^\circ$). Substituting this into our master equation:

$$I_1 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\cos(0^\circ)$$

Since $\cos(0^\circ) = 1$, the equation simplifies to:

$$I_1 = 2I_0 + 2I_0 = 4I_0$$

This is the condition for perfect constructive interference, yielding the maximum possible intensity.

### Case 2: Incoherent Sources ($I_2$)

Now, let's consider the incoherent sources. Because the sources are incoherent, the phase difference $\phi$ does not stay constant; it varies randomly and incredibly fast.

When we measure intensity, we are actually measuring the time-averaged energy flow. Because $\phi$ takes on all possible random values between $0$ and $2\pi$, the average value of $\cos\phi$ over any observable time interval is exactly zero ($\langle \cos\phi \rangle = 0$).

Consequently, the entire interference term vanishes! The resultant intensity is simply the algebraic sum of the individual intensities:

$$I_2 = I_0 + I_0 = 2I_0$$

Notice how the lack of coherence means the energies simply add up, without forming any spatial interference pattern (no bright or dark fringes).

### Final Calculation

We are asked to find the ratio of the intensity in the first case to the intensity in the second case, $\frac{I_1}{I_2}$.

$$\frac{I_1}{I_2} = \frac{4I_0}{2I_0}$$

The $I_0$ terms cancel out perfectly, leaving us with:

$$\frac{I_1}{I_2} = 2$$

This elegant result shows that at the central maximum, coherent sources produce twice the intensity compared to incoherent sources of the same individual strength.

Select Answer:

Visualized Solution

$Understanding the Setup$

$Intensity for Coherent Sources$

$Calculating I_{1}$

$Intensity for Incoherent Sources$

$Calculating I_{2}$

$Finding the Ratio$

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

Analyzing the Setup

The Master Equation for Interference

Case 1

Case 2

Final Calculation

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