In the given figure, P and Q are two equally intense coherent sources emitting radiation of waveleng

Q: Why is the phase difference due to path negative for point A?

Because we defined our total phase difference as $\phi_P - \phi_Q$. Since the wave from $P$ travels a longer distance to reach $A$, it accumulates a larger negative phase term ($-kx$). Therefore, the contribution of the path to $\phi_P - \phi_Q$ is negative.

Q: Why is the intensity at B not maximum even though it's on the perpendicular bisector?

In a standard Young's Double Slit Experiment, the sources are in phase, so the perpendicular bisector (central maximum) has zero phase difference and maximum intensity. Here, the sources have an intrinsic initial phase difference of $90^\circ$. So even with zero path difference, they arrive $90^\circ$ out of phase.

Q: How do we know the path difference at A is exactly 5m?

The problem states that $A$ is on the line passing through $P$ and $Q$. The distance from $P$ to $A$ is $PQ + QA$, and the distance from $Q$ to $A$ is just $QA$. Therefore, the difference $x_{PA} - x_{QA}$ is exactly the distance $PQ$, which is given as $5\text{ m}$.

Q: What if Q was ahead of P initially?

If $Q$ was ahead of $P$ by $90^\circ$, the initial phase difference $\phi_{0P} - \phi_{0Q}$ would be $-\frac{\pi}{2}$. This would flip the intensities at $A$ and $C$, making $I_A = 0$ and $I_C = 4I_0$.

Q: Why do we add the initial phase difference and path phase difference?

The phase of a wave at any point is given by $\phi = \omega t - kx + \phi_0$. When you find the difference between two waves, $\phi_1 - \phi_2$, you naturally subtract their path terms ($-k(x_1 - x_2)$) and add their initial phase terms ($\phi_{01} - \phi_{02}$). They linearly superimpose.

Question

In the given figure, $P$ and $Q$ are two equally intense coherent sources emitting radiation of wavelength $20	ext{ m}$. The separation between $P$ and $Q$ is $5	ext{ m}$ and the phase of $P$ is ahead of that of $Q$ by $90^\circ$. $A, B$ and $C$ are three distinct points of observation, each equidistant from the mid-point of $PQ$. The intensities of radiation at $A, B$ and $C$ will be in the ratio

Options:

(A) $0 : 1 : 4$

(B) $2 : 1 : 0$

(C) $0 : 1 : 2$

(D) $4 : 1 : 0$

Sigma Percentile · Accepted Answer

### The Dance of Phase and Path Difference

Interference is not just about waves meeting; it is about *how* they meet. The resultant intensity at any point in space depends entirely on the total phase difference between the arriving waves. In this beautiful problem, we explore a scenario where the sources themselves are not perfectly in sync. Source $P$ is given a head start—an initial phase lead of $90^\circ$ or $\frac{\pi}{2}$ over source $Q$.

### The Master Equation

To find the intensity at any point, we rely on the fundamental interference equation:

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

The critical term here is $\Delta \phi$, the total phase difference. It is the sum of two distinct components: the phase difference arising from the path difference, and the intrinsic initial phase difference between the sources. Mathematically, if we define the phase of the waves as $\phi_P$ and $\phi_Q$, the phase difference is:

$$\Delta \phi = \phi_P - \phi_Q = \frac{2\pi}{\lambda}(x_Q - x_P) + (\phi_{0P} - \phi_{0Q})$$

Here, $\phi_{0P} - \phi_{0Q} = \frac{\pi}{2}$ is our given initial condition.

### Point A: The Perfect Compensation

Let's look at point $A$. It lies on the axis, closer to $Q$. The wave from $P$ must travel an extra $5	ext{ m}$ to reach $A$ compared to the wave from $Q$. This extra distance creates a phase lag for $P$. Using our formula, the phase difference due to the path is:

$$\Delta \phi_{path} = \frac{2\pi}{20}(-5) = -\frac{\pi}{2}$$

Fascinatingly, this path lag of $-\frac{\pi}{2}$ perfectly cancels out $P$'s initial lead of $+\frac{\pi}{2}$. The total phase difference at $A$ becomes exactly $0$. The waves arrive perfectly in phase, resulting in constructive interference!

$$I_A = I_0 + I_0 + 2I_0\cos(0) = 4I_0$$

### Point B: The Pure Initial Phase

Point $B$ lies on the perpendicular bisector of the line joining $P$ and $Q$. By symmetry, the path lengths $x_{PB}$ and $x_{QB}$ are identical. The path difference is zero, meaning the geometry contributes nothing to the phase difference.

The only surviving term is the initial phase difference of $\frac{\pi}{2}$. Plugging this into our intensity equation:

$$I_B = I_0 + I_0 + 2I_0\cos\left(\frac{\pi}{2}ight) = 2I_0$$

### Point C: The Double Lead

Finally, we examine point $C$, which is closer to $P$. Here, the wave from $Q$ must travel the extra $5	ext{ m}$. This means $P$ effectively gains an additional phase lead due to the shorter path it takes. The phase difference due to the path is:

$$\Delta \phi_{path} = \frac{2\pi}{20}(5) = +\frac{\pi}{2}$$

When we add this path lead to $P$'s initial lead, the total phase difference becomes $\frac{\pi}{2} + \frac{\pi}{2} = \pi$. A phase difference of $\pi$ means the crest of one wave meets the trough of the other. This is perfect destructive interference.

$$I_C = I_0 + I_0 + 2I_0\cos(\pi) = 0$$

### The Final Ratio

We have successfully mapped the intensities at all three points. The ratio of intensities $I_A : I_B : I_C$ is simply $4I_0 : 2I_0 : 0$, which simplifies beautifully to **$2 : 1 : 0$**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Master Interference Equation

Analyzing Point A

Intensity at Point A

Analyzing Point B

Analyzing Point C

Final Ratio

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

The Dance of Phase and Path Difference

The Master Equation

Point A

Point B

Point C

The Final Ratio

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