Question

Two points of monochromatic and coherent sources of light of wavelength \( \lambda \) each, are placed as shown in figure. The initial phase difference between the sources is zero, (\( D \gg d \)). Mark the correct statement(s).

• A. If \( d = \frac{7\lambda}{2} \), O will be a minima
• B. If \( d = \lambda \), only one maxima can be observed on the screen
• C. If \( d = 4.8\lambda \), then total 5 minima would be there on the screen
• D. If \( d = \lambda \), the intensity at O would be minimum

Hint:

For circular interference fringes produced by coaxial sources, the path difference decreases as you move away from the center. To find the number of minima, simply find the number of half-integral wavelengths that are less than or equal to the separation distance \( d \). For example, if \( d = 4.8\lambda \), the half-integers are \( 0.5, 1.5, 2.5, 3.5, 4.5 \), immediately giving 5 minima.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The diagram shows two coherent sources \( S_1 \) and \( S_2 \) aligned along the axis perpendicular to the screen. This coaxial arrangement produces circular interference fringes centered at point \( O \) on the screen. We need to evaluate the positions of constructive (maxima) and destructive (minima) interference.

Step 2: Key Formula or Approach:
1. At any point \( P \) on the screen making an angle \( \theta \) with the axis of the sources:
The path difference between the waves is:
\[ \Delta x \approx d \cos\theta \]
2. At the central point \( O \) (\( \theta = 0^\circ \)), the path difference is maximum:
\[ \Delta x(O) = d \]
3. As we move radially outwards from \( O \) on the screen, \( \theta \) increases from \( 0^\circ \), so \( \cos\theta \) decreases. Therefore, the path difference decreases from \( d \) at \( O \) to \( 0 \) at infinity:
\[ 0 < \Delta x \le d \]
4. Condition for a minimum (destructive interference):
\[ \Delta x = (m - 0.5)\lambda \quad \text{for } m = 1, 2, 3, \dots \]
5. Condition for a maximum (constructive interference):
\[ \Delta x = m\lambda \quad \text{for } m = 0, 1, 2, \dots \]

Step 3: Detailed Explanation:
Let's analyze each statement:
- Option (A): If \( d = \frac{7\lambda}{2} = 3.5\lambda \).
The path difference at \( O \) is \( \Delta x(O) = 3.5\lambda \). Since this is an odd multiple of \( \frac{\lambda}{2} \), destructive interference occurs at \( O \). Thus, \( O \) is a minimum. Statement (A) is correct.
- Option (B): If \( d = \lambda \).
The path difference on the screen lies in the range \( 0 < \Delta x \le \lambda \).
Within this range, the only constructive interference value is \( \Delta x = \lambda \), which occurs uniquely at point \( O \). (The value \( \Delta x = 0 \) is at infinity). Thus, only one maximum (at \( O \)) can be observed. Statement (B) is correct.
- Option (C): If \( d = 4.8\lambda \).
The path difference on the screen is in the range \( 0 < \Delta x \le 4.8\lambda \).
The possible values for destructive interference (minima) within this interval are:
\[ \Delta x \in \{ 0.5\lambda, 1.5\lambda, 2.5\lambda, 3.5\lambda, 4.5\lambda \} \]
Each of these 5 distinct path differences corresponds to a concentric circular minimum ring on the screen. Therefore, a total of 5 circular minima are observed. Statement (C) is correct.
- Option (D): If \( d = \lambda \).
The path difference at \( O \) is \( \Delta x(O) = d = \lambda \), which is an integer multiple of \( \lambda \). This leads to constructive interference, so the intensity at \( O \) is maximum. Statement (D) is incorrect.

Step 4: Final Answer:
The correct statements are (A), (B), and (C).