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).
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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.
If \( d = \frac{7\lambda}{2} \), O will be a minima
If \( d = \lambda \), only one maxima can be observed on the screen
If \( d = 4.8\lambda \), then total 5 minima would be there on the screen
If \( d = \lambda \), the intensity at O would be minimum
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The Correct Option isA
Solution and Explanation
Step 1: Understanding the Concept:
Interference of light from two sources depends on the path difference (\(\Delta x\)) at a point on the screen.
For maxima: \(\Delta x = n\lambda\).
For minima: \(\Delta x = (2n-1)\lambda/2\).
In this specific geometry, the sources are on the same line as the point \(O\). The path difference at \(O\) is exactly \(d\). Step 2: Key Formula or Approach:
1. Path difference at \(O\): \(\Delta x_O = d\).
2. Maxima condition: \(d = n\lambda\).
3. Minima condition: \(d = (n+1/2)\lambda\). Step 3: Detailed Explanation:
Step A: Checking (A). If \(d = 7\lambda/2 = 3.5\lambda\).
At point \(O\), the path difference is \(3.5\lambda\), which corresponds to destructive interference (minima). Correct.
Step B: Checking (B) and (D). If \(d = \lambda\).
At \(O\), \(\Delta x = \lambda\), so \(O\) is a maxima. Correct.
For any other point on the screen, the path difference \(\Delta x\) will be less than \(d\). Since \(d = \lambda\), then \(\Delta x<\lambda\).
No other maxima (integral multiple of \(\lambda\)) can exist because \(\Delta x\) decreases from \(\lambda\) to 0 as we move away from \(O\).
So, only one maxima exists (at \(O\)). (B) is Correct.
Step C: Checking (C). If \(d = 4.8\lambda\).
Path difference at \(O\) is \(4.8\lambda\). As we move along the screen, \(\Delta x\) decreases from \(4.8\lambda\) to 0.
Minima occur when \(\Delta x\) is \(4.5\lambda, 3.5\lambda, 2.5\lambda, 1.5\lambda\), and \(0.5\lambda\).
Counting these: there are exactly 5 minima. Correct. Step 4: Final Answer:
By evaluating the path difference limits in the axial geometry, the interference conditions for various separations \(d\) are verified, confirming options (A), (B), and (C).
