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).
Let's analyze the problem of interference from two coherent sources, \(S_1\) and \(S_2\), placed at a distance \(d\) apart with the screen located at a distance \(D\) from the sources (where \(D \gg d\)). We will use the condition for constructive and destructive interference to find the correct option.
For a point on the screen to be a point of minima (destructive interference), the path difference between the waves from \(S_1\) and \(S_2\) must be:
\(\Delta x = (n + \frac{1}{2})\lambda\), where \(n\) is an integer.
The path difference at point \(O\) is equal to distance \(d\) because \(D \gg d\). Hence, \(\Delta x = d\).
To have a minimum at \(O\):
\(d = (n + \frac{1}{2})\lambda\).
For \(d = \frac{7\lambda}{2}\):
\(\frac{7\lambda}{2} = (n + \frac{1}{2})\lambda \Rightarrow n = 3\).
This confirms there will be a minimum at \(O\).
Let's consider other options:
- If \(d = \lambda\):
- For \(\lambda = 2(n + \frac{1}{2})\lambda\), which is not possible for any integer \(n\), O cannot have constructive interference or a single maxima.
- If \(d = 4.8\lambda\):
- The number of minima would be less than 5, as \(4.8\) does not satisfy the total number of half wavelengths for 5 minima correctly.
Thus, the correct statement is "\(d = \frac{7\lambda}{2}\), O will be a minima" because the condition for destructive interference is satisfied.
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