A triple slit experiment (with basic conditions similar to a YDSE) is performed as shown in the figure with monochromatic light of amplitude $a_{0}$ . Consider light waves from $S_{1}, S_{2}, S_{3}$ reaching a point P (shown by straight lines) on screen. $S_{2} M, S_{1} N$ are perpendicular to $S_{3} P, S_{2} P$ respectively. $S_{3} M \approx S_{2} N = δ$ . The graph between resultant amplitude of light (A) versus $δ$ is plotted. Choose the CORRECT graph between options (A) and (B) and appropriate relation between $A_{10}$ and $A_{20}$ between options (C) and (D).

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$\frac{A_{10} + A_{20}}{A_{10} - A_{20}} = 5$ , where $A_{10}$ and $A_{20}$ are the values of local maxima of A in the graph $(A_{10} > A_{20})$

$\frac{A_{10} + A_{20}}{A_{10} - A_{20}} = 2$ , where $A_{10}$ and $A_{20}$ are the values of local maxima of A in the graph $(A_{10} > A_{20})$

answer is A, C.

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Detailed Solution

$\begin{array}{l} ϕ = \frac{2 π}{λ} δ \\ L e t s a y y_{2} = a_{0} \sin (ω t - k x) \\ \Rightarrow y_{1} = a_{0} \sin (ω t - k x - ϕ) a n d y_{3} = a_{0} \sin (ω t - k x + ϕ) \\ Resultant Amplitude= A = | a_{0} + 2 a_{0} \cos ϕ | = a_{0} | (1 + 2 \cos ϕ) | \\ \Rightarrow A = a_{0} | (1 + 2 \cos (\frac{2 π δ}{λ})) | \\ F o r δ = 0, A_{10} = a_{0} | (1 + 2 \cos (0)) | = 3 a_{0} \\ F o r δ = \frac{λ}{3}, A = a_{0} | (1 + 2 \cos (\frac{2 π}{3})) | = 0 \\ F o r δ = \frac{λ}{2}, A_{20} = a_{0} | (1 + 2 \cos (π)) | = a_{0} \\ F o r δ = \frac{2 λ}{3}, A = a_{0} | (1 + 2 \cos (\frac{4 π}{3})) | = 0 \\ ∴ \frac{A_{10} + A_{20}}{A_{10} - A_{20}} = \frac{3 a_{0} + a_{0}}{3 a_{0} - a_{0}} = 2 \end{array}$

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