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).

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{A_{10} + A_{20}}{A_{10} - A_{20}} = 5$ , 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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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