In Young's double slit experiment, following figure shows that \(Q\) is the position of second bright fringe on the right side of point \(O\). \(P\) is the eleventh bright fringe on the other side measured from point \(Q\). If the wavelength of light used is \(6000\) Å, then what will be the value of \(S_1B\)?
Show Hint
Find the order of fringe P: it is 11 fringes from Q (order 2) on the other side, so order 9. Path difference is order times wavelength.
Step 1: Understand the figure:
The figure shows the two slits S1 and S2, a screen, and a point P on the screen. S2B is drawn perpendicular to the line S1P, so S1B is the path difference at P (the screen is far away, so S2P is almost equal to BP).
Step 2: Fix the order of the fringes:
Take the central bright fringe O as order 0. Q is the second bright fringe on the right, so it has order +2. P is the eleventh bright fringe counted from Q, on the other side of O. Moving 11 fringes to the left from Q lands at order \(2 - 11 = -9\). So P is the 9th bright fringe on the left of O.
Step 3: Path difference at P:
For a bright fringe the path difference is a whole number of wavelengths. At order 9:
\[ S_1B = 9\lambda \]
Step 4: Put in the numbers:
\(\lambda = 6000\ \text{\AA} = 6\times10^{-7}\) m.
\[ S_1B = 9\times 6\times10^{-7} = 54\times10^{-7} = 5.4\times10^{-6}\ \text{m} \]
Step 5: Why the other options are wrong:
The values \(3.142\times10^{-7}\) and \(3.138\times10^{-7}\) m are about half a wavelength, nowhere near 9 wavelengths. \(6.6\times10^{-6}\) m equals \(11\lambda\), which comes from counting 11 fringes from O and forgetting that the count starts at Q.
Final Answer:
The path difference at P is \(9\lambda = 5.4\times10^{-6}\) m, option (D).
\[ \boxed{5.4\times10^{-6}\ \text{m}} \]
