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: Set up:
Let the fringe order be measured from the centre O, with right side positive. Q has $n=2$. Moving 11 fringes to the left of Q gives $n = 2-11 = -9$, so P is the 9th bright fringe on the left.
Step 2: Use the bright fringe condition:
A bright fringe needs $|S_2P - S_1P| = n\lambda$, so the path difference at P has magnitude $9\lambda$. In the figure, B is the foot of the perpendicular from S2 on S1P, so $S_1B$ equals this path difference.
Step 3: Compute:
With $\lambda = 6000\times10^{-10}$ m, $S_1B = 9\times6000\times10^{-10} = 5.4\times10^{-6}$ m.
Step 4: Check:
$11\lambda = 6.6\times10^{-6}$ m would be wrong because the 11 fringes are counted from Q, not from O. The two options near $3.1\times10^{-7}$ m are less than one wavelength, so they cannot be a path difference of several fringes.
Final Answer:
The answer is $5.4\times10^{-6}$ m.
\[ \boxed{5.4\times10^{-6}\ \text{m}} \]
