Question

In the YDSE as shown in figure, $Q$ is the position of the first bright fringe on the right side and $P$ is the 1 second fringe on the other side as measured from $Q$. If wavelength of light used is $6000\mathring{A}$, then $S,B$ will be equal to

• A. $6\times {10}^{-6}m$
• B. $6.6\times {10}^{-6}m$
• C. $3.13\times {10}^{-7}m$
• D. $3.14\times {10}^{-7}m$

Answer 1

The correct option is A $6\times {10}^{-6}m$

Given, $P$ is the position of $11$th fringe measured from $Q$. So, it is 10th fringe as measured from $O$.
$\therefore$ The path difference $S$, $B$ should contain $10$ full waves
i.e., $S,B=\lambda =10\times 6\times {10}^{-7}m=6\times {10}^{-6}m$