Physics - Wave Optics Question with Solution | TestHub

Let the central maximum be at point $O$ with path difference $\Delta x_O=0$.

The point $Q$ is the second bright fringe on one side of $O$. The path difference at $Q$ is:
$\Delta x_Q=2\lambda$

The point $P$ is the eleventh bright fringe on the other side, measured from point $Q$. This means the distance between $Q$ and $P$ corresponds to $11$ fringe widths.
Since $P$ is on the other side of the central maximum, its position in terms of path difference is:
$\Delta x_P=2\lambda -11\lambda =-9\lambda$
The magnitude of the path difference at $P$ is $9\lambda$.

From the given figure, $S_{2}B$ is perpendicular to $S_{1}P$. For a point $P$ on the screen at an angular position $\theta$, the path difference is given by $\Delta x=S_{1}P-S_{2}P$.
In the right-angled triangle $S_{1}B{S}_2$, the side $S_{1}B$ represents this path difference because $S_{1}P\approx S_{1}B+S_{2}P$ (since $D\gg d$).
Therefore, $S_{1}B=\Delta x_P=9\lambda$.

Given the wavelength $\lambda =6000\text{ A˚}=6000\times 10^{-10}\text{ m}=6\times 10^{-7}\text{ m}$.

Substituting the value of $\lambda$:
$S_{1}B=9\times 6\times 10^{-7}\text{ m}$
$S_{1}B=54\times 10^{-7}\text{ m}=5.4\times 10^{-6}\text{ m}$