Answer (4)
Sol. Path difference for destructive interference in YDSE
$
\begin{array}{c}
\Rightarrow \Delta X_{n}=\frac{(2 n-1)}{2} \lambda \quad n=1,2,3 \ldots . \\
\Delta X_{5^{k+}}=\left(\frac{2 \times 5-1}{2}\right) \lambda=\frac{9 \lambda}{2}
\end{array}
$
In a Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference (1) $11 \frac{\lambda}{2}$ (2) $5 \frac{\lambda}{2}$ (3) $10 \frac{\lambda}{2}$ (4) $9 \frac{\lambda}{2}$
In a Young’s double slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference (1) $11 \frac{\lambda}{2}$ (2) $5 \frac{\lambda}{2}$ (3) $10 \frac{\lambda}{2}$ (4) $9 \frac{\lambda}{2}$