In Yong's double slit experiment, in interference pattern, a minimum is observed exactly in front of one…
In Yong's double slit experiment, in interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent source is ' $\mathrm{d}$ ' and ' $\mathrm{D}$ ' is the distance between the source and screen. The possible wavelength used are inversely proportional to
$\mathrm{D}, 5 \mathrm{D}, 9 \mathrm{D} \ldots$
$\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D} \ldots$
$3\mathrm{D}, 4 \mathrm{D}, 5 \mathrm{D} \ldots$
$3\mathrm{D}, 7 \mathrm{D}, 10 \mathrm{D} \ldots$
Solution
If $x$ is the fringe width, then three will be a minimum in front of the slit if
$\begin{aligned}
& \frac{\mathrm{d}}{2}=\frac{\mathrm{x}}{2}, \frac{3 \mathrm{x}}{2}, \frac{5 \mathrm{x}}{2}, \ldots \ldots \\
& \text { or } \mathrm{d}=\mathrm{x}, 3 \mathrm{x}, 5 \mathrm{x} \ldots . \\
& \therefore \mathrm{x}=\mathrm{d}, \frac{\mathrm{d}}{3}, \frac{\mathrm{d}}{5}, \ldots \ldots \\
& \because \lambda=\frac{\mathrm{xd}}{\mathrm{D}} \\
& \therefore \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}, \frac{\mathrm{d}^2}{3 \mathrm{D}}, \frac{\mathrm{d}^2}{5 \mathrm{D}}
\end{aligned}$
$\therefore \lambda$ is inversely proportional to $\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D} \ldots$
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