Question

A beam of light consists of two wavelengths, $6300 \, \mathring{A} $ and $5600 \, \mathring{A} $ . This beam of light is used to obtain an interference pattern in YDSE. If $4^{th}$ bright fringe of $6300 \, \mathring{A} $ coincides with the $n^{th}$ dark fringe of $5600 \, \mathring{A} $ from the central line, then find the value of $n$ .

Answer 1

Fringe width $\Delta \mathrm{w}_1=\frac{\left(\begin{array}{ll}6300 \AA & \AA\end{array}\right) \mathrm{D}}{\mathrm{d}}$ and for $5600 \AA \AA \Delta \mathrm{w}_2=\frac{\left(\begin{array}{ll}5600 \AA & \AA\end{array}\right) \mathrm{D}}{\mathrm{d}}$
So $4^{th}$ bright fringe position of $6300 \, \mathring{A} $ is at $4\times \frac{\left(6300 \, \mathring{A} \right) D}{d}$ .
So for $n^{th}$ dark fringe of $5600 \, \mathring{A} $ is at $\left(n - \frac{1}{2}\right)\frac{5600 \times D}{d}$ .
$\Rightarrow 4\times \frac{6300 \times D}{d}=\left(n - \frac{1}{2}\right)\times \frac{5600 D}{d}$
$\Rightarrow \frac{4 \times 63}{56}=n-\frac{1}{2}$
$\Rightarrow n-\frac{1}{2}=\frac{9}{2}$
$\Rightarrow n=5$