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

Answer: 5

Fringe width ${\left(\Delta w\right)}_1=\frac{\left(6300\mathring{A}\right)D}{d}$ and for $5600\mathring{A}$ ; ${\left(\Delta w\right)}_2=\frac{\left(5600\mathring{A}\right)D}{d}$
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 $\left(n-\frac{1}{2}\right)\frac{5600\times D}{d}$ .
$\Rightarrow 4\times \frac{6300\times D}{d}=\left(n-\frac{1}{2}\right)\times \frac{5600D}{d}$
$\Rightarrow \frac{4\times 63}{56}=n-\frac{1}{2}$
$\Rightarrow n-\frac{1}{2}=\frac{9}{2}$
$\Rightarrow n=5$