In an interference experiment as shown in the figure, the source and the screen are separated by a distance of 1m. At a certain position of source the fringe-width is   $\frac{1}{4}\mathrm{mm}$ and by moving the source away from the mirror along the line AB by 0.6 mm the fringe width changed to   $\frac{1}{6}\mathrm{mm}$ . The wavelength of light used is $\mathrm{n}\times {10}^{-7}\mathrm{m}$ . Find $\mathrm{n}$?

 

Let AS = h
Now, $\beta =\frac{\lambda D}{d}$
In the first case, $d=2h$ 
$\therefore \beta =\frac{\lambda D}{2h}$ ...(i)
In the second case, $d=2(h+\mathrm{\Delta }x)$ , where $\mathrm{\Delta }x=$ shift in the source away from the mirror along AB. 
$\therefore {\beta }^{\mathrm{\prime }}=\frac{\lambda D}{2(h+\mathrm{\Delta }x)}$..(ii)
Dividing equation (i) by equation (ii), we have, 
$\frac{\beta }{{\beta }^{\mathrm{\prime }}}=\frac{h+\mathrm{\Delta }x}{h}=1+\frac{\mathrm{\Delta }x}{h}$
Or $\frac{\beta -{\beta }^{\mathrm{\prime }}}{{\beta }^{\mathrm{\prime }}}=\frac{\mathrm{\Delta }x}{h}\Rightarrow h=\left(\frac{\mathrm{\Delta }x\times {\beta }^{\mathrm{\prime }}}{\beta -{\beta }^{\mathrm{\prime }}}\right)=\frac{.6\times \frac{1}{6}}{\frac{1}{4}-\frac{1}{6}}=1.2\mathrm{mm}$ 
Putting the value of h in equation (1), we get 
$\lambda =\frac{2\times \beta \times \mathrm{h}}{\mathrm{D}}=\frac{2\times 1/4\times {10}^{-3}\times 1.2\times {10}^{-3}}{1}=6\times {10}^{-7}\mathrm{m}$