The path difference between the two waves ${\mathrm{Y}}_1={\mathrm{A}}_1\sin (\mathrm{\omega t}-\frac{2\mathrm{\pi x}}{\mathrm{\lambda }})$ and ${\mathrm{Y}}_2={\mathrm{A}}_2\cos (\mathrm{\omega t}-\frac{2\mathrm{\pi x}}{\mathrm{\lambda }}+\mathrm{\varphi })$ is $\mathrm{N}\left[\frac{\mathrm{\lambda }}{2\mathrm{\pi }}(\mathrm{\varphi }+\frac{\mathrm{\pi }}{2})\right]$ then N is _____  

For the given wave equations 

${\mathrm{\varphi }}_1=(\mathrm{\omega t}-\frac{2\mathrm{\pi x}}{\mathrm{\lambda }}) \mathrm{and} {\mathrm{\varphi }}_2=(\mathrm{\omega t}-\frac{2\mathrm{\pi x}}{\mathrm{\lambda }}+\mathrm{\varphi }+\frac{\mathrm{\pi }}{2})$

The phase difference is 

$\therefore {\mathrm{\varphi }}_2-{\mathrm{\varphi }}_1=\left(\mathrm{\varphi }+\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow \mathrm{\Delta \varphi }=\left(\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\right)\mathrm{\Delta x},$

$\therefore \mathrm{\Delta x}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}\left(\mathrm{\varphi }+\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow \mathrm{N}=1$