In Young's double slit experiment, the slits are horizontal. The intensity at a point P (as shown in figure) is   where   the maximum intensity. If the value of   is found to be  . Then the value of N+M is ___

 

Here,  ---------
$\left[\because I_R=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)=I_m{\cos }^2\left(\frac{\varphi }{2}\right)\right]$

$\Rightarrow \cos \frac{\varphi }{2}=\frac{\sqrt{3}}{2}\Rightarrow \frac{\varphi }{2}=\frac{\pi }{6}\Rightarrow \varphi =\frac{\pi }{3}$ 
From figure, Path difference, d cos  
Phase difference $=\frac{2\pi }{\lambda }\times \text{ path difference }$ 

$\therefore \frac{\pi }{3}=\frac{2\pi }{\lambda }\times 2\lambda \cos \theta \Rightarrow \cos \theta =\frac{1}{12}$
 

Comparing with $\theta ={\cos }^{-1}\left(\frac{N}{M}\right)$ 

$\therefore \mathrm{N}=1,\mathrm{M}=12\therefore \mathrm{N}+\mathrm{M}=13$