In Young’s double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit.  If  $I_m$  be the maximum intensity, the resultant intensity I when they interfere at phase difference $\varphi$  is given by  $I=\frac{I_{\max }}{9}[1+n{\cos }^2\frac{\varphi }{2}]$  then n = ____

It is given,  $A_2=2A_1$
 We know, Intensity $\infty {\left(Amplitude\right)}^2$ 
Hence   $\frac{I_2}{I_1}={\left(\frac{A_2}{A_1}\right)}^2={\left(\frac{2A_1}{A_1}\right)}^2=4$
  $\Rightarrow I_2=4I_1$
Maximum intensity,  $I_m={(\sqrt{I_1}+\sqrt{I_2})}^2$
    $={(\sqrt{I_1}+\sqrt{2})}^2 ={\left(3\sqrt{I_1}\right)}^2=9I_1$
Hence  $I_1=\frac{I_m}{9}$
Resultant intensity,  $I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
$=I_1+4I_1+2\sqrt{I_1\left(4I_1\right)}\cos \varphi =I_1+4I_1+4I_1\cos \varphi =I_1+4I_1(1+\cos \varphi )$

$=I_1+8I_1{\cos }^2\left(\frac{\varphi }{2}\right) (\because 1+\cos \varphi =2{\cos }^2\frac{\varphi }{2})$

$I=\frac{I_{\max }}{9}(1+n{\cos }^2\frac{\theta }{2})$

  Putting the value of  $I_1$  from Eq. (i), we get  $I=\frac{I_{\max }}{9}(1+n{\cos }^2\frac{\theta }{2})$