In a young’s double -slit experiment, the separation between the two slits is d and wavelength of light is $\lambda$ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Find the correct choices:

(A) the maximum path difference will be less than $\lambda$ for $\mathrm{d}=\lambda$ Thus, only maximum will be observed
(B) for $\lambda <\mathrm{d}<2\lambda$ maximum path difference will be less than $2\lambda$ thus, 3 maximum will be observed
(C) Initially ${\mathrm{I}}_1=4{\mathrm{I}}_0\text{ and }{\mathrm{I}}_2={\mathrm{I}}_0$
So,  ${\mathrm{I}}_{max}=9{\mathrm{I}}_0;{\mathrm{I}}_{min}={\mathrm{I}}_0$
$\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+2\sqrt{{\mathrm{I}}_1{\mathrm{I}}_2}\cos \left(\mathrm{\Delta }\varphi \right)$
For maximum intensity  $\cos \left(\mathrm{\Delta }\varphi \right)=1$
Minimum intensity  $\cos \left(\mathrm{\Delta }\varphi \right)=-1$
${\mathrm{I}}_1^{\mathrm{\prime }}={\mathrm{I}}_0\text{ and }{\mathrm{I}}_2^{\mathrm{\prime }}={\mathrm{I}}_0\text{ (given) }$
Thus, $I_{max}^{\mathrm{\prime }}=4I_0\text{ and }{I}_{min}^{\mathrm{\prime }}=0$ (decreases)
So, 
$\left(\mathrm{d}\right) {\mathrm{I}}_1^{\mathrm{\prime }}=4{\mathrm{I}}_0$ and ${\mathrm{I}}_2^{\mathrm{\prime }}=4{\mathrm{I}}_0$  (given)
So, ${\mathrm{I}}_{max}^{\mathrm{\prime }}=16{\mathrm{I}}_0\text{ and }{\mathrm{I}}_{min}^{\mathrm{\prime }}=0$