In a Young’s double slit experiment, the separation between two slits is 𝑑 and the wavelength of the light used is 𝜆. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s)

For maximum to be obtained on the screen at position y we have d sin $\mathrm{\theta }$ - n$\mathrm{\lambda }$
$=>\mathrm{d}=\mathrm{\lambda }\Rightarrow \sin \mathrm{\theta }=1=>={90}^{\circ }$
=> only one central maximum (n = 0) is possible and $\mathrm{n}=1=>\mathrm{y}=\mathrm{\infty }\left(\frac{\mathrm{y}}{\mathrm{D}}=\tan \mathrm{\theta }\right)$
$=>\mathrm{d}=2\mathrm{\lambda }=>\sin \mathrm{\theta }=\frac{\mathrm{n}}{2}$
There will be at least 2 more maximum (n = 1) possible on either side of central maximum (n = 0) Ratio of minimum to maximum intensity is given by $\frac{{\mathrm{I}}_{min}}{{\mathrm{I}}_{min}}={\left(\frac{\sqrt{{\mathrm{I}}_1}-\sqrt{{\mathrm{I}}_2}}{\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_2}}\right)}^2={\left(\frac{\sqrt{4\mathrm{I}}-\sqrt{\mathrm{I}}}{\sqrt{4\mathrm{I}}+\sqrt{\mathrm{I}}}\right)}^2=\frac{1}{9}$
If the intensity at slit 1 is reduced to make it equal to intensity at slit 2, dark fringe becomes completely dark (Intensity becomes zero from I) and intensity of bright fringe also reduced to 4I from 9I