A concave mirror of short aperture and radius of curvature 10cm is cut into two equal parts, slightly separated and placed in front of a monochromatic line source S of wavelength  2000 A° as shown in figure. Consider only interference in reflected light after first reflection from mirrors. Assume direct light from S is blocked from reaching the screen.If half of reflecting surface of upper part of mirror is painted black then choose correct option:

Due to mirrors real images S1 and S2 are formed.Now, Half Part of Upper Mirror is only covered as shown in figure.This will lead to decrease in intensity of image S1 , but no change in positions of any image. Thus , values of d and D will remain same.Position of first order minima will also remain same. y=0.5λDd Using the formulae   Imax=I1+I22,Imin=I1−I22Before covering the mirror, Imin was zero. But now, Imin becomes a non zero value . Thus we can say Imin increases.On the other hand, Imax decreases.