For a single slit of width a ,the first minimum of the interference pattern of a monochromatic ligth of wavlength `lambda`.occurs at an angle of `lambda//a`.At the same angle of `lambda//a`,we get a maximum for two narrow sits separated bby a distance a.Explain.

In Young's double slit experiment,ligth of wavelnegth 6,000 overset @A is used to get an interference pattern on a screen.The fringe width changes by 1.5 mm,when the screen is brougnt towards the double slit by 50 cm.Find the distance between the two slits.

Consider a two slit interference arrangements (Fig. EP 10.18) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of lambda such that the first minima on the screen falls at a distance D from the center O.

In figure S is a monochromatic point source emitting light of wavelength lambda=500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L_(1) and L_(2) by a plane passing through a doameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from A to L_(1) and L_(2) is 0.15 m , while that from L_(1) and L_(2) to O is 1.30 m. The screen at O is normal to SO. (a) If the 3^(rd) intensity maximum occurs at point P on screen, find distance OP.