In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P . If D = 0 . 2 m , the minimum distance between the slits S 1 and S 2 is α mm . Write the value of 10 α to the nearest integer.

Step-by-step solution

According to the given diagram, the path difference for minima at P is given by ∆ x = 2 D 2 + d 2 - 2 D . . . 1 And, from the condition of minima, it follows that ∆ x = λ 2 . . . 2 Equations (1) and (2) imply that 2 D 2 + d 2 - 2 D = λ 2 ⇒ D 2 + d 2 - D = λ 4 ⇒ D 2 + d 2 = λ 4 + D ⇒ D 2 + d 2 = D 2 + λ 2 16 + D λ 2 ⇒ d 2 = D λ 2 + λ 2 16 ⇒ d 2 = 0 . 2 × 400 × 10 - 9 2 + 16 × 10 - 14 16 ⇒ d 2 ≈ 400 × 10 - 10 ∴ d = 20 × 10 - 5 m = 0 . 20 mm Therefore, 10 α = 2 .