In a special arrangement of Young's double-slit experiment, the distance between the slits d is twice the distance between the screen and the slits D , i.e d = 2 D . For this setup, the value of D such that the first minima on the screen fall at a distance D from the centre O is found to be λ N . What is the value of N ? [Take 5 = 2 . 24 ]

Step-by-step solution

From diagram as provided in question, OP = x CO = D S 1 C = S 2 C = D T 1 P = T 1 O - OP = D - x T 2 P = T 2 O + OP = D + x Now, S 1 p = S 1 T 1 2 + T 1 P 2 = D 2 + D - x 2 S 2 P = S 2 T 2 2 + T 2 P 2 = D 2 + D + x 2 For fist minimum to occur, Path difference S 2 P - S 1 P = λ 2 ⇒ D 2 + D + x 2 - D 2 + D - x 2 = λ 2 The first minimum falls at a distance D from the centre, i.e., x = D D 2 + 4 D 2 1 2 - D = λ 2 ∴ D 5 - 1 = λ 2 ⇒ D = λ 2 2 . 24 - 1 = λ 2 . 48