A screen is at a distance D = 80 cm from a diaphragm having two narrow slits S 1 and S 2 which are d = 2 mm apart. The slit S 1 is covered by a transparent sheet of thickness t 1 = 2.5 μm and the slit S 2 by another sheet of thickness t 2 = 1.25 μm ​as shown in the figure. Both sheets are made of the same material having a refractive index μ m = 1.40 . Water is filled in space between the diaphragm and the screen. A

Step-by-step solution

Optical path = (Refractive Index) × (Geometrical Path Length) Path Difference at point C on the screen Δ x = μ m t 1 + μ w S 1 C - t 1 - μ m t 2 + μ w S 2 C - t 2 Δ x = μ m t 1 - t 2 - μ w t 1 + μ w t 2 ( ∵ s 1 c = s 2 c ) Δ x = μ m t 1 - t 2 - μ w t 1 - t 2 Δ x = μ m - μ w t 1 - t 2 Δ x = 1.4 - 4 3 2.5 × 10 - 6 - 1.25 × 10 - 6 Δ x = 0.2 3 × 1.25 × 10 - 6 Δ x = 2.5 3 × 1 0 - 7 m Δ x = 2500 3 Å = 5000 6 Å Δ x = λ 6 , Phase difference ϕ = π 3 = 60 ο at any point on the screen I n e t = I 1 + I 2 + 2 I 1 I 2 cos ϕ Res