) By passing light of wavelength 5000 A.U. through two pin holes 0.5 mm apart, an interference pattern is formed on a screen kept parallel to the plane of the pin holes and 100 cm away from them. Find the distance along the screen between third and seventh dark bands on the same side.
Answers
Given info : By passing light of wavelength 5000 A.U. through two pin holes 0.5 mm apart, an interference pattern is formed on a screen kept parallel to the plane of the pin holes and 100 cm away from them.
To find : The distance along screen between third and seventh dark band on the same side.
solution : wavelength, λ = 5000 A.U = 5000 × 10¯¹⁰ m = 5 × 10^-7 m [ A.U = Anstrom unit ]
width of slits , d = 0.5 mm = 0.5 × 10¯³ m
distance between slits and screen, D = 100 cm = 1 m
we know, distance of nth dark band is given by, x = (2n - 1)λD/2d
∴ distance of 3rd dark band, x = (2 × 3 - 1) × λD/2d
= 5λD/2d
distance of 7th dark band, X = (2 × 7 - 1) × λD/2d
= 13λD/2d
so the distance between 7th and 3rd dark band, ∆x = X - x = 13λD/2d - 5λD/2d = 8λD/2d = 4λD/d
now putting values of D, λ and d
we get, ∆x = 4 × 5 × 10^-7 × 1/(0.5 × 10^-3)
= 4 × 10¯³ m = 4mm
Therefore the distance along the screen between 3rd and 7th dark band on the same side is 4 mm.