Advertisements
Advertisements
Question
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
Options
2.87 nm
2 nm
1.27 μm
2.05 μm
Solution
1.27 μm
Explanation:
Given: In a double slit experiment the distance between two slits is d = 1 mm, wavelength of incident light is λ = 632.8 nm, distance between the slits and the screen is D = 100 cm, on screen the distance between a bright fringe and the central bright fringe is y = 1.27 mm.
To find: Δ the path difference between the waves forming the bright fringe at y = 1.27 mm.
Δ = `"dy"/"D"`
= `(1xx10^-3xx1.27xx10^-3)/(100xx10^-2)`
= 1.27 μm