Question

Two slits 0.200 mm apart. Light of wavelength 488 nm impinges on the slits and the interference pattern is observed on a surface 1.00 m from the slits. How far from the central axis is the fourth maxima? (9.8 mm)

Answer 1

Answer:

Explanation:

distance between two slits d = .2 x 10⁻³ m = 2 x 10⁻⁴

Distance of screen D = 1.00 m

wave length of light λ = 488 x 10⁻⁹ m

distance of fourth maxima

= 4 x  λ D / d

= 4 x 488 x 10⁻⁹ x 1 / 2 x 10⁻⁴

= 976 x 10⁻⁵ m .

= 9.76 x 10⁻³ m

= 9.8 mm .