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)

Question

Physics

Guerra

26 March, 11:36

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)

+2

Answers (1)

Know the Answer?

Caiden Long · Answer 1

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.