A light beam ( 557.1 nm) illustrates a single slit 0.85mm wide. (a) How far is the screen from the slit if there is 0.85 mm between the first minimum and the central minimum on the screen? (b) what is the width of the central maximum on the screen?

Answer 1

(a) b•sinφ = k•λ,
sinφ = k•λ/b = 1•557.1•10^9/0.85•10^-3 =
= 6.56•10^-4 = tan φ
tan φ = x/L ,
L = 0.85•10^-3/6.56•10^-4 = 1.3 m.
(b) 2•0.85 mm = 1.7 mm