Question

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit

Answer 1

The width of the slit is 0.4 mm (0.00040 m).

Step-by-step explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×
`10^(-9)` m)

d = (3.30 × 563 ×
`10^(-9)` ) ÷ (0.0047)

= 1.8579 ×
`10^(-6)` ÷ 0.0047

= 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).