Question

A screen is placed 50.0 cm from a single slit, which is illuminated with light of wavelength 690 nm. If the distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit?

Answer 1

Answer:

d = 2.3 x 10⁻⁴ m = 0.23 mm

Explanation:

In a Young's Double Slit experiment the distance between two consecutive minima is given as:

Δx = λL/d

for, the distance between the 1st and 3rd minima it will become:

2Δx = 2λL/d

where,

2Δx = distance between 1dt and 4rd minima = 3 mm = 0.003 m

λ = wavelength = 690 nm = 6.9 x 10⁻⁷ m

L = Distance of Screen = 50 cm = 0.5 m

d = width of slit = ?

Therefore,

0.003 m = 2(6.9 x 10⁻⁷ m)(0.5 m)/d

d = 6.9 x 10⁻⁷ m/0.003 m

d = 2.3 x 10⁻⁴ m = 0.23 mm