Light from an LED with a wavelength of 4.90 ✕ 102 nm is incident on (and perpendicular to) a pair of slits separated by 0.310 mm. An interference pattern is formed on a screen 2.20 m from the slits. Find the distance (in mm) between the first and second dark fringes of the interference pattern.

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hamzaahmeds hamzaahmeds · Answer 1 · 2020-04-23T06:56:09+02:00

Answer:

Δx = 3.477 x 10⁻³ m = 3.477 mm

Explanation:

The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:

Δx = λL/d

where,

Δx = distance between two consecutive dark fringes = ?

λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m

L = Distance between slits and screen = 2.2 m

d = slit separation = 0.31 mm = 0.31 x 10⁻³ m

Therefore,

Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)

Δx = 3.477 x 10⁻³ m = 3.477 mm

adefioyelaoye adefioyelaoye · Answer 2 · 2021-11-18T18:53:46+01:00

The distance (in mm) between the first and second dark fringes of the

interference pattern is 3.477 mm

This is calculated by using the formula in Young's Double Slit experiment:

Δx = λL/d

where,

Δx = distance between two consecutive dark fringes which is unknown

λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m

L which is distance between slits and screen = 2.2 m

d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m

We then substitute them into the equation

Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)

Δx = 3.477 x 10⁻³ m = 3.477 mm

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