Question

In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?

Answer 1

Δy = 1 10⁻⁴ m

Step-by-step explanation:

In double-slit experiments the constructive interference pattern is described by the equation

d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

tan θ = y / L

as the angles are small,

tan θ = sin θ / cos θ = sin θ

substituting

sin θ = y / L

d y / L = m λ

y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)

y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)

y₂= 4.21 10⁻⁴ m

the separation of these two lines is

Δy = y₂ - y₁

Δy = (4.21 - 3.21) 10⁻⁴ m

Δy = 1 10⁻⁴ m