Question

A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm

Answer 1

Step-by-step explanation:

The double slit interference phonemene is described for the case of constructive interference

d sin θ= m λ (1)

let's use trigonometry to find the sinus

tan θ = y / L

in general in interference phenomena the angles are small

tan θ = sin θ / cos θ = sin θ

The double slit interference phonemene is described for the case of constructive interference

d sin θ = m lam (1)

let's use trigonometry to find the sinus

tan θ = y / L

in general in interference phenomena the angles are small

tan θ = sin θ / cos θ = sin θ

we substitute

sin θ = y / L

we substitute in equation 1

d y / L = m λ

λ = dy / L m

let's reduce the magnitudes to the SI system

d = 0.44 mm = 0.44 10⁻³ m

y = 5.5 mm = 5.5 10⁻³ m

L = 4.2m

m = 1

let's calculate

λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)

λ = 5.76190 10-7 m

let's reduce to num

lam = 5.56190 10-7 m (109 nm / 1m)

lam = 556,190 nmtea

we substitute

without tea = y / L

we substitute in equation 1

d y / L = m lam

lam = dy / L m

let's reduce the magnitudes to the SI system

d = 0.44 me = 0.44 10-3 m

y = 5.5 mm = 5.5 10-3

L = 4.2m

m = 1

let's calculate

lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)

lam = 5.76190 10⁻⁷ m

let's reduce to num

lam = 5.56190 10⁻⁷ m (109 nm / 1m)

lam = 556,190 nm