Young's Double-slit

Question

the question :

In a Young's double-slit experiment, the fringe separation when violet light of 400nm is used is 1.2 mm. The distance from the slits to the screen is D. If red light of 680nm is used while the screen is moved towards the slits by 1/2D, what is the new fringe separation?

Answer 1

Hi Puteri,

For Young's Double Slit we have d*sin(θ) = m*λ, where d = distance between slits, θ = the angle between the central fringe and the m constructive fringe, m = a whole number, and λ = light wavelength.

For the violet light well have d*sin(θ_v) = m*λ_v, where θ_v = angle between the central fringe and the first constructive fringe (where m = 1), and λ_v = violet light wavelength. (subscripts v for violet)

For small θ, sin(θ_v) = θ_v, therefore d*θ_v = m*λ_v. D*θ_v = x_v, where x_v is the violet fringe separation.

For the red light well have d*θ_r = m*λ_r, and D*θ_r/2 = x_r.

From the violet light equations we get:

D*θ_v = x_v, and θ_v = m*λ_v/d

D = d*x_v/(m*λ_v)

For the red light:

θ_r = m*λ_r/d

So for D*θ_r/2 = x_r, we substitute D and θ_r:

[d*x_v/(m*λ_v)]*[m*λ_r/d]/2 = x_r

x_r = [x_v/2]*[λ_r/λ_v]

x_r = [1.2/2]*[680/400] = 1.02 mm

I hope this helps, Joe.