In a double-slit experiment,the slit separation is 2.0 mm,and two wavelengths,750 nm and 900 nm,illuminate the slits.A screen is placed 2.0 m from the slits.At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?
A)1.5 mm
B)3.0 mm
C)4.5 mm
D)6.0 mm

Question

Quizplus · Accepted Answer

The condition for constructive interference at a point on the screen for the double-slit experiment is:

d sin θ = mλ,

where d is the slit separation, θ is the angle between the line passing through the point and the central maximum, λ is the wavelength, and m is an integer (the order of the fringes).

For the central maximum (m = 0), sin θ = 0, so the bright fringes from both wavelengths will coincide at the central maximum.

For the first order fringe (m = 1), we have:

d sin θ = λ1  and d sin θ = λ2

where λ1 = 750 nm and λ2 = 900 nm.

Solving for sin θ in each equation and then taking the difference gives:

sin θ = (λ2 - λ1)/d

sin θ = (900 nm - 750 nm)/2.0 mm = 0.075

θ ≈ 4.3°

The distance from the central maximum to the first order fringe is given by:

y = L tan θ ≈ 4.5 mm

where L is the distance from the slits to the screen (2.0 m in this case). Therefore, the answer is (C) 4.5 mm.

In a Double-Slit Experiment,the Slit Separation Is 2