Question

A double-slit interference experiment is done in a ripple tank (a water tank using a vibrating rod to produce a plane wave on the surface of the water). The slits are 4.50 cm apart, and a viewing screen is 1.60 m from the slits. The wave speed of the ripples in water is 0.024 m/s, and the frequency of the rod producing the ripples is 6.05 Hz. How far from the centerline of the screen will a first-order minimum be found? The first-order minimum is the second time that destructive interference happens, since the first destructive spot we find is the zeroth-order minimum.

Answer 1

Distance from slit to screen D = 1.60m

Distance between the silts d = 4.50 cm = 4.50 x 10^-2m

Wave speed v = 0.024 m/s

frequency of wave f = 6.05 hz

wavelength of the wave = (v/f) m

Condition for destructive in double slit experiment is

since is small , we can write

where y = distance of destructive spot from central axis

We want first order destructive spot , i,e n = 1

Distance y = 0.21m from the centerline of the screen fisrt order minimum is found.