Question

In a double-slit interference experiment, the slit separation is 2.29 μm, the light wavelength is 532 nm, and the separation between the slits and the screen is 4.42 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.8% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move?

Answer 1

given

d = 2.29 micro m = 2.29*10^-6 m
lamda = 532 nm = 532*10^-9 m
R = 4.42 m

a) for third bright fringe(m=3), d*sin(theta) = m*lamda

sin(theta) = m*lamda/d

sin(theta) = 3*532*10^-9/(2.29*10^-6)

theta = sin^-1(0.6969)

= 44.2 degrees

b) if the frequency decreases wavelength increases.

so, the third bright fringe moves away from the pattern's center.

c) actual place of third bright fringe from the center, y3 = m*lamda*R/d

= 3*532*10^-9*4.42/(2.29*10^-6)

= 3.080 m

when frequency decreases to 94.8%,

wavelegth, lamda' = lamda/0.948

y3' = m*lamda'*R/d

= 3*(532*10^-9/0.948)*4.42/(2.29*10^-6)

= 3.249 m

so, y3' - y3 = 3.249 - 3.080

= 0.169 m <<<<<<<<<<<<---------------Answer