Answer:
a) The width of the central bright fringe is 1.41492×10^-2 m
b) The width of the first bright field on either side of the bright fringe is 7.075×10^-3 m
Explanation:
Let d be the width of the slit, s be the screen distance, m the order of diffraction. And x be the distance from the center to the edge of the central bright fringe and W be the width of the central bright fringe . And let y be the distance from the center of the central bright fringe to the edge of the first bright fringe on either side of the central bright fringe and w be the width of the first bright fringe. Let θ1 and θ2 respectively be the scattering angles of the central bright fringe and the first bright fringe.
a) According to Bragg's law
d×sinθ1 = m×λ
(0.340×10^-3)×sinθ1 = 1×(633×10^-9)
θ1 = 10.667×10^-2°
x = s×tanθ1
x = 3.80×tan(10.667×10^-2°)
x = 7.0741×10^-3 m
Now to get the width of the fringe you need to add x and x together
W = 2×x
W = 2×7.0741×10^-3
W= 1.41492×10^-2 m
b) For the second bright fringe
d×sinθ2 = m×λ
3.8×sinθ2 = 2×(633×10^-9)
θ2 = 21.33×10^-2°
y = s×tanθ2
y = 3.80×tan( 21.33×10^-2°)
y = 1.4149×10^-2 m
The width of the first bright fringe on either side of the central bright fringe is given by:
w = y - x
w = 7.075×10^-3 m
