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Bright and Dark Fringe Spacing
Relevant equation for the following scenarios: d sin θ = m λ and d sin θ = (m + ½) λ
Scenario: You shoot a laser beam
with a wavelength of 400 nm to
illuminate a double slit, with a
spacing of 0.002 cm, and produce
an interference pattern on a
screen 75.0 cm away.
Q1) What is the angle θ
between the m = 0 fringe and
the m = 1 fringe?
Q2) What is the distance y
between the m = 0 fringe and
the m = 1 fringe?
Q3) What is the distance between the m = 1 bright fringe and the dark fringe between m
=1 and m = 2?
Solutions:
Q1) Using equation dsin θ = m λ , we know:
d = 0.002 cm = 0.002 x 10-2
m
λ = 400 nm= 400 x 10-9
m
Rearrangingthe equation, θ= arcsin(λ/d) =arcsin(400 x 10-9
m/0.002 x 10-2
m) = 1.15°
Q2) To obtaina value fory, we can use trigonometry.
tan θ = y/D => y = D*tan θ = (75 x 10-2
m)*tan (1.15°) = 0.015 m
Q3) First we have to solve forthe angle betweenthe m= 0 fringe andthe dark fringe betweenm= 1
and m = 2, thenfindthe difference betweenyfoundinQ2 andthe positionof the darkfringe.
Usingequation d sin θ = (m+ ½) λ , we know:
d = 0.002 cm = 0.002 x 10-2
m
λ = 400 nm= 400 x 10-9
m
(m+ ½) = 1.5
Rearrangingthe equation, θ1/2 = arcsin(((m+½)λ)/d)
= arcsin(1.5*400 x 10-9
m/0.002 x 10-2
m) = 1.72°
Positionof the darkfringe: y1/2 = D*tan θ1/2 = 0.0225 m
The dark fringe andthe adjacentbrightfringe are 0.0075m apart.

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Bright Fringes and Dark Fringes

  • 1. Bright and Dark Fringe Spacing Relevant equation for the following scenarios: d sin θ = m λ and d sin θ = (m + ½) λ Scenario: You shoot a laser beam with a wavelength of 400 nm to illuminate a double slit, with a spacing of 0.002 cm, and produce an interference pattern on a screen 75.0 cm away. Q1) What is the angle θ between the m = 0 fringe and the m = 1 fringe? Q2) What is the distance y between the m = 0 fringe and the m = 1 fringe? Q3) What is the distance between the m = 1 bright fringe and the dark fringe between m =1 and m = 2?
  • 2. Solutions: Q1) Using equation dsin θ = m λ , we know: d = 0.002 cm = 0.002 x 10-2 m λ = 400 nm= 400 x 10-9 m Rearrangingthe equation, θ= arcsin(λ/d) =arcsin(400 x 10-9 m/0.002 x 10-2 m) = 1.15° Q2) To obtaina value fory, we can use trigonometry. tan θ = y/D => y = D*tan θ = (75 x 10-2 m)*tan (1.15°) = 0.015 m Q3) First we have to solve forthe angle betweenthe m= 0 fringe andthe dark fringe betweenm= 1 and m = 2, thenfindthe difference betweenyfoundinQ2 andthe positionof the darkfringe. Usingequation d sin θ = (m+ ½) λ , we know: d = 0.002 cm = 0.002 x 10-2 m λ = 400 nm= 400 x 10-9 m (m+ ½) = 1.5 Rearrangingthe equation, θ1/2 = arcsin(((m+½)λ)/d) = arcsin(1.5*400 x 10-9 m/0.002 x 10-2 m) = 1.72° Positionof the darkfringe: y1/2 = D*tan θ1/2 = 0.0225 m The dark fringe andthe adjacentbrightfringe are 0.0075m apart.