SME
University Physics III: Waves, Optics & Modern
Interference and diffraction
Use this for “Light of 520 nm passes two slits 0.30 mm apart onto a screen 1.5 m away. At what angle, and how far from the centre, is the second bright fringe?”
Diagram
Δy = λL/d
Δy
=
600 ×
10
-9
m
×
2
m/(0.25
×
10
-3
m)
Δy = 4.8 mm
Bright where the paths from the two slits differ by a whole number of wavelengths (mλ), dark halfway between.
This is an example. Type your own values to start a new problem.
d (mm)
sin(
θ
°)
=
m
×
λ (nm)
L
Screen distance
example · m
sin θ
Sine of the angle
worked out (not typed)
y
Position on the screen
calculated · mm
Δy
Fringe spacing
worked out (not typed) · mm
Formulas
sin
θ
=
m
×
λ
d
×
10
6
0.0072
=
3
×
600
0.25 ×
10
6
θ
=
sin⁻¹(sin
θ
)
0.4125 = sin⁻¹(0.0072)
y
=
L
×
tan(
θ
)
×
1,000
14.4004 = 2 × tan(0.4125°) × 1,000
Δy
=
λ
×
L
d
÷
1,000
4.8
=
600
×
2
0.25
÷
1,000
Assumptions
•
One wavelength reaches both slits in step, and the screen is far compared with d, so the two rays to one point are parallel.
•
Bright where the path difference d sin θ is a whole number of wavelengths: d sin θ = mλ, with m = 0, ±1, ±2, …
•
The fringe is at y = L tan θ from the centre at any angle; for small angles y ≈ mλL ÷ d, so fringes are Δy = λL ÷ d apart.
•
λ is in nm and d in mm, so d is d × 10⁶ nm; y comes out in mm (× 1000).
Step-by-step
WE KNOW
λ = 600 nm, d = 0.25 mm, L = 2 m, m = 3
FIND
sine of the angle (sin θ), angle to the fringe (θ), position on the screen (y), fringe spacing (Δy)
Step 1 · Find sine of the angle (sin θ)
Use sin θ = m × λ ÷ (d × 10⁶)
From
d
sin
θ
=
mλ:
the
path
difference
holds
m
wavelengths.
d
in
nm
is
d
×
10
6
.
sin
θ
=
3
×
600
0.25 ×
10
6
sin
θ
=
1,800
0.25 ×
10
6
sin
θ
=
0.0072
Step 2 · Find angle to the fringe (θ)
Use θ = sin⁻¹(sin θ)
The angle whose sine that is.
θ
=
sin⁻¹(0.0072)
θ
=
0.412533°
Step 3 · Find position on the screen (y)
Use y = L × tan(θ) × 1,000
Across
the
screen:
L
times
tan
θ
,
in
m
,
then
×
1000
for
mm.
y
=
2
×
tan(0.412533°)
×
1,000
y
=
2
×
0.0072
×
1,000
y
=
0.0144
×
1,000
y
=
14.4004
mm
Step 4 · Find fringe spacing (Δy)
Use Δy = λ × L ÷ d ÷ 1,000
For
small
angles
the
bright
fringes
are
evenly
spaced,
λL
d
apart;
nm
×
m
mm
gives
μm,
so
1000
gives
mm.
Δy
=
600
×
2
0.25
÷
1,000
Δy
=
1,200
0.25
÷
1,000
Δy
=
4,800
1,000
Δy = 4.8 mm
Check
Put the numbers back in. Both sides should match.
0.0072
=
3
×
600
0.25 ×
10
6
✓
0.412533 = sin⁻¹(0.0072) ✓
14.4004 = 2 × tan(0.412533°) × 1,000 ✓
4.8
=
600
×
2
0.25
÷
1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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