An unlabelled laser pointer turns up in the optics drawer, and the class sets out to identify it with three experiments run in sequence. First the pointer is aimed at a transmission grating whose rulings are 1.500 μm apart: on the screen behind, bright spots flank the straight-through beam, and careful work with a protractor places the first-order beam 25.0° off the straight-through direction. The same laser is then shone on a pair of slits 0.180 mm apart with a screen 2.400 m away, where a row of evenly spaced fringes appears in the dark. Finally the beam is pointed at a soap film of refractive index 1.33 held in a wire loop in air, its surface crawling with slow bands of colour as it drains. Find the laser's wavelength, its frequency, the spacing of the fringes on the screen, and the thinnest film that reflects this colour brightly.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
d = 1.5 μm — Grating ruling spacing
θ = 25 ° — First-order angle off the straight-through
s = 0.18 mm — Double-slit separation
L = 2.4 m — Slits-to-screen distance
n = 1.33 — — Refractive index of the soap film
Determine
(a)the laser's wavelength
(b)its frequency
(c)the fringe spacing at the double slit
(d)the thinnest film that reflects this colour brightly
Step 1 of 4(a) · solve for Wavelength
A grating is the precise instrument in the room, which is why the wavelength is measured here and reused everywhere else. Its angles are large and readable to a fraction of a degree.
Rearranged for λ
λ=mdsinθ
Your values, in your units
λ=(1)(1.5μm)sin((25∘))
Converted to base units
λ=(1)(0.0000015m)sin((25∘))
Answer
λ=633.93nm
Carried onward at full precision, not this rounded figure.
Frequency from c = fλ. Nothing later needs it, but it is the property that survives a change of medium — the wavelength shortens in glass, the frequency does not, and neither does the colour.
Rearranged for f
f=λv
633.93 nmcarried from step 1
Your values, in your units
f=(6.33927e−07m)(299,792,000m/s)
Answer
f=472.91THz
Carried onward at full precision, not this rounded figure.
Now the same wavelength at a double slit. Slit separation on the bottom, screen distance on top — swap them and the answer comes out in kilometres, which is at least an obvious kind of wrong.
Rearranged for Δy
Δy=dλL
633.93 nmcarried from step 1
Your values, in your units
Δy=(0.18mm)(6.33927e−07m)(2.4m)
Converted to base units
Δy=(0.00018m)(6.33927e−07m)(2.4m)
Answer
Δy=8.4524mm
Carried onward at full precision, not this rounded figure.
The thinnest bright film, order m = 0. Air on both sides of a soap film means exactly one reflection flips by half a wave, so the bright condition carries the extra half: 2nt = (m + ½)λ.
Rearranged for t
t=2n(m+21)λ
633.93 nmcarried from step 1
Your values, in your units
t=2×(1.33)((0)+21)(6.33927e−07m)
Converted to base units
t=2×(1.33)((0)+21)(633.927nm)
Answer
t=119.16nm
Carried onward at full precision, not this rounded figure.
Therefore the pointer runs at 633.9 nm — a red diode — oscillating at 4.73 × 10¹⁴ Hz; the same light lays fringes 8.45 mm apart on the screen, and the thinnest soap film that sends it back brightly is a mere 119 nm thick.
Why this order
Three interference experiments, one number joining them. The grating goes first because it is the only one of the three that measures a wavelength well: d sin θ = mλ puts the first order 25° off axis, where a protractor is plenty, while a double slit crowds its fringes into millimetres and a soap film gives you a colour rather than a number. Once λ is pinned down, the other two become predictions — and that is the honest order of a real lab, calibrate on the instrument that resolves best, then carry the result forward.
Step 4 is where the marks go missing. Both thin-film conditions look like 2nt = (something)λ, and choosing between them is not a question of bright versus dark, it is a question of counting reflections that invert. Light bouncing off a denser medium flips by half a cycle; light bouncing off a rarer one does not. A soap film in air inverts at the top surface only — one flip, so the extra half-wave lands on the bright condition and 2nt = (m + ½)λ. Lay the same film on glass and both surfaces invert, the flips cancel, and the two conditions swap places; that single bookkeeping step is the difference between a quarter-wave anti-reflection coating that works and one that makes a lens shine. Notice how thin the answer is — 119 nm, about a fifth of a wavelength. Let a film drain until 2nt approaches zero and it satisfies no bright order at all, which is why the top of a bubble goes black just before it bursts. Thomas Young pointed at Newton's own rings in 1801 and said this is interference; it took most of a century for anyone to be believed saying it.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
