One wavelength, three experiments: grating, double slit, soap film

SPH4U Grade 12 Physics · The Wave Nature of Light

An unlabelled laser pointer is aimed at a transmission grating whose rulings are 1.500 μm apart, and its first-order beam lands 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, and finally on a soap film of refractive index 1.33 held in a wire loop in air. Find the laser's wavelength, its frequency, the spacing of the fringes on the screen, and the thinnest film that reflects this colour brightly.

Step 1 of 4 · 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=dsinθm\lambda = \tfrac{d \sin\theta}{m}
Your values, in your units
λ=(1.5 μm)sin((25 ))(1)\lambda = \tfrac{\left( 1.5\ \mu\text{m} \right) \sin(\left( 25\ ^{\circ} \right))}{\left( 1 \right)}
Converted to base units
λ=(0.0000015 m)sin((25 ))(1)\lambda = \tfrac{\left( 0.0000015\ \text{m} \right) \sin(\left( 25\ ^{\circ} \right))}{\left( 1 \right)}
Answer
λ=634\lambda = 634

Carried onward at full precision, not this rounded figure.

Open the Diffraction Grating Equation solver →

Step 2 of 4 · solve for Frequency

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=vλf = \tfrac{v}{\lambda}
634 nmcarried from step 1
Your values, in your units
f=(299,792,000 m/s)(6.33927e07 m)f = \tfrac{\left( 299{,}792{,}000\ \text{m/s} \right)}{\left( 6.33927e-07\ \text{m} \right)}
Answer
f=473,000,000,000,000f = 473{,}000{,}000{,}000{,}000

Carried onward at full precision, not this rounded figure.

Open the Wave Speed (v = fλ) solver →

Step 3 of 4 · solve for Fringe spacing

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=λLd\Delta y = \tfrac{\lambda L}{d}
634 nmcarried from step 1
Your values, in your units
Δy=(6.33927e07 m)(2.4 m)(0.18 mm)\Delta y = \tfrac{\left( 6.33927e-07\ \text{m} \right) \, \left( 2.4\ \text{m} \right)}{\left( 0.18\ \text{mm} \right)}
Converted to base units
Δy=(6.33927e07 m)(2.4 m)(0.00018 m)\Delta y = \tfrac{\left( 6.33927e-07\ \text{m} \right) \, \left( 2.4\ \text{m} \right)}{\left( 0.00018\ \text{m} \right)}
Answer
Δy=8.45\Delta y = 8.45

Carried onward at full precision, not this rounded figure.

Open the Double-Slit Fringe Spacing solver →

Step 4 of 4 · solve for Film thickness

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=(m+12)λ2nt = \frac{\left(m + \tfrac{1}{2}\right)\lambda}{2n}
634 nmcarried from step 1
Your values, in your units
t=((0)+12)(6.33927e07 m)2×(1.33)t = \frac{\left(\left( 0 \right) + \tfrac{1}{2}\right) \left( 6.33927e-07\ \text{m} \right)}{2 \times \left( 1.33 \right)}
Converted to base units
t=((0)+12)(633.927 nm)2×(1.33)t = \frac{\left(\left( 0 \right) + \tfrac{1}{2}\right) \left( 633.927\ \text{nm} \right)}{2 \times \left( 1.33 \right)}
Answer
t=119t = 119

Carried onward at full precision, not this rounded figure.

Open the Thin-Film Constructive Interference (Bright Reflection) solver →

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 spend the result.

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.