The Laser Pointer That Was 'Faulty'
A Leaving Certificate student is measuring the wavelength of a red laser pointer using a transmission diffraction grating stamped "300 lines per mm". She mounts the grating in a holder on an optical bench, aims the laser at it and places a screen 1.85 m from the grating, measured from the front face of the holder. She marks the centres of the two first-order spots and measures the distance between them as 0.72 m. Using the grating formula with n = 1 she obtains a wavelength of 1.2 × 10⁻⁶ m and concludes that the laser is faulty, because "that is infrared, not red". Looking more carefully at the screen she notices the left-hand first-order spot is 0.34 m from the central maximum and the right-hand spot is 0.38 m from it. The laboratory is lit by fluorescent tubes; the room light falls on the screen, but the spot pattern is still clearly visible.
1. Reproduce her calculation, identify the error and obtain the correct wavelength. Decide whether the laser is faulty.
Read the worked answer
Grating spacing: d = 1/(300 mm⁻¹) = 1/300 mm = 3.33 × 10⁻⁶ m.
Her calculation used x = 0.72 m as the distance from the central maximum to one spot: tan θ = 0.72/1.85 = 0.389 → θ = 21.3° → sin θ = 0.3626 λ = d sin θ = 3.33 × 10⁻⁶ × 0.3626 = 1.21 × 10⁻⁶ m.
That value is the clue: 1210 nm is beyond the visible range (roughly 400–700 nm), so the measurement, not the laser, must be wrong. 0.72 m is the separation of the two first-order spots, i.e. twice the distance from the centre. Using x = 0.36 m:
tan θ = 0.36/1.85 = 0.1946 → θ = 11.0° → sin θ = 0.1910 λ = 3.33 × 10⁻⁶ × 0.1910 = 6.37 × 10⁻⁷ m ≈ 637 nm.
That is exactly where a red laser pointer sits (~630–670 nm), so the laser is fine. Note also that for small angles the approximation λ ≈ d·x/L works well here (it gives 649 nm) but the sine form is the correct one and should be used when θ exceeds a few degrees.
2. The two first-order spots are not equidistant from the centre. What does this indicate, and how should she change the procedure and the reported uncertainty?
Read the worked answer
The grating equation d sin θ = nλ assumes the beam strikes the grating normally. If the beam is at an angle to the grating normal (grating rotated in its holder, or the laser not square to the bench), the geometry is asymmetric and the +1 and −1 orders are displaced by different amounts on the screen - exactly the 0.34 m / 0.38 m pattern seen.
Procedure fix: rotate the grating (or the laser) until both first-order spots are equidistant from the central maximum, and re-measure. Also measure the grating-to-screen distance perpendicular to the screen from the grating plane itself, not from the front face of a holder several millimetres thick; at L = 1.85 m a 5 mm error is small, but it is systematic and easy to avoid.
Uncertainty: the spread between the two sides (0.34 vs 0.38, about ±6%) is a direct indication of the misalignment, so quoting λ = 637 nm to four significant figures would be false precision. Report something like (6.4 ± 0.2) × 10⁻⁷ m, and state that the dominant uncertainty is alignment, not the ruler.
3. Explain why the fluorescent room light does not destroy the measurement, and predict what she would see if the laser were replaced by a white-light source.
Read the worked answer
Room light is incoherent: its phase varies randomly across the beam and it is not collimated, so the light arriving at any point on the screen comes from a huge range of path differences with no fixed relationship. The grating still diffracts it, but the maxima for each wavelength and each source point fall in different places and average out into a faint, uniform background rather than sharp spots. The laser, by contrast, is monochromatic and spatially coherent, so all of its light at a given order adds in phase at one place, giving an intense sharp spot that is far brighter than the background.
With white light, each wavelength would be diffracted by a different angle (λ ∝ sin θ), so the first order would be a continuous spectrum spread out from violet (smallest θ) to red (largest θ), with a white central maximum. Higher orders would overlap, and because the source is an extended filament rather than a point, each colour would also be smeared into a broad band unless a narrow slit and collimator were placed in front of it.