Diffraction Grating with a Green Laser
A student uses a diffraction grating labelled 600 lines per millimetre to measure the wavelength of a green laser. The grating is mounted 1.50 m from a screen. The two first-order spots are measured to be 1.18 m apart, centre to centre. A faint second-order pair is visible farther out. The room is not completely dark, and the spot centres are broad, so the student estimates each position to ±5 mm. The student then repeats the measurement with a red laser. The grating label is accurate to ±1%. The laser beam is incident normally on the grating. The screen is vertical and parallel to the grating. The student wonders whether a small-angle approximation would be adequate and whether a third-order spot should appear.
1. Calculate the wavelength of the green laser. State whether the small-angle approximation is acceptable, and estimate its percentage error.
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The grating spacing is d = 1/(600 mm^-1) = 1.667×10^-6 m. The first-order spot is half the measured separation from the centre: y = 1.18/2 = 0.590 m. The angle satisfies tanθ = y/L = 0.590/1.50 = 0.3933, so θ = 21.47°. The grating equation is d sinθ = mλ with m = 1, so
λ = d sinθ = (1.667×10^-6)(0.3660) = 6.10×10^-7 m = 610 nm.
The small-angle approximation sinθ ≈ tanθ would give λ ≈ d y/L = (1.667×10^-6)(0.3933) = 6.56×10^-7 m = 656 nm. That is about 7% too high, far larger than the 1% grating tolerance, so the approximation is not acceptable here.
2. Predict whether the second-order spots are farther from the centre, and determine whether a third-order spot can appear. Justify with the grating equation.
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For order m, sinθ_m = mλ/d. With λ/d = 0.366, the second order has sinθ_2 = 0.732, so θ_2 = 47.1°. Its distance from the centre is y_2 = L tanθ_2 = 1.50 tan47.1° = 1.61 m, compared with 0.590 m for the first order, so it is farther out. The third order would require sinθ_3 = 3(0.366) = 1.098, which is greater than 1. No real angle satisfies this, so no third-order spot can appear. The faintness of the second order is an intensity effect, not a change in the grating equation.
3. The student repeats with red light. Predict how the first-order separation changes. Describe one change to the apparatus or procedure that would most improve the wavelength determination, and justify it using the uncertainty.
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Red light has a longer wavelength than green. Since sinθ = λ/d for first order, a larger λ gives a larger angle and therefore a larger separation on the screen, so the red first-order spots are farther apart. To improve the wavelength determination, increase the screen distance L substantially, for example to 3 m, and measure the spot positions with a travelling microscope or a ruler with a fiducial mark. A larger L increases y for the same angle, so the fixed ±5 mm position uncertainty becomes a smaller fractional uncertainty in y and hence in tanθ. Measuring several orders and plotting sinθ against m would also average random errors and check consistency; the slope gives λ/d without relying on a single spot pair.