Questions: Diffraction Gratings: Construction, Spectra, and Spectroscopy

Question 1

You replace a diffraction grating with 500 slits with one having 1000 slits, keeping the slit spacing d the same. What changes in the diffraction pattern?

• A. New bright maxima appear at new angles, and the peaks get sharper
• B. The angles of the bright maxima are unchanged, but the peaks become sharper and more intense per unit angle
• C. The grating equation changes, so the bright maxima shift to new angles
• D. The pattern becomes dimmer overall because more slits divide the light into more beams

Answer: (B)

The positions of bright maxima are determined by d sin θ = mλ. Since d and λ are unchanged, the angles of the maxima don't change. What changes is the sharpness: with more slits, destructive interference from the many slit pairs eliminates intensity almost immediately on either side of each maximum, producing much narrower peaks. The resolving power R = mN also doubles (from 500N to 1000N for order m), meaning the grating can now separate wavelengths twice as close together. More slits = sharper peaks at the same locations, not new peaks.

Question 2

A diffraction grating with N = 800 slits is used in the third diffraction order. What is its resolving power?

• A. 800
• B. 2400
• C. 267
• D. It depends on the wavelength being resolved

Answer: (B)

Resolving power R = mN = 3 × 800 = 2400. This means the grating can just resolve two wavelengths λ and λ + Δλ when λ/Δλ = 2400 — it can distinguish wavelengths differing by as little as 1 part in 2400. Resolving power depends on the order m and total number of slits N, not directly on the wavelength. Higher order and more slits both increase resolving power, which is why spectrometers are designed to use many slits and sometimes higher orders.

Question 3

Using a diffraction grating in a higher diffraction order (m = 2 rather than m = 1) increases angular dispersion between wavelengths, but the intensity of the maxima decreases.

• T. True
• F. False

Answer: True

Both statements are correct. The grating equation d sin θ = mλ shows that at higher order m, the same wavelength difference Δλ produces a larger angle difference Δθ — better angular separation. Resolving power R = mN also increases with m. However, the intensity falls off in higher orders because the total energy is being distributed across more maxima, and single-slit diffraction envelope effects reduce intensity at larger angles. First-order is typically the most intense; spectrometers often use it exclusively, accepting lower resolving power in exchange for intensity.

Question 4

A diffraction grating with more slits produces more bright maxima at more angles than a grating with fewer slits.

• T. True
• F. False

Answer: False

This is a common misconception. The positions of the bright maxima are determined solely by d sin θ = mλ — only the slit spacing d and wavelength λ matter, not the number of slits. More slits produce the same maxima at the same angles, but sharper. The effect of adding more slits is to increase destructive interference between the maxima, narrowing each peak without adding new ones. The number of possible orders is limited by the condition sin θ ≤ 1, not by the number of slits.

Question 5

Why do more slits in a diffraction grating produce sharper peaks rather than more peaks?

Model answer: Each additional slit adds a new source of destructive interference between the bright maxima. With N slits, there are N-1 zeros between consecutive orders. A small deviation from the maximum angle causes waves from pairs of slits to be slightly out of phase, and with many slits, these partial cancellations sum to nearly complete destructive interference very quickly on either side of the maximum. With only 2 slits, the cancellation is gradual; with 1000 slits, it is rapid, leaving a very narrow bright line at each maximum. Peak positions (from d sin θ = mλ) are unchanged; only peak width narrows.

The mathematical reason is that the intensity pattern for N slits is proportional to sin²(Nδ/2)/sin²(δ/2), where δ is the phase difference between adjacent slits. At the bright maximum δ = 2πm, this expression equals N². Just off the maximum, sin(Nδ/2) oscillates rapidly through zero as N increases, producing the sharp destructive minima. The width of the central peak scales as 1/N — exactly why resolving power R = mN grows with N. More slits narrow the peaks without shifting them.