Skip to main content

Minimum lines to resolve sodium lines

Short answer

Resolving power R = λ/dλ = nN. With mean λ=5893 Å, dλ=6 Å, n=2, required N = 491.1 → 492 lines. Lines per cm = N/width = 491.1/2.5 = 196.4 → 197 lines/cm.

Calculate the minimum no of lines per cm in a 2.5 cm wide grating which will just resolve the sodium lines 5890 Å and 5896 Å in second order spectrum.

Resolving power of a grating: R = λ/dλ = nN, where n is the order and N the total number of lines.

Given: λ₁ = 5890 Å, λ₂ = 5896 Å, n = 2, width = 2.5 cm.

  1. Mean wavelength λ = (5890 + 5896)/2 = 5893 Å; dλ = 6 Å
  2. Required resolving power:
R = λ/dλ = 5893 / 6 = 982.2
  1. Number of lines needed:
N = R / n = 982.2 / 2 = 491.1  →  at least 492 lines
  1. Lines per cm in the 2.5 cm wide grating:
N / width = 491.1 / 2.5 = 196.4 lines/cm  →  at least 197 lines/cm

Answer: the minimum is about 197 lines per cm (196.4 rounded up), i.e. about 492 lines across the 2.5 cm width.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Physics chapter 4 questions →