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A diffraction grating is an optical component with many equally spaced parallel slits. It produces sharp, well-defined maxima and is far more useful than a double slit for precise wavelength measurements and spectroscopy.
A typical diffraction grating consists of a glass or plastic slide with thousands of equally spaced parallel lines ruled on its surface. The unruled gaps between the lines act as slits.
Gratings are specified by the number of lines per unit length, N. For example:
The slit spacing (also called the grating spacing or grating element), d, is:
d = 1/N
For a grating with 300 lines per mm: d = 1/(3.0 × 10⁵) = 3.33 × 10⁻⁶ m = 3.33 μm
When monochromatic light passes through a diffraction grating, sharp bright maxima (principal maxima) are observed at specific angles. The condition for a maximum is:
d sin θ = nλ
where:
Adjacent slits are separated by distance d. For constructive interference at angle θ, the path difference between light from adjacent slits must equal a whole number of wavelengths:
Path difference = d sin θ = nλ
This is the same principle as for a double slit, but with many more slits producing much sharper maxima.
At n = 0: d sin θ = 0, so θ = 0°. The zero-order maximum is always straight ahead, regardless of wavelength. This is simply the undeviated beam.
With a double slit, the maxima are broad because only two waves are interfering. With a grating containing thousands of slits, the waves from all the slits must arrive in phase for a maximum. Even a slight departure from the exact angle causes significant cancellation across the many slit pairs. The result is extremely narrow, intense maxima separated by wide dark regions.
The more slits, the sharper the maxima.
Worked Example 1 — Light of wavelength 550 nm is incident on a diffraction grating with 400 lines per mm. Calculate the angle of the first-order maximum.
d = 1/N = 1/(4.00 × 10⁵) = 2.50 × 10⁻⁶ m
For n = 1: d sin θ = λ
sin θ = λ/d = (550 × 10⁻⁹)/(2.50 × 10⁻⁶) = 0.220
θ = arcsin(0.220) = 12.7°
Worked Example 2 — Using the same grating as above, calculate the angle of the second-order maximum.
For n = 2: sin θ = 2λ/d = 2 × 0.220 = 0.440
θ = arcsin(0.440) = 26.1°
Worked Example 3 — Determine the maximum order visible for this grating and wavelength.
The maximum possible value of sin θ is 1. So:
d sin θ = nλ → n_max = d/λ = (2.50 × 10⁻⁶)/(550 × 10⁻⁹) = 4.55
Since n must be an integer, the maximum order is n = 4.
Check: sin θ = 4 × 550 × 10⁻⁹/(2.50 × 10⁻⁶) = 0.880, so θ = 61.6° — this is valid.
For n = 5: sin θ = 5 × 0.220 = 1.10 > 1 — impossible. So n = 4 is confirmed as the highest order.
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