recall and use d sin θ = nλ

Diffraction Grating

Objective

Recall and apply the grating equation

\$d\sin\theta = n\lambda\$

where:

• d – grating spacing (distance between adjacent slits)
• θ – diffraction angle measured from the normal to the grating
• n – order of the diffracted maximum (integer, n = 0, ±1, ±2, …)
• λ – wavelength of the incident light

Principle of Operation

A diffraction grating consists of a large number of equally spaced parallel slits. When monochromatic light of wavelength λ is incident on the grating, each slit acts as a source of secondary wavelets. Constructive interference occurs at angles that satisfy the grating equation, producing bright maxima of order n.

Derivation (Brief)

1. Consider two adjacent slits separated by distance d.
2. The path difference between rays from these slits at an angle θ is d sin θ.
3. Constructive interference requires the path difference to be an integer multiple of the wavelength: d sin θ = nλ.
4. Because the grating contains many slits, the maxima are sharp and well‑defined.

Using the Grating Equation

Typical tasks include:

• Finding the wavelength of an unknown light source.
• Determining the grating spacing d from a known wavelength and measured angle.
• Predicting the angles of higher orders.

Worked Example

Problem: A diffraction grating with 5000 lines cm⁻¹ is illuminated by a monochromatic source. The first‑order maximum (n = 1) is observed at an angle of 20.0° from the normal. Calculate the wavelength of the light.

Solution:

1. Convert the line density to spacing:

   \$d = \frac{1}{\text{lines per metre}} = \frac{1}{5.0\times10^{5}\ \text{m}^{-1}} = 2.0\times10^{-6}\ \text{m}\$

2. Use the grating equation for n = 1:

   \$\lambda = d\sin\theta = (2.0\times10^{-6}\,\text{m})\sin20.0^{\circ}\$

3. Calculate:

   \$\lambda = 2.0\times10^{-6}\times0.342 = 6.84\times10^{-7}\ \text{m} = 684\ \text{nm}\$

Thus the wavelength of the light is 684 nm, which lies in the red region of the visible spectrum.

Table of Common Grating Parameters

Key Points to Remember

• The grating equation is valid for any order n, but higher orders may overlap if λ is large.
• The zero‑order maximum (n = 0) always occurs at θ = 0° and corresponds to the undeviated beam.
• Maximum observable order is limited by the condition |sin θ| ≤ 1, giving n_{\max} = \lfloor d/λ \rfloor.
• For a given grating, shorter wavelengths diffract at smaller angles than longer wavelengths.

Practice Questions

1. A grating with 2000 lines cm⁻¹ is used with light of wavelength 500 nm. Calculate the angles for the first three observable orders.
2. If the second‑order maximum for a wavelength of 650 nm appears at 30°, what is the grating spacing?
3. Explain why the intensity of higher‑order maxima generally decreases compared with the first order.