recall and use d sin θ = nλ

Diffraction Grating

Objective

Students will be able to recall and use the grating equation

\[ d\sin\theta = n\lambda \]

and to understand its origin from the principle of superposition, its limits, and the concept of resolving power.

1. What a Diffraction Grating Is

• A flat plate that contains a very large number (10⁴–10⁶) of equally spaced, parallel slits (or rulings).
• Each slit acts as a source of secondary wavelets (Huygens’ principle). The overall pattern is the result of superposition of all these wavelets.
• For monochromatic, collimated light the superposition gives a series of very sharp bright maxima – the diffraction orders.

2. Connection to the Superposition Principle

The intensity distribution of a grating is the product of two factors:

1. Single‑slit envelope – determined by the width a of each slit (≈ \(\displaystyle I_{\text{single}} \propto \bigl[\frac{\sin(\pi a\sin\theta/\lambda)}{\pi a\sin\theta/\lambda}\bigr]^2\)).
2. Multi‑slit interference term – arising from the path‑difference between the N slits, which leads to the condition for constructive interference: \[ d\sin\theta = n\lambda\qquad (n=0,\pm1,\pm2,\dots) \]

The sharpness of the maxima is a direct consequence of the large number of slits (intensity ∝ N² at the maxima).

3. Derivation of the Grating Equation

3.1 Two‑slit argument (quick reminder)

1. Two adjacent slits are separated by the grating spacing \(d\).
2. A ray emerging at angle \(\theta\) from the normal travels an extra distance \(d\sin\theta\) from the lower slit.
3. Constructive interference ⇔ path difference = integer × wavelength: \[ d\sin\theta = n\lambda . \]

3.2 General N‑slit derivation

1. Consider N equally spaced slits, each separated by \(d\).
2. The phase difference between light from adjacent slits is \[ \delta = \frac{2\pi}{\lambda}\,d\sin\theta . \]
3. The resultant amplitude is the sum of a geometric series: \[ A = A_0\frac{\sin(N\delta/2)}{\sin(\delta/2)} . \]
4. Maximum intensity occurs when the numerator is maximised, i.e. when \[ N\frac{\delta}{2}=m\pi \quad\Longrightarrow\quad \delta = 2m\pi . \]
5. Substituting \(\delta\) gives the familiar grating condition \[ d\sin\theta = m\lambda \;(m=n). \]
6. At a maximum the intensity is \[ I_{\max}=N^{2}I_{\text{single}}, \] which explains why the peaks are much sharper and brighter than for a double‑slit.

4. Limits of the Grating Equation

4.1 Maximum observable order

Because \(|\sin\theta|\le 1\), the highest integer order that can appear is

\[ n_{\max}= \left\lfloor\frac{d}{\lambda}\right\rfloor . \]

4.2 Order overlap (spectral crowding)

When two different wavelengths satisfy the grating equation for different orders, they can appear at the same angle, making the spectrum ambiguous.

Example: For a grating with \(d=1.00\;\mu\text{m}\)

\[ \begin{aligned} \text{For }\lambda_1 &= 600\;\text{nm},\; n_1=2: &\; d\sin\theta &= 2(600\text{ nm}) = 1.20\;\mu\text{m} \\ \text{For }\lambda_2 &= 400\;\text{nm},\; n_2=3: &\; d\sin\theta &= 3(400\text{ nm}) = 1.20\;\mu\text{m} \end{aligned} \] Both give the same \(\sin\theta = 1.20\) → \(\theta\approx 50.0^{\circ}\). Thus the second‑order red line overlaps the third‑order violet line. High‑density gratings (small \(d\)) reduce the chance of overlap because the angular separation between orders increases.

4.3 Single‑slit envelope

If the slit width \(a\) is not negligible compared with \(d\), the envelope \(\displaystyle \bigl[\frac{\sin(\pi a\sin\theta/\lambda)}{\pi a\sin\theta/\lambda}\bigr]^2\) suppresses higher orders that fall outside its central lobe.

5. Resolving Power

The ability of a grating to separate two close wavelengths \(\lambda\) and \(\lambda+\Delta\lambda\) is quantified by the resolving power

\[ R=\frac{\lambda}{\Delta\lambda}=nN , \] where

• \(n\) = order of the observed maximum.
• \(N\) = total number of illuminated slits (or grooves). For a grating of width \(W\) and spacing \(d\), \(N = W/d\).

Derivation (brief)

1. Two wavelengths \(\lambda\) and \(\lambda+\Delta\lambda\) produce maxima at angles \(\theta\) and \(\theta+\Delta\theta\).
2. Using the grating condition and a first‑order Taylor expansion, \[ d\cos\theta\,\Delta\theta = n\Delta\lambda . \]
3. The Rayleigh criterion for just‑resolved lines requires the angular separation to be at least the angular width of a principal maximum, \(\Delta\theta_{\text{min}} = \lambda/(Nd\cos\theta)\).
4. Combining the two relations gives \[ \frac{\lambda}{\Delta\lambda}=nN . \]

Numerical example

A 1200 lines mm⁻¹ grating (so \(d = 8.33\times10^{-7}\,\text{m}\)) is illuminated over a width of 2.0 cm.

\[ N = \frac{W}{d}= \frac{0.020\ \text{m}}{8.33\times10^{-7}\ \text{m}} \approx 2.4\times10^{4}. \] For the third order (\(n=3\)): \[ R = nN \approx 3 \times 2.4\times10^{4}=7.2\times10^{4}. \] Thus the grating could, in principle, separate wavelengths that differ by \[ \Delta\lambda = \frac{\lambda}{R} \approx \frac{600\ \text{nm}}{7.2\times10^{4}} \approx 0.0083\ \text{nm}. \]

6. Laboratory Set‑up (Spectrometer)

In a typical Cambridge A‑Level experiment the grating is mounted on a rotating table (goniometer) inside a spectrometer.

• Collimated monochromatic source – a slit illuminated by a lamp or a laser.
• Rotating table / goniometer – carries the grating and has a calibrated angular scale (usually to 0.1°).
• Detector – a screen with a movable telescope or a photodiode that records the position of each order.
• Read‑out procedure – rotate the table until a chosen order appears, read \(\theta\) from the scale, then apply the grating equation.

Typical sources of experimental error: mis‑alignment of the normal, finite width of the slit, reading error on the scale, and temperature‑induced expansion of the grating.

7. Using the Grating Equation – Problem‑Solving Template

1. Identify what is given – values of \(d\), \(\lambda\), \(\theta\), or the order \(n\).
2. Choose the unknown – rearrange the equation accordingly:
   • Find \(\lambda\): \(\displaystyle \lambda = \frac{d\sin\theta}{n}\)
   • Find \(d\): \(\displaystyle d = \frac{n\lambda}{\sin\theta}\)
   • Find \(\theta\): \(\displaystyle \theta = \sin^{-1}\!\left(\frac{n\lambda}{d}\right)\)
3. Check feasibility – ensure \(|n\lambda/d|\le 1\); if not, the order is impossible.
4. Calculate the required quantity** using a calculator with sufficient significant figures.
5. State the answer with appropriate units and sign** (positive \(\theta\) measured from the normal; \(n\) may be negative for the opposite side of the normal).

8. Worked Examples

Example 1 – Find the wavelength

A grating with 5000 lines cm⁻¹ produces a first‑order maximum at \(\theta = 20.0^{\circ}\). Find \(\lambda\).

1. Convert line density to spacing: \[ d = \frac{1}{5.0\times10^{5}\ \text{m}^{-1}} = 2.0\times10^{-6}\ \text{m}. \]
2. Use the template for \(\lambda\) (with \(n=1\)): \[ \lambda = d\sin20.0^{\circ}=2.0\times10^{-6}\times0.342 = 6.84\times10^{-7}\ \text{m}=684\ \text{nm}. \]

Example 2 – Determine the grating spacing

The second‑order maximum of a 650 nm source is observed at \(\theta = 30.0^{\circ}\). Find \(d\).

\[ d = \frac{n\lambda}{\sin\theta}= \frac{2\times650\times10^{-9}}{\sin30^{\circ}} = \frac{1.30\times10^{-6}}{0.500}=2.60\times10^{-6}\ \text{m}. \]

Example 3 – Maximum observable order and order overlap

For a grating with \(d = 1.00\;\mu\text{m}\) and light of \(\lambda = 550\;\text{nm}\):

• Maximum order: \(n_{\max}= \bigl\lfloor d/\lambda \bigr\rfloor = \lfloor 1.00/0.55 \rfloor = 1\). Only the 0‑ and ±1‑orders can appear.
• Suppose a second wavelength \(\lambda' = 440\;\text{nm}\) is also present. The third‑order of \(\lambda'\) gives \[ d\sin\theta = 3(440\text{ nm}) = 1.32\;\mu\text{m}, \] which is impossible (\(\sin\theta>1\)). Hence no overlap occurs in this case.

9. Table of Common Grating Parameters

Lines per mm	Grating spacing \(d\) (µm)	Typical use in the Cambridge syllabus
300	3.33	Low‑resolution colour analysis; introductory experiments.
600	1.67	Medium‑resolution work – identification of emission lines.
1200	0.833	High‑resolution spectroscopy; resolving closely spaced lines, investigating resolving power.

10. Key Points to Remember

• The grating equation \(d\sin\theta=n\lambda\) follows directly from the superposition of wavelets from many slits.
• Zero‑order (\(n=0\)) always appears at \(\theta=0^{\circ}\) (the undeviated beam).
• Higher orders give larger angles but are limited by \(|\sin\theta|\le1\); use \(n_{\max}= \lfloor d/\lambda\rfloor\) to check feasibility.
• Shorter wavelengths diffract at smaller angles for a given order; longer wavelengths give larger angles.
• Intensity of higher orders falls because the single‑slit envelope suppresses them and because the total intensity is divided among many orders.
• Resolving power \(R=nN\) improves with higher order and with a larger number of illuminated grooves.
• Order overlap can make a spectrum ambiguous; choosing a grating with a smaller \(d\) (higher line density) reduces overlap.
• In the laboratory, careful alignment and accurate reading of \(\theta\) are essential for reliable results.

11. Practice Questions

1. A grating with 2000 lines cm⁻¹ is illuminated by light of wavelength 500 nm. Calculate the diffraction angles for the first three observable orders.
2. If the second‑order maximum for a wavelength of 650 nm appears at 30°, determine the grating spacing \(d\).
3. Explain why the intensity of higher‑order maxima generally decreases compared with the first order.
4. A 1200 lines mm⁻¹ grating is illuminated over a width of 2.0 cm. Determine the theoretical resolving power for the third order (n = 3). (Assume \(N = \text{width} / d\).)
5. Two wavelengths, 580 nm and 460 nm, are incident on a grating with \(d = 1.20\;\mu\text{m}\). Show whether any order overlap occurs for orders up to \(n=4\).