describe the use of a diffraction grating to determine the wavelength of light (the structure and use of the spectrometer are not included)

Diffraction Grating: Finding the Wavelength of Light 🔬

What is a Diffraction Grating?

A diffraction grating is like a giant “mirror” made of many, many tiny slits (usually thousands per centimetre). When light hits these slits, it spreads out and interferes with itself, creating bright and dark spots. Think of it as a musical instrument where each slit is a note; when the notes line up, you hear a clear tone (a bright spot). This property lets us measure the colour (wavelength) of light.

How Light Bounces Off the Grating

When a light wave passes through two adjacent slits, the waves travel slightly different distances before they meet. The extra distance is called the path difference and is given by:

\$\,d\,\sin\theta = m\,\lambda\,\$

• \$d\$ = distance between adjacent slits (the grating spacing)
• \$\theta\$ = angle of the bright spot relative to the normal (straight‑up)
• \$m\$ = order of the maximum (1st, 2nd, 3rd …)
• \$\lambda\$ = wavelength of the light (what we want to find)

If you know any three of these values, you can solve for the fourth. In a typical experiment we measure the angle \$\theta\$, we know \$d\$ from the grating’s specification, and we pick an order \$m\$ (usually \$m=1\$). Then we calculate \$\lambda\$.

Using the Grating to Measure Wavelength

1. Place the light source so that it shines straight onto the grating.
2. Measure the angle \$\theta\$ of the first bright spot (the 1st order) with a protractor or a screen with a ruler.
3. Look up the grating spacing \$d\$ (e.g. 2000 lines mm⁻¹ → \$d = 1/2000\$ mm = 5 µm).
4. Insert the values into \$\,\lambda = \dfrac{d\,\sin\theta}{m}\,\$ and compute.
5. Check your answer: visible light wavelengths are between 400 nm (violet) and 700 nm (red).

Example Problem

Suppose a grating has 1200 lines mm⁻¹ and the first‑order bright spot appears at an angle of \$12^\circ\$. What is the wavelength of the light?

First, find \$d\$:

\$\,d = \dfrac{1}{1200}\,\text{mm} = 8.33\times10^{-4}\,\text{mm} = 8.33\,\mu\text{m}\,\$

Now use the formula with \$m=1\$:

\$\,\lambda = d\,\sin12^\circ = 8.33\,\mu\text{m}\,\times 0.2079 \approx 1.73\,\mu\text{m}\,\$

This wavelength (1730 nm) is in the infrared, so the light source is not visible. If we had measured a smaller angle, we would get a visible colour.

Exam Tips 📚

• Always write down the formula \$d\sin\theta = m\lambda\$ before starting.
• Check units: convert mm to µm or nm as needed.
• Remember that \$m\$ is an integer; \$m=1\$ is simplest.
• For small angles, \$\sin\theta \approx \theta\$ (in radians) – useful for quick estimates.
• State the range of visible wavelengths (400–700 nm) when discussing your result.
• Use a clear diagram (drawn with text) to show the grating, incident beam, and diffracted angles.

Quick Reference Table 📊